XI Mathematics IIT JEE Advanced Study Package 2014 159

8/20/2019 XI Mathematics IIT JEE Advanced Study Package 2014 159
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Class-XI
Ph.06226-252314 , Mobile:9431636758, 9931610902
2 Trigonometric Equations 14 3
3 Properties of Triangle 24 5
4 Functions 40 5
5 Complex Numbers 37 5 6 Quadratic Equations 23 6
7 Permutations and Combinations 19 5
8 Binomial Theorem 24 8
9 Probability 36 5
10 Progressions 25 5
12 Circles 23 5
13 Parabola, Ellipse and Hyperbola 68 15 14 Highlights on Conic Sections 25
15 Vector Algebra and 3-D Geometry 62 8
16 Limits 18 5
17 Differentiation 17 2
Ratio and Identity
3. Answer Key
1
Trigonometric Ratios & Identities
1.1 .1 .1 . B as ic Tri gonome tri c I dentit ies :Basic Trigonometric Identities:Basic Trigonometric Identities:Basic Trigonometric Identities:
(a) sin²
θ + cos²
θ = 1; −1 ≤ sin θ ≤ 1; −1 ≤ cos θ ≤ 1 ∀ θ ∈ R
(b) sec² θ − tan² θ = 1 ; sec θ ≥ 1 ∀ θ ∈ R – ( )
Ι∈
π + n,
2 1n2
(c) cosec² θ − cot² θ = 1 ; cosec θ ≥ 1 ∀ θ ∈ R – { Ι∈π n,n
Solved Example # 1
(ii) 1AsecAtan
= cos2A – sin2A + 1 [∴ cos2A + sin2A = 1]
= 2 cos2A
(ii) 1AsecAtan
Asin1+
If sin x + sin2x = 1, then find the value of
cos12x + 3 cos10x + 3 cos8x + cos6x – 1
Solution
= (cos 4
= 1 – 1 = 0
Solved Example # 3
1 , then show that sec θ – tan θ = – 2m or
m2
1
Solution
Depending on quadrant in which θ falls, sec θ can be ± m4
1m4 2 +

Self Practice Problem
(iii) sec2A cosec2A = tan2A + cot2A + 2
(iv) (tan α + cosec β)2 – (cot β – sec α)2 = 2 tan α cot β (cosec α + sec β)
(v)
2
n2mn2m
mn2m
2
n2mn2
mn2m
+ +
2 .2 .2 .2 . C ir cu la r D ef in it io n O f T ri go no me tr ic Fu nc ti on s:Ci rcular Def ini t ion Of Tr igonometr ic Funct ions:Ci rcular Def ini t ion Of Tr igonometr ic Funct ions:Ci rcular Def ini t ion Of Tr igonometr ic Funct ions:
sin θ = OP
θsin
1 , sin θ ≠ 0
3 .3 .3 .3 . T ri go no me tr ic F un ct io ns Of A ll ied A ng le s:Tr igonometr ic Funct ions Of Al l ied Angles :Tr igonometr ic Funct ions Of Al l ied Angles :Tr igonometr ic Funct ions Of Al l ied Angles :
If θ is any angle, then − θ, 90 ± θ, 180 ± θ, 270 ± θ, 360 ± θ etc. are called ALLIED ANGLES.
(a) sin (− θ) = − sin θ ; cos (− θ) = cos θ (b) sin (90° − θ) = cos θ ; cos (90°− θ) = sin θ (c) sin (90° + θ) = cos θ ; cos (90° + θ) = − sin θ (d) sin (180°− θ) = sin θ ; cos (180°− θ) = − cos θ (e) sin (180° + θ) = − sin θ ; cos (180° + θ) = − cos θ (f) sin (270°− θ) = − cos θ ; cos (270°− θ) = − sin θ (g) sin (270° + θ) = − cos θ ; cos (270° + θ) = sin θ (h) tan (90° − θ) = cot θ ; cot (90°− θ) = tan θ
Solved Example # 4
Prove that
(i) cot A + tan (180º + A) + tan (90º + A) + tan (360º – A) = 0 (ii) sec (270º – A) sec (90º – A) – tan (270º – A) tan (90º + A) + 1 = 0
Solution
(i) cot A + tan (180º + A) + tan (90º + A) + tan (360º – A)
= cot A + tan A – cot A – tan A = 0
(ii) sec (270º – A) sec (90º – A) – tan (270º – A) tan (90º + A) + 1
= – cosec2A + cot2A + 1 = 0
(i) sin 420º cos 390º + cos (–300º) sin (–330º) = 1
(ii) tan 225º cot 405º + tan 765º cot 675º = 0
3
4444 .... G raphs o f Trigonome tr ic funct ions:Graphs of Trigonometric functions:Graphs of Trigonometric functions:Graphs of Trigonometric functions:
(a) y = sin x x ∈ R; y ∈ [–1, 1]
(b) y = cos x x ∈ R; y ∈ [ – 1, 1]
(c) y = tan x x ∈ R – (2n + 1) π /2, n ∈ Ι ; y ∈ R
(d) y = cot x x ∈ R – nπ , n ∈ Ι; y ∈ R
(e) y = cosec x x ∈ R – nπ , n ∈ Ι ; y ∈ (−−−− ∞ , −−−− 1] ∪∪∪∪ [1, ∞∞∞∞)
(f) y = sec x x ∈ R – (2n + 1) π /2, n ∈ Ι ; y ∈∈∈∈ (− ∞, − 1] ∪ [1, ∞)
4
Solved Example # 5 Find number of solutions of the equation cos x = |x |
Solution
Clearly graph of cos x & |x| intersect at two points. Hence no. of solutions is 2
Solved Example # 6
Find range of y = sin2x + 2 sin x + 3 ∀ x ∈ R
Solution
⇒ 0 ≤ sin x +1 ≤ 2
⇒ 2 ≤ (sin x +1)2 + 2 ≤ 6
Hence range is y ∈ [2, 6]
xy4
5. Find range of the followings.
(i) y = 2 sin2x + 5 sin x +1∀ x ∈ R Answer [–2, 8]
(ii) y = cos2x – cos x + 1 ∀ x ∈ R Answer

3 ,1
5555 .... Tr igonometric Functions of Sum or Difference of Two Angles:Trigonometric Functions of Sum or Difference of Two Angles:Trigonometric Functions of Sum or Difference of Two Angles:Trigonometric Functions of Sum or Difference of Two Angles:
(a) sin (A ± B) = sinA cosB ± cosA sinB
(b) cos (A ± B) = cosA cosB sinA sinB
(c) sin²A − sin²B = cos²B − cos²A = sin (A+B). sin (A− B)
(d) cos²A − sin²B = cos²B − sin²A = cos (A+B). cos (A − B)
(e) tan (A ± B) = BtanAtan1
BtanAtan
CtanBtanAtanCtanBtanAtan
Prove that
(i) sin (45º + A) cos (45º – B) + cos (45º + A) sin (45º – B) = cos (A – B)
(ii) tan
Solution
(i) Clearly sin (45º + A) cos (45º – B) + cos (45º + A) sin (45º – B)
= sin (45º + A + 45º – B)
= sin (90º + A – B)
3 , cos β =
65
33 ,
65
63
13 +
A = tan A cot
A – 1 = sec A
6666 .... F ac t or is at io n of t he Su m or D if fe re nc e o f Tw o S in es o rFactor isat ion of the Sum or Di f ference of Two Sines orFactor isat ion of the Sum or Di f ference of Two Sines orFactor isat ion of the Sum or Di f ference of Two Sines or
C o s i n e s :C o s i n e s :C o s i n e s :C o s i n e s :
(a) sinC + sinD = 2 sin 2
DC+ cos
2
DC+ cos
Prove that sin 5A + sin 3A = 2sin 4A cos A
Solution
L.H.S. sin 5A + sin 3A = 2sin 4A cos A = R.H.S.
[ sin C + sin D = 2 sin 2
DC + cos
Solved Example # 9
Find the value of 2 sin 3θ cos θ – sin 4θ – sin 2θ Solution
2 sin 3θ cos θ – sin 4θ – sin 2θ = 2 sin 3θ cos θ – [2 sin 3θ cos θ ] = 0
x13 sin
−+− = cot 4A
7777.... Transf ormat ion o f P roduc ts i nto Sum or D if fe rence o f S ines &Transformation of Products into Sum or Difference of Sines &Transformation of Products into Sum or Difference of Sines &Transformation of Products into Sum or Difference of Sines &
C o s i n e sC o s i n e sC o s i n e sC o s i n e s ::::
(a) 2 sinA cosB = sin(A+B) + sin(A−B) (b) 2 cosA sinB = sin(A+B) − sin(A−B)
(c) 2 cosA cosB = cos(A+B) + cos(A−B) (d) 2 sinA sinB = cos(A−B) − cos(A+B)
Solved Example # 10
4sin3sincos2cos
3tan5tan
Solution
4sin3sin2cos2cos2
3cos6sin2cos8sin2
7coscoscos3cos
3sin9sin7sin9sin =
3tan5tan
3tan5tan =
θ sin
11θ = sin 2θ sin 5θ
12. Prove that cos A sin (B – C) + cos B sin (C – A) + cos C sin (A – B) = 0
13. Prove that 2 cos 13
π cos
5π = 0
8888.... Mul tip le and Sub-mult iple Angle s :Multiple and Sub-multiple Angles :Multiple and Sub-multiple Angles :Multiple and Sub-multiple Angles :
(a) sin 2A = 2 sinA cosA ; sin θ = 2 sin θ 2
cos θ 2
(b) cos 2A = cos²A − sin²A = 2cos²A − 1 = 1 − 2 sin²A; 2 cos² 2
θ = 1 + cos θ, 2 sin²
2
(e) sin 3A = 3 sinA − 4 sin3A (f) cos 3A = 4 cos3A − 3 cosA
(g) tan 3A = Atan31
(iii) )BAcos(BcosAcos1
Atan1 2+ = 2
(iii) L.H.S. )BAcos(BcosAcos1
θ+θ 2coscos1
2sinsin = tan θ
15. Prove that sin 20º sin 40º sin 60º sin 80º = 16
3
16. Prove that tan 3A tan 2A tan A = tan 3A – tan 2A – tan A
17. Prove that tan
9999 .... Important Trigonomet ri c Rat ios:Important Trigonometric Ratios:Important Trigonometric Ratios:Important Trigonometric Ratios:
(a) sin n π = 0 ; cos n π = (−1)n ; tan n π = 0, where n ∈ Ι 8
π =
22
12
5π ;
π =
22
12
5π ;
13
13
15− & cos 36° or cos 5
π = 4
15+
1 0 .1 0 .1 0 .1 0 . C o nd it io na l I de nt it ie sCondi t ional Ident i t iesCondi t ional Ident i t iesCondi t ional Ident i t ies ::::
If A + B + C = π then :
(i) sin2A + sin2B + sin2C = 4 sinA sinB sinC
(ii) sinA + sinB + sinC = 4 cos 2
A cos
B + cos 2
C = − 1 − 4 cos A cos B cos C
(iv) cos A + cos B + cos C = 1 + 4 sin 2
A sin
(vi) tan 2
2
C
(viii) cot A cot B + cot B cot C + cot C cot A = 1
(ix) A + B + C =
π 2 then tan A tan B + tan B tan C + tan C tan A = 1
Solved Example # 12
If A + B + C = 180°, Prove that, sin2A + sin2B + sin2C = 2 + 2cosA cosB cosC.
Solution.
so that 2S = 2sin2A + 1 – cos2B +1 – cos2C
= 2 sin2A + 2 – 2cos(B + C) cos(B – C)
= 2 – 2 cos2A + 2 – 2cos(B + C) cos(B – C)
∴ S = 2 + cosA [cos(B – C) + cos(B+ C)]
since cosA = – cos(B+C)
Solved Example # 13
x2
− + 2y1
y2
so that we have
tanA + tanB + tanC = tanA tanB tanC ⇒ A + B + C = nπ, where n ∈ Ι Hence
L.H.S.
9
= tan2A tan2B tan2C
(i) sin(B + 2C) + sin(C + 2A) + sin(A + 2B) = 4sin 2
CB − sin
19. If A + B + C = 2S, prove that
(i) sin(S – A) sin(S – B) + sinS sin (S – C) = sinA sinB.
(ii) sin(S – A) + sin (S – B) + sin(S – C) – sin S = 4sin 2
A sin
1 11 11 11 1 .... RangeRangeRangeRange oooof Trigonometric Expressionf Trigonometric Expressionf Trigonometric Expressionf Trigonometric Expression::::
E = a sin θ + b cos θ
E = 22 ba + sin (θ + α), where tan α = a
b
a
Solved Example # 14
(i) 3sinx + 4cosx
– 5 ≤ 3sinx + 4cosx ≤ 5
(ii) 1+ 2sinx + 3cos2x
(i) 3 + (sinx – 2)2 Answer max = 12, min = 4.
(ii) 10cos2x – 6sinx cosx + 2sin2x Answer max = 11, min = 1.
(iii) cosθ + 3 2 sin

1111 2222 .... SineSineSineSine aaaand Cosine Seriesnd Cosine Seriesnd Cosine Seriesnd Cosine Series::::
sin α + sin (α + β) + sin (α + 2β ) +...... + sin ( )β−+α 1n =
2
1n
cos α + cos (α + β) + cos (α + 2β ) +...... + cos ( )β−+α 1n =
2
(i) cos 7
7 4π + cos
21. cos 1n2 +
1
22. sin2α + sin3α + sin4α + ..... + sin nα, where (n + 2)α = 2π Answer 0.
12
−1 ≤ cos θ ≤ 1 ∀ θ ∈ R
(b)sec2θ − tan2θ = 1 ; sec θ ≥ 1 ∀ θ ∈ R
(c)cosec2θ − cot2θ = 1 ; cosec θ ≥ 1 ∀ θ ∈ R
2. IMPORTANT T′ RATIOS: (a)sin n π = 0 ; cos n π = (-1)n ; tan n π = 0 where n ∈ I
(b)sin 2
(c)sin 15° or sin 12
π =
22
12
5π ;
π =
22
12
5π ;
13
5
π =
4
15+
3. TRIGONOMETRIC FUNCTIONS OF ALLIED ANGLES : If θ is any angle, then − θ, 90 ± θ, 180 ± θ, 270 ± θ, 360 ± θ etc. are called ALLIED ANGLES.
(a) sin (− θ) = − sin θ ; cos (− θ) = cos θ (b) sin (90°- θ) = cos θ ; cos (90° − θ) = sin θ (c) sin (90°+ θ) = cos θ ; cos (90°+ θ) = − sin θ (d)sin (180°− θ) = sin θ; cos (180°− θ) = − cos θ (e) sin (180°+ θ) = − sin θ ; cos (180°+ θ) = − cos θ (f) sin (270°− θ) = − cos θ ; cos (270°− θ) = − sin θ (g) sin (270°+ θ) = − cos θ ; cos (270°+ θ) = sin θ
4. TRIGONOMETRIC FUNCTIONS OF SUM OR DIFFERENCE OF TWO ANGLES :
(a) sin (A ± B) = sinA cosB ± cosA sinB (b) cos (A ± B) = cosA cosB sinA sinB
(c) sin²A − sin²B = cos²B − cos²A = sin (A+B) . sin (A− B)
(d) cos²A − sin²B = cos²B − sin²A = cos (A+B) . cos (A − B)
(e) tan (A ± B) = tan tan
tan tan
A B
A B
cot cot
A B
B A
1
± 5. FACTORISATION OF THE SUM OR DIFFERENCE OF TWO SINES OR COSINES :
(a) sinC + sinD = 2 sin 2
DC+ cos
2
DC+ cos
2
2
DC−
6. TRANSFORMATION OF PRODUCTS INTO SUM OR DIFFERENCE OF SINES & COSINES : (a) 2 sinA cosB = sin(A+B) + sin(A−B) (b) 2 cosA sinB = sin(A+B) − sin(A−B) (c) 2 cosA cosB = cos(A+B) + cos(A−B) (d) 2 sinA sinB = cos(A−B) − cos(A+B)
7. MULTIPLE ANGLES AND HALF ANGLES :
(a) sin 2A = 2 sinA cosA ; sin
θ = 2 sin
cos θ = cos2 θ 2
− sin² θ
2 = 2cos2 θ
2 .
2 cos2A = 1 + cos 2A , 2sin2A = 1 − cos 2A ; tan2A = A2cos1
A2cos1
2
(e) sin 3A = 3 sinA − 4 sin3A
(f) cos 3A = 4 cos3A − 3 cosA (g) tan 3A = Atan31
AtanAtan3 2
CtanBtanAtanCtanBtanAtan
−−− −++
NOTE IF : (i) A+B+C = π then tanA + tanB + tanC = tanA tanB tanC
(ii) A+B+C = 2
π then tanA tanB + tanB tanC + tanC tanA = 1
(b) If A + B + C = π then : (i) sin2A + sin2B + sin2C = 4 sinA sinB sinC
(ii) sinA + sinB + sinC = 4 cos 2
A cos
2
C
9. MAXIMUM & MINIMUM VALUES OF TRIGONOMETRIC FUNCTIONS: (a) Min. value of a2tan2θ + b2cot2θ = 2ab where θ ∈ R
(b) Max. and Min. value of acosθ + bsinθ are 22 ba + and – 22 ba +(c) If f(θ) = acos(α + θ) + bcos(β + θ) where a, b, α and β are known quantities then
– )cos(ab2ba 22 β−α++ < f(θ) < )cos(ab2ba
22 β−α++
, π

and α + β = σ (constant) then the maximum values of the expression
cosα cosβ, cosα + cosβ, sinα + sinβ and sinα sinβ
occurs when α = β = σ /2.
(e) If α,β ∈ 0 2
, π

and α + β = σ(constant) then the minimum values of the expression
secα + secβ, tanα + tanβ, cosecα + cosecβ occurs when α = β = σ /2. (f) If A, B, C are the angles of a triangle then maximum value of
sinA + sinB + sinC and sinA sinB sinC occurs when A = B = C = 600
(g) In case a quadratic in sinθ or cosθ is given then the maximum or minimum values can be interpretedby making a perfect square. 10. Sum of sines or cosines of n angles,
sin α + sin (α + β) + sin (α + 2β ) + ...... + sin ( )α β+ −n 1 = 2
2
n
sin
sin
β
β
sin
1n
cos α + cos (α + β) + cos (α + 2β ) + ...... + cos ( )α β+ −n 1 =
2
2
n
sin
sin
β
β
cos
1n
EXERCISE–I Q.1 Prove that cos²α + cos² (α + β) − 2cos α cos β cos (α + β) = sin²β Q.2 Prove that cos 2α = 2 sin²β + 4cos (α + β) sin α sin β + cos 2(α + β) Q.3 Prove that , tan α + 2 tan 2α + 4 tan 4α + 8 cot 8 α = cot α . Q.4 Prove that : (a) tan 20° . tan 40° . tan 60° . tan 80° = 3
(b) tan 9° − tan 27° − tan 63° + tan 81° = 4 . (c) 2
3
16
Q.5 Calculate without using trigonometric tables :
(a) cosec 10° − 3 sec 10° (b) 4 cos 20° − 3 cot 20° (c) °
°−° 20sin
20cos40cos2
(d)
Q.6(a) If X = sin
X − = 2 tan2θ.
(b) Prove that sin²12° + sin² 21° + sin² 39° + sin² 48° = 1+ sin² 9° + sin² 18° .
Q.7 Show that : (a) cot 7 1
2
2 3 6− − .
Q.8 If m tan (θ - 30°) = n tan (θ + 120°), show that cos 2 θ = )nm(2
nm
5 ; sin (α - β) =
4 π , then find the value of tan 2α.
Q.11 Prove that if the angles α & β satisfy the relation ( )
( )nm m
.
Q.12 (a) If y = 10 cos²x − 6 sin x cos x + 2 sin²x , then find the greatest & least value of y . (b) If y = 1 + 2 sin x + 3 cos2 x , find the maximum & minimum values of y ∀ x ∈ R . (c) If y = 9 sec2x + 16 cosec2x, find the minimum value of y ∀ x∈ R.
(d) Prove that 3 cos θ π
+

3 + 5 cos θ + 3 lies from - 4 & 10 .
(e) Prove that ( )2 3 4+ sin θ + 4 cos θ lies between − ( )522 + & 522 + .
Q.13 If A + B + C = π, prove that ∑

Ctan.Btan Atan = ∑ (tan A)A) − 2∑ (cot A).
Q.14 If α + β = c where α, β > 0 each lying between 0 and π /2 and c is a constant, find the maximum or minimum value of (a) sin α + sin β (b) sin α sin β (c) tan α + tan β (d) cosec α + cosec β
Q.15 Let A1 , A2 , ...... , An be the vertices of an n-sided regular polygon such that ;
413121 AA
1 += . Find the value of n.
Q.16 Prove that : cosec θ + cosec 2 θ + cosec 22 θ + ...... + cosec 2 n − 1 θ = cot (θ /2) − cot 2 n - 1θ Q.17 For all values of α , β , γ prove that;
cos α + cos β + cos γ + cos (α + β + γ ) = 4 cos 2
β+α .cos 2
γ +β . cos 2
tan . tan
sin sin
sin . sin
2 2
+ .
Q.20 If α + β = γ , prove that cos² α + cos² β + cos² γ = 1 + 2 cos α cos β cos γ .
Q.21 If α + β + γ = π 2
, show that ( ) ( )( )( )
+ + 1
.
Q.22 If A + B + C = π and cot θ = cot A + cot B + cot C, show that , sin (A − θ) . sin (B − θ) . sin (C − θ) = sin3 θ .
Q.23 If P = 19
π , then find P – Q.
Q.24 If A, B, C denote the angles of a triangle ABC then prove that the triangle is right angled if and only if sin4A + sin4B + sin4C = 0.
Q.25 Given that (1 + tan 1°)(1 + tan 2°)......(1 + tan 45°) = 2n, find n.
EXERCISE–II Q.1 If tan α = p/q where α = 6β, α being an acute angle, prove that;
2
1 (p cosec 2 β − q sec 2 β) = 22
qp + .
Q.2 Let A1 , A2 , A3 ............ An are the vertices of a regular n sided polygon inscribed in a circle of radius R. If (A
1
A n
)2 = 14 R2 , find the number of sides in the polygon. 15
= (cosθ + cosφ) cos(θ + φ) – (sinθ + sinφ) sin(θ + φ)
Q.4 Without using the surd value for sin 180 or cos 360 , prove that 4 sin 360 cos 180 = 5
Q.5 Show that , 2
x1 · x2 = 64
Q.7 If θ = 7π
, prove that tan θ . tan 2 θ + tan 2 θ . tan 4 θ + tan 4 θ . tan θ = − 7.
Q.8 For 0 < x < π 4
prove that , )xsinx(cosxsin
xcos 2 − > 8.
2π prove that, sin α + sin 2α + sin 4α =
2
88
1 =
k cos 2
Q.11 Prove that the value of cos A + cos B + cos C lies between 1 & 2
3 where AA + B + C = π.
Q.12 If cosA = tanB, cosB = tanC and cosC = tanA , then prove that sinA = sinB = sinC = 2 sin18°.
Q.13 Show that xsin
xcos3+ Rx∈∀ can not have any value between 22− and 22 . What inference
can you draw about the values of xcos3
xsin
5 . Find the value of (1 – sin t)(1 – cos t).
Q.15 Prove that from the equality ba
1
b
cos
a
α +
α .
Q.16 Prove that the triangle ABC is equilateral iff , cot A + cot B + cot C = 3 . Q.17 Prove that the average of the numbers n sin n°, n = 2, 4, 6, ......., 180, is cot 1°.
Q.18 Prove that : 4 sin 27° = ( ) ( ) 2 / 12 / 1 5355 −−+ .
Q.19 If A+B+C = π; prove that tan2 2
A + tan2
2
C
≥ 1.
Q.20 If A+B+C = π (A , B , C > 0) , prove that sin 2
A . sin
2
C ≤
8
1 .
Q.21 Show that elliminating x & y from the equations , sin x + sin y = a ;
cos x + cos y = b & tan x + tan y = c gives ( ) 2222
a4ba
ba8
−+ = c.
Q.22 Determine the smallest positive value of x (in degrees) for which tan(x + 100°) = tan(x + 50°) tan x tan (x – 50°).
Q.23 Evaluate : ∑ =

1 + cos α + cos β + cos γ = 0.
Q.25 ∀ x ∈ R, find the range of the function, f (x) = cos x (sin x + sin sin2 2x + α ) ; α ∈ [0, π]
EXERCISE–III Q.1 sec2θ =
+ is true if and only if : [JEE ’96, 1]
(A) x + y ≠ 0 (B) x = y , x ≠ 0 (C) x = y (D) x ≠ 0 , y ≠ 0
Q.2 (a) Let n be an odd integer. If sin nθ = r
n
0 br sinr θ, for every value of θ, then :
(A) b0 = 1, b1 = 3 (B) b0 = 0, b1 = n 16
1 = n (D) b
1 A
2 A
3 A
4 A
5 be a regular hexagon inscribed in a circle of unit radius .
Then the product of the lengths of the line segments A0 A1, A0 A2 & A0 A4 is :
(A) 3
3 3
2 (c) Which of the following number(s) is/are rational ? [ JEE '98, 2 + 2 + 2 = 6 out of 200 ]
(A) sin 15º (B) cos 15º (C) sin 15º cos 15º (D) sin 15º cos 75º

tan (1+ sec θ) (1+ sec 2θ) (1+ sec 4θ) .... (1 + sec2nθ) Then
(A) f 2 π
32
64
128
= 1 [JEE '99,3]
Q.4(a) Let f (θ) = sin θ (sin θ + sin 3 θ) . Then f (θ) : [ JEE 2000 Screening. 1 out of 35 ] (A) ≥ 0 only when θ ≥ 0 (B) ≤ 0 for all real θ (C) ≥ 0 for all real θ (D) ≤ 0 only when θ ≤ 0 .
(b) In any triangle ABC, prove that, cot 2
A + cot
C . [JEE 2000]
Q.5(a) Find the maximum and minimum values of 27cos 2x · 81sin 2x.
(b) Find the smallest positive values of x & y satisfying, x − y = 4
π , cot x + cot y = 2. [REE 2000, 3]
Q.6 If α + β = π 2
and β + γ = α then tanα equals [ JEE 2001 (Screening), 1 out of 35 ]
(A) 2(tanβ + tanγ ) (B) tanβ + tanγ (C) tanβ + 2tanγ (D) 2tanβ + tanγ
Q.7 If θ and φ are acute angles satisfying sinθ = 2
1 , cos φ =
(A)
, 6
5
Q.8 In an equilateral triangle, 3 coins of radii 1 unit each are kept so that they touch each other and also the sides of the triangle. Area of the triangle is
(A) 4 + 2 3 (B) 6 + 4 3
(C) 12 + 4
37 (D) 3 +
π 4
,0 and t1 = (tanθ)tanθ, t2 = (tanθ)cotθ, t3 = (cotθ)tanθ , t4 = (cotθ)cotθ, then
(A) t1 > t2 > t3 > t4 (B) t4 > t3 > t1 > t2 (C) t3 > t1 > t2 > t4 (D) t2 > t3 > t1 > t4 [JEE 2006, 3]
ANSWER SHEET ( EXERCISE–I)
Q 5. (a) 4 (b) −1 (c) 3 (d) 4 (e) 4
5 (f) 3 Q 10.
5
33
Q 12. (a) ymax = 11 ; ymin = 1 (b) ymax = 13
3 ; ymin = − 1, (c) 49
Q14. (a) max = 2 sin (c/2), (b) max. = sin2 (c/2), (c) min = 2 tan (c/2), (d) min = 2 cosec (c/2) Q 15. n = 7 Q23. 1 Q.25 n = 23
EXERCISE –II
− Q.25 – 1 2+ sin α ≤ y ≤ 1 2+ sin α
EXERCISE–III Q.1 B Q.2 (a) B, (b) C, (c) C Q.3 A, B, C, D Q.4 (a) C
Q.5 (a) max. = 35 & min. = 3–5 ; (b) x = 12
5π ; y =
Q.8 B Q.9 B
1 .1 .1 .1 .
when simplified reduces to:
(A) sin x cos x (B) − sin2x (C) − sin x cos x (D) sin2x
2. The expression 3
sin 66 is equal to
(A) 0 (B) 1 (C) 3 (D) sin 4α + sin 6α 3. If tan A & tan B are the roots of the quadratic equation x2 − ax + b = 0, then the value of sin2 (A + B).
(A) 22
)a1(b a −
4. The value of log 2 [cos2 (α + β) + cos2 (α − β) − cos 2α. cos 2β] :
(A) depends on α & β both (B) depends on α but not on β (C) depends on β but not on α (D) independent of both α & β.
5. ° °°°+°
80sin
10sin50sin70sin820cos 2 is equal to:
(A) 1 (B) 2 (C) 3/4 (D) none 6. If cos A = 3/4, then the value of 16cos2 (A/2) – 32 sin (A/2) sin (5A/2) i s
(A) – 4 (B) – 3 (C) 3 (D) 4 7. If y = cos2 (45º + x) + (sin x − cos x)2 then the maximum & minimum values of y are:
(A) 2 & 0 (B) 3 & 0 (C) 3 & 1 (D) none
8. The value of cos 19
π + cos
17π is equal to:
(A) 1/2 (B) 0 (C) 1 (D) none 9. The greatest and least value of ( )23xcosxsinlog
2 +− are respectively:
(A) 2 & 1 (B) 5 & 3 (C) 7 & 5 (D) 9 & 7
10. In a right angled triangle the hypotenuse is 2 2 times the perpendicular drawn from the opposite vertex. Then the other acute angles of the triangle are
(A) 3
12. If 4
3π < α < π, then α+α 2sin 1cot2 is equal to
(A) 1 + cot α (B) – 1 – cot α (C) 1 – cot α (D) – 1 + cot α
13. If x ∈
4 + x2sinxsin4 24 + is always equal to
(A) 1 (B) 2 (C) – 2 (D) none of these 14. If 2 cos x + sin x = 1, then value of 7 cos x + 6 sin x is equal to
(A) 2 or 6 (B) 1 or 3 (C) 2 or 3 (D) none of these
15. If cosec A + cot A = 2
11 , then tan A is
(A) 22
21 (B)
1
– cos α = n, then (A) m (mn2)1/3 – n(nm2)1/3 = 1 (B) m(m2n)1/3 – n(nm2)1/3 = 1 (C) n(mn2)1/3 – m(nm2)1/3 = 1 (D) n(m2n)1/3 m(mn2)1/3 = 1
17. The expression xcos10x3cos5x5cos
is equal to
(A) cos 2x (B) 2 cos x (C) cos2 x (D) 1 + cos x
18. If Bsin
Bcos
Acos =
2
5 , 0 < A, B < π /2, then tan A + tan B is equal to
(A) 5 / 3 (B) 3 / 5 (C) 1 (D) 5 / )35( +
19. If sin 2θ = k, then the value of θ+
θ 2
Part : (B) May have more than one options correct 20. Which of the following is correct ?
(A) sin 1° > sin 1 (B) sin 1° < sin 1 (C) cos 1° > cos 1 (D) cos 1° < cos 1 21. If 3 sin β = sin (2α + β), then tan (α + β) – 2 tan α is
(A) independent of α (B) independent of β (C) dependent of both α and β (D) independent of α but dependent of β
22. It is known that sin β = 5
4 & 0 < β < π then the value of
α
sin
cos
2
is:
(A) independent of α for all β in (0, π) (B) 3
5 for tan β > 0
(C) 15

β < 0 (D) none 23. If the sides of a right angled triangle are {cos2α + cos2β + 2cos(α + β)} and
{sin2α + sin2β + 2sin(α + β)}, then the length of the hypotenuse is:
(A) 2[1+cos(α − β)] (B) 2[1 − cos(α + β)] (C) 4 cos2
2
β+α
24. If x = sec φ − tan φ & y = cosec φ + cot φ then:
(A) x = 1y
(D) xy + x − y + 1 = 0
25. (a + 2) sin α + (2a – 1) cos α = (2a + 1) i f tan α =
(A) 4
3 (B)
, (a ≠ c)
y = a cos2x + 2b sin x cos x + c sin2x z = a sin2x – 2b sin x cos x + c cos2x, then
(A) y = z (B) y + z = a + c (C) y – z = a – c (D) y – z = (a – c)2 + 4b2
27.
n
BsinAsin
BcosAcos
2
BA − : n is even (C) 0 : n is odd (D) none
28. The equation sin6x + cos6x = a2 has real solution if
(A) a ∈ (–1, 1) (B) a ∈

1,
2
1
EXERCISE–IV (Subjective) 1. The minute hand of a watch is 1.5 cm long. How far does its tip move in 50 minutes?(Use π = 3.14). 2. If the arcs of the same length in two circles subtend angles 75° and 120° at the centre, find the ratio of
their radii. 3. Sketch the following graphs :
(i) y = 3 sin 2x (ii) y = 2 tan x (iii) y = sin 2
x
θ – cos 3 θ cos
2
2
3 , π < x <
2
9α = cosec 4α.
8. Prove that, sin 3 x. sin3 x + cos 3 x. cos3 x = cos3 2 x.
9. If tan α = q
p where α = 6 β, α being an acute angle, prove that;
2
1
(p cosec 2 β − q sec 2 β) = 22 qp + .
10. If tan β = γ α+
γ +α
γ α+
γ +α
1° or tan 82
2 / 12 / 1
12. Prove that, tan α + 2 tan 2α + 4 tan 4α + 8 cot 8 α = cot α.
13. If cos (β − γ ) + cos (γ − α) + cos (α − β) = 2
3− , prove that
cos α + cos β + cos γ = 0, sin α + sin β + sin γ = 0.
14. Prove that from the equality ba
1
b
cos
a
α +
α
15. Prove that: cosec θ + cosec 2 θ + cosec 22 θ +... + cosec 2 n − 1θ = cot (θ /2) − cot 2n − 1 θ. Hence or
otherwise prove that cosec 15
4π + cosec
32π = 0
16. Let A1, A2,......, An be the vertices of an n−sided regular polygon such that; 413121 AA
1
AA
1
AA
1 += .
Find the value of n. 17. If A + B + C = π, then prove that
(i) tan² 2
3
n = cosnθ – sinnθ, then show that
P n – P
n – 4 Q
n – 4 and hence show that
P 4 = 1 – 2 sin2θ cos2θ , Q
4 = cos2θ – sin2θ
20. If sin (θ + α) = a & sin (θ + β) = b (0 < α, β, θ < π /2) then find the value of cos2 (α − β) − 4 ab cos(α − β)
21. If A + B + C = π, prove that tan B tan C + tan C tan A + tan A tan B = 1 + sec A. sec B. sec C.
22. If tan2α + 2tanα. tan2β = tan2β + 2tanβ. tan2α, then prove that each side is equal to 1 or tan α = ± tan β.
EXERCISE–IV 1. D 2. B 3. A 4. D 5. B 6. C 7. B
8. A 9. B 10. B 11. B 12. B 13. B 14. A
15. C 16. A 17. B 18. D 19. B 20. BC
21. AB 22. BC 23. AC 24. BCD 25. BD 26. BC
27. BC 28. BD
EXERCISE–V 1. 7.85 cm 2. r1 : r2 = 8 : 5
6. sin 2
20
Equations
Index:
3. Answer Key
1
P a g e : 2 o f 1 5 T R I G . E Q U A T I O N S
1.1 .1 .1 . Trigonome tric E quatio n :Trigonometric Equation :Trigonometric Equation :Trigonometric Equation : An equation involving one or more trigonometric ratios of an unknown angle is called a trigonometric equation.
2.2 .2 .2 . Solut ion of Tr igonometric Equat ion :Solution of Trigonometric Equation :Solution of Trigonometric Equation :Solution of Trigonometric Equation : A solution of trigonometric equation is the value of the unknown angle that satisfies the equation.
e.g. if sinθ = 2
4
π ,
4
3π ,
4
9π ,
4
11π , ...........
Thus, the trigonometric equation may have infinite number of solutions (because of their periodic nature) and can be classified as : (i) Principal solution (ii) General solution.
2.12.12.12 .1 Princ ipal solut ions:Principal solutions:Principal solutions:Principal solutions: The solutions of a trigonometric equation which lie in the interval [0, 2π) are called Principal solutions.
e.g Find the Principal solutions of the equation sinx = 2
1 .
Solution.
5π which lie in [0, 2π) and whose sine is
2
1
1 are
2.22.22.22 .2 General Solution :General Solution :General Solution :General Solution :
The expression involving an integer 'n' which gives all solutions of a trigonometric equation is called General solution.
General solution of some standard trigonometric equations are given below.
3.3 .3 .3. General Solution of Some Standard Trigonometric Equations :General Solution of Some Standard Trigonometric Equations :General Solution of Some Standard Trigonometric Equations :General Solution of Some Standard Trigonometric Equations :

2 , n ∈ Ι.
(ii) If cos θ = cos α ⇒ θ = 2 n π ± α where α ∈ [0, π], n ∈ Ι.

2 , n ∈ Ι.
(iv) If sin² θ = sin² α ⇒ θ = n π ± α, n ∈ Ι.
(v) If cos² θ = cos² α ⇒ θ = n π ± α, n ∈ Ι.
(vi) If tan² θ = tan² α ⇒ θ = n π ± α, n ∈ Ι. [ Note: α is called the principal angle ] Some Important deductions :Some Important deductions :Some Important deductions :Some Important deductions :
(i) sinθ = 0 ⇒ θ = nπ, n ∈ Ι
(ii) sinθ = 1 ⇒ θ = (4n + 1) 2
π , n ∈ Ι
π , n ∈ Ι
π , n ∈ Ι
(v) cosθ = 1 ⇒ θ = 2nπ, n ∈ Ι (vi) cosθ = – 1 ⇒ θ = (2n + 1)π, n ∈ Ι (vii) tanθ = 0 ⇒ θ = nπ, n ∈ Ι
Solved Example # 1
sin θ = 2
Solved Example # 2
5ππππ , n ∈ Ι∈ Ι∈ Ι∈ Ι Ans.
Solved Example # 3
Solve tanθθθθ = 2
Solution.
tanθ = 2 ............(i)Let 2 = tanα ⇒ tanθ = tanα ⇒⇒⇒⇒ θ θ θ θ = nππππ + αααα, where αααα = tan –1(2), n ∈ Ι∈ Ι∈ Ι∈ Ι
Self Practice Problems:
π , n ∈ Ι (2)
Solved Example # 5
For equation (i) to be defined θ ≠ (2n + 1) 2
π , n ∈ Ι
θ θ
⇒ sin2θ =
2
2
3
⇒ sin2θ = sin2 3
Self Practice Problems :
Ans. (1) nπ ± 3
2
4
Types of Trigonometric Equations :Types of Trigonometric Equations :Types of Trigonometric Equations :Types of Trigonometric Equations :
Type -1Type -1Type -1Type -1
Trigonometric equations which can be solved by use of factorization.
Solved Example # 6
Solve (2sinx – cosx) (1 + cosx) = sin2x.
Solution. (2sinx – cosx) (1 + cosx) = sin2x ⇒ (2sinx – cosx) (1 + cosx) – sin2x = 0 ⇒ (2sinx – cosx) (1 + cosx) – (1 – cosx) (1 + cosx) = 0 ⇒ (1 + cosx) (2sinx – 1) = 0 ⇒ 1 + cosx = 0 or 2sinx – 1 = 0
⇒ cosx = – 1 or sinx = 2
1
⇒ x = (2n + 1)π, n ∈ Ι or sin x = sin 6
π ⇒ x = nπ + (– 1)n
∴ Solution of given equation is
(2n + 1)ππππ, n ∈ Ι∈ Ι∈ Ι∈ Ι or nππππ + (–1)n 6
ππππ , n ∈ Ι∈ Ι∈ Ι∈ Ι Ans.
2
Ans. (1) (2n + 1)π, n ∈ Ι (2) 2nπ –
2
6
Type - 2Type - 2Type - 2Type - 2
Trigonometric equations which can be solved by reducing them in quadratic equations.
Solved Example # 7
Solve 2 cos2x + 4cosx = 3sin2x
⇒

∴ cosx ≠ 5
cosx = 5
192 +−

4
Self Practice Problems : 1. Solve cos2θ – ( 2 + 1)

Ans. (1) 2nπ ± 3
4

Type - 3Type - 3Type - 3Type - 3
Trigonometric equations which can be solved by transforming a sum or difference of trigonometric ratios into their product.
Solved Example # 8
Solution. cos3x + sin2x – sin4x = 0 ⇒ cos3x + 2cos3x.sin(– x) = 0
⇒ cos3x – 2cos3x.sinx = 0 ⇒ cos3x (1 – 2sinx) = 0 ⇒ cos3x = 0 or 1 – 2sinx = 0
⇒ 3x = (2n + 1) 2
2
1
6
π
(2n + 1) 6
6
2. Solve 5sinx + 6sin2x +5sin3x + sin4x = 0
3. Solve cosθ – sin3θ = cos2θ
Ans. (1) 3
2π , n ∈ Ι
2
4
π , n ∈ Ι
Type - 4 Trigonometric equations which can be solved by transforming a product of trigonometric ratios into their sum or difference.
Solved Example # 9
Solve sin5x.cos3x = sin6x.cos2x
⇒ sin2x = 0 or 2cos2x – 1 = 0
⇒ 2x = nπ, n ∈ Ι or cos2x = 2
1
3
2
6
Type - 5
Trigonometric Equations of the form a sinx + b cosx = c, where a, b, c ∈ R, can be solved by dividing
both sides of the equation by 22 ba + . 5
P a g e : 6 o f 1 5 T R I G . E Q U A T I O N SSolved Example # 10
Solve sinx + cosx = 2
∴ divide both sides of equation (i) by 2 , we get
sinx . 2
1 + cosx.
2nππππ + 4
Note : Trigonometric equation of the form a sinx + b cosx = c can also be solved by changing sinx and cosx into their corresponding tangent of half the angle.
Solved Example # 11
cosx =
2
⇒ t = 2

6
P a g e : 7 o f 1 5 T R I G . E Q U A T I O N S2. Solve sinx + tan
2
Type - 6
Trigonometric equations of the form P(sinx ± cosx, sinx cosx) = 0, where p(y, z) is a polynomial, can be solved by using the substitution sinx ± cosx = t.
Solved Example # 12
Solve sinx + cosx = 1 + sinx.cosx
Solution. sinx + cosx = 1 + sinx.cosx ........(i) Let sinx + cosx = t ⇒ sin2x + cos2x + 2 sinx.cosx = t2
⇒ sinx.cosx = 2
1t2 − in (i), we get
t = 1 + 2
1t2 −
⇒ t2 – 2t + 1 = 0 ⇒ t = 1 t = sinx + cosx ⇒ sinx + cosx = 1 .........(ii)
divide both sides of equation (ii) by 2 , we get
⇒ sinx. 2
1 + cosx.
x = 2nππππ + 2
(ii) i f we take negative sign, we get x = 2nππππ, n ∈ Ι∈ Ι∈ Ι∈ Ι Ans.
Self Practice Problems:
3. Solve (1 – sin2x) (cosx – sinx) = 1 – 2sin2x.
Ans. (1) nπ – 4
4
4
Type - 7
Trigonometric equations which can be solved by the use of boundness of the trigonometric ratios sinx and cosx.
Solved Example # 13
⇒ sin 4 x5 + cosx = 2 ........(ii)
7
Now equation (ii) will be true if
sin 4
⇒ 4
⇒ x = 5
)2n8( π+ , n ∈ Ι ........(iii) and x = 2mπ, m ∈ Ι ........(iv)
Now to find general solution of equation (i)
5
⇒ n = 4
......... ......... .........
......... ......... ......... if m = 4p – 3, p ∈ Ι then n = 5p – 4, p ∈ Ι
∴ general solution of given equation can be obtained by substituting either m = 4p – 3 in equation (iv) or n = 5p – 4 in equation (iii)
∴ general solution of equation (i) is (8p – 6)ππππ, p ∈ Ι∈ Ι∈ Ι∈ Ι Ans.
2. Solve 3xcosx5sin3 2 −− = 1 – sinx
Ans. (1) (4p – 3) 2
π , p ∈ Ι (2) 2mπ +
2
THINGS TO REMEMBER :
1. If sin θ = sin α ⇒ θ = n π + (−1)n α where α ∈ −
π π
2 2 , , n ∈ I .
2. If cos θ = cos α ⇒ θ = 2 n π ± α where α ∈ [0 , π] , n ∈ I .

2 2 , , n ∈ I .
4. If sin² θ = sin² α ⇒ θ = n π ± α.
5. cos² θ = cos² α ⇒ θ = n π ± α.
6. tan² θ = tan² α ⇒ θ = n π ± α. [ Note : α is called the principal angle ]
7. TYPES OF TRIGONOMETRIC EQUATIONS :
(a) Solutions of equations by factorising . Consider the equation ;
(2 sin x − cos x) (1 + cos x) = sin² x ; cotx – cosx = 1 – cotx cosx
(b) Solutions of equations reducible to quadratic equations. Consider the equation :
3 cos² x − 10 cos x + 3 = 0 and 2 sin2x + 3 sinx + 1 = 0
(c) Solving equations by introducing an Auxilliary argument . Consider the equation :
sin x + cos x = 2 ; 3 cos x + sin x = 2 ; secx – 1 = ( 2 – 1) tanx
(d) Solving equations by Transforming a sum of Trigonometric functions into a product.
Consider the example : cos 3 x + sin 2 x − sin 4 x = 0 ;
sin2x + sin22x + sin23x + sin24x = 2 ; sinx + sin5x = sin2x + sin4x
(e) Solving equations by transforming a product of trigonometric functions into a sum.
8
P a g e : 9 o f 1 5 T R I G . E Q U A T I O N SConsider the equation :
sin 5 x . cos 3 x = sin 6 x .cos 2 x ; 8 cosx cos2x cos4x = xsin
x6sin ; sin3θ = 4sinθ sin2θ sin4θ
(f) Solving equations by a change of variable :
(i) Equations of the form of a . sin x + b . cos x + d = 0 , where a , b & d are real
numbers & a , b ≠ 0 can be solved by changing sin x & cos x into their corresponding
tangent of half the angle. Consider the equation 3 cos x + 4 sin x = 5.
(ii) Many equations can be solved by introducing a new variable . eg. the equation
sin4 2 x + cos4 2 x = sin 2 x . cos 2 x changes to
2 (y + 1) y−
1
2 = 0 by substituting , sin 2 x . cos 2 x = y..
(g) Solving equations with the use of the Boundness of the functions sin x & cos x or by
making two perfect squares. Consider the equations :
sin x cos sin x
x 4
sin2x + 2tan2x + 3
4 tanx – sinx +
11 = 0
8. TRIGONOMETRIC INEQUALITIES :There is no general rule to solve a Trigonometric inequations
and the same rules of algebra are valid except the domain and range of trigonometric functions should be
kept in mind.
Consider the examples :
2
1
2
15 +
Q.2 Find all the values of θ satisfying the equation; sin θ + sin 5 θ = sin 3 θ such that 0 ≤ θ ≤ π.
Q.3 Find all value of θ, between 0 & π, which satisfy the equation; cos θ . cos 2 θ . cos 3 θ = 1/4.
Q.4 Solve for x , the equation 13 18− tanx = 6 tan x – 3, where – 2π < x < 2π.
Q.5 Determine the smallest positive value of x which satisfy the equation, 1 2 2 3 0+ − =sin cosx x .
Q.6 2 3 4
π
Q.7 Find the general solution of the trigonometric equation 223 )xsinx(coslog
)xsinx(coslog 2
=− −

++ .
Q.8 Find all values of θ between 0° & 180° satisfying the equation;
cos 6 θ + cos 4 θ + cos 2 θ + 1 = 0 .
Q.9 Find the solution set of the equation,
10
x62x log
−− (sin 2x).
Q.10 Find the value of θ, which satisfy 3 − 2 cosθ − 4 sinθ − cos 2θ + sin 2θ = 0.
Q.11 Find the general solution of the equation, sin πx + cos πx = 0. Also find the sum of all solutions
in [0, 100].
Q.12 Find the least positive angle measured in degrees satisfying the equation
sin3x + sin32x + sin33x = (sinx + sin2x + sin3x)3. 9
P
a g e : 1 0 o f 1 5 T R I G . E Q U A T I O N SQ.13 Find the general values ofθ for which the quadratic function
(sinθ) x2 + (2cosθ)x + 2
sincos θ+θ is the square of a linear function.
Q.14 Prove that the equations (a) sin x · sin 2x · sin 3x = 1 (b) sin x · cos 4x · sin 5x = – 1/2
have no solution.
Q.15 Let f (x) = sin6x + cos6x + k(sin4x + cos4x) for some real number k. Determine
(a) all real numbers k for which f (x) is constant for all values of x.
(b) all real numbers k for which there exists a real number 'c' such that f (c) = 0.
(c) If k = – 0.7, determine all solutions to the equation f (x) = 0.
Q.16 If α and β are the roots of the equation, a cos θ + b sin θ = c then match the entries of column-I
with the entries of column-II.
Column-I Column-II
b2
ac
+ −
Q.17 Find all the solutions of, 4 cos2x sin x − 2 sin2x = 3 sin x.
Q.18 Solve for x, (− π ≤ x ≤ π) the equation; 2 (cos x + cos 2 x) + sin 2 x (1 + 2 cos x) = 2 sin x.
Q.19 Solve the inequality sin2x > 2 sin2x + (2 – 2 )cos2x.
Q.20 Find the set of values of 'a' for which the equation, sin4 x + cos4 x + sin 2x + a = 0 possesses solutions.
Also find the general solution for these values of 'a'.
Q.21 Solve: tan22 x + cot22 x + 2 tan 2 x + 2 cot 2 x = 6.
Q.22 Solve: tan2 x . tan23 x . tan 4 x = tan2 x − tan23 x + tan 4 x.
Q.23 Find the set of values of x satisfying the equality
sin
x2cos2 3sin3cos
+ .
Q.24 Let S be the set of all those solutions of the equation,
(1 + k)cos x cos (2x − α) = (1 + k cos 2x) cos(x − α) which are independent of k & α. Let H be the set of all such solutions which are dependent on k & α. Find the condition on k & α such that H is a
non-empty set, state S. If a subset of H is (0, π) in which k = 0, then find all the permissible values of α.
Q.25 Solve for x & y, x y x y y
x y x y y
cos cos sin
sin cos sin
+ =
+ =
Q.26 Find the value of α for which the three elements set S = {sin α, sin 2α, sin 3α} is equal to the three
element set T = {cos α, cos 2α, cos 3α}.
Q.27 Find all values of 'a' for which every root of the equation, a cos 2x + a cos 4x + cos 6x = 1
is also a root of the equation, sin x cos 2 x = sin 2x cos 3x − 1 2
sin 5x , and conversely, every root 10
P
a g e : 1 1 o f 1 5 T R I G . E Q U A T I O N Sof the second equation is also a root of the first equation.
Q.28 Solve the equations for 'x' given in column-I and match with the entries ofcolumn-II.
Column-I Column-II
(A) cos 3x . cos3 x + sin 3x . sin3 x = 0 (P) n π ± π 3
(B) sin 3α = 4 sin α sin(x + α) sin(x − α) (Q) nπ + 4
π , n ∈ I
where α is a constant ≠ nπ.
(C) | 2 tan x – 1 | + | 2 cot x – 1 | = 2. (R) nπ π4 8+ , n ∈ I
(D) sin10 x + cos10 x = 29
16 cos42 x. (S)
± π 4
EXERCISE–II
Q.1 Solve the following system of equations for x and y [REE ’2001(mains), 3]
5 (cos sec )ec x y
2 2 3−
= 64.
Q.2 The number of integral values of k for which the equation 7cosx + 5sinx = 2k + 1 has a solution is
(A) 4 (B) 8 (C) 10 (D) 12 [JEE 2002 (Screening), 3]
Q.3 cos(α – β) = 1 and cos(α + β) = 1/e, where α, β ∈ [– π, π], numbers of pairs of α, β which satisfy
both the equations is
(A) 0 (B) 1 (C) 2 (D) 4 [JEE 2005 (Screening)]
Q.4 If 0 < θ < 2π, then the intervals of values of θ for which 2sin2θ – 5sinθ + 2 > 0, is
(A)
Q.5 The number of solutions of the pair of equations
2 sin2θ – cos2θ = 0
2 cos2θ – 3 sin θ = 0
in the interval [0, 2π] is
(A) zero (B) one (C) two (D) four [JEE 2007, 3]
ANSWER EXERCISE–I
π 6
, π 3
8
7 ,
3
2 ,
8
5 ,
8
3 ,
3 ,
8
ππππππ
Q.4 α − 2 π ; α − π , α , α + π , where tan α = 2
3 Q.5 x = π /16
Q.6 In; 12
12
Q.9 x = − 3
5π Q.10 θ = 2 n π or 2 n π +
π 2
1 , n ∈ I; sum = 5025Q.12 72°
Q.13 4
n2 π
+π or (2n+1)π – tan–12 , In∈ Q.15 (a) – 2
3 ; (b) k ∈
2
nπ ±
6
π
Q.16 (A) R; (B) S; (C) P; (D) Q Q.17 nπ ; n π + (−1)n π 10
or n π + (−1)n −
P
a g e : 1 2 o f 1 5 T R I G . E Q U A T I O N S
Q.20 1
where n ∈ I and a ∈ −

+ (−1)n π 8
Q.23 x = 2nπ + 4
3π , In∈
Q.24 (i) k sin α ≤ 1 (ii) S = n π , n ∈ I (iii) α ∈ (− m π , 2 π − m π) m ∈ I
Q.25 x = ± 5 5 & y = n π + tan−1 1
2 Q.26
π
Q.27 a = 0 or a < − 1 Q.28 (A) S; (B) P; (C) Q; (D) R
EXERCISE–II
Q.2 B Q.3 D Q.4 A Q.5 C
Part : (A) Only one correct option
1. The solution set of the equation 4sinθ.cosθ – 2cosθ – 2 3 sinθ + 3 = 0 in the interval (0, 2π) is
(A)
ππ
4
7 ,
4
πππ
6
11 ,
6
5 ,
6
2. All solutions of the equation, 2 sinθ + tanθ = 0 are obtained by taking all integral values of m and n in:
(A) 2nπ + 2
2
3
(C) nπ or m π ± π 3
where n, m ∈ Ι (D) nπ or 2m π ± π 3
where n, m ∈ Ι
3. If 20 sin2 θ + 21 cos θ − 24 = 0 & 7
4
θ 2
4. The general solution of sinx + sin5x = sin2x + sin4x is:
(A) 2 nπ ; n ∈ Ι (B) nπ ; n ∈ Ι (C) nπ /3 ; n ∈ Ι (D) 2 nπ /3 ; n ∈ Ι
5. A triangle ABC is such that sin(2A + B) = 2 1 . If A, B, C are in A.P. then the angle A, B, C are
respectively.
6. The maximum value of 3sinx + 4cosx is
(A) 3 (B) 4 (C) 5 (D) 7
7. If sin θ + 7 cos θ = 5, then tan (θ /2) is a root of the equation
(A) x2 − 6x + 1 = 0 (B) 6x2 − x − 1 = 0 (C) 6x2 + x + 1 = 0 (D) x2 − x + 6 = 0
12
P
8. sin cos
(A) θ ∈ 0 2

9. The number of integral values of a for which the equation cos 2x + a sin x = 2a − 7 possesses a solution
is
(A) 2 (B) 3 (C) 4 (D) 5
10. The principal solution set of the equation, 2 cos x = 2 2 2+ sin x is
(A)
ππ
8
13 ,
11. The number of all possible triplets (a 1 , a
2 , a
(A) 0 (B) 1 (C) 2 (D) infinite

n ,0 , n ∈ N, then greatest value of n is
(A) 8 (B) 10 (C) 13 (D) 15
13. The solution of |cosx| = cosx – 2sinx is
(A) x = nπ, n ∈ Ι (B) x = nπ + 4
π , n ∈ Ι
π , n ∈ Ι (D) (2n + 1)π +
4
π , n ∈ Ι
14. The arithmetic mean of the roots of the equation 4cos3x – 4cos2x – cos(π + x) – 1 = 0 in the interval
[0, 315] is equal to
(A) 49π (B) 50π (C) 51π (D) 100π
15. Number of solutions of the equation cos 6x + tan 2 x + cos 6x . tan2 x = 1 in the interval [0, 2π] is :
(A) 4 (B) 5 (C) 6 (D) 7
Part : (B) May have more than one options correct
16. sinx − cos2x − 1 assumes the least value for the set of values of x given by:
(A) x = nπ + (−1)n+1 (π /6) , n ∈ Ι (B) x = nπ + (−1)n (π /6) , n ∈ Ι (C) x = nπ + (−1)n (π /3), n ∈ Ι (D) x = nπ − (−1)n (π /6) , n ∈ Ι
17. cos4x cos8x − cos5x cos9x = 0 if
(A) cos12x = cos 14 x (B) sin13 x = 0
(C) sinx = 0 (D) cosx = 0
18. The equation 2sin 2
x . cos2x + sin2x = 2 sin
2
(A) sin2x = 1 (B) sin2x = – 1 (C) cosx = 2
1
19. sin2x + 2 sin x cos x − 3cos2x = 0 if
(A) tan x = 3 (B) tanx = − 1
(C) x = nπ + π /4, n ∈ Ι (D) x = nπ + tan−1 (−3), n ∈ Ι
20. sin2x − cos 2x = 2 − sin 2x if
(A) x = nπ /2, n ∈ Ι (B) x = nπ − π /2, n ∈ Ι (C) x = (2n + 1) π /2, n ∈ Ι (D) x = nπ + (−1)n sin−1 (2/3), n ∈ Ι
13
P
1. Solve cotθ = tan8θ
5. Solve the equation: sin 6x = sin 4x − sin 2x .
6. Solve: cos θ + sin θ = cos 2 θ + sin 2 θ .
7. Solve 4 sin x . sin 2x . sin 4x = sin 3x .
8. Solve sin2nθ – sin2(n – 1)θ = sin2θ, where n is constant and n ≠ 0, 1
9. Solve tanθ + tan2θ + 3 tanθ tan2θ = 3 .
10. Solve: sin3 x cos 3 x + cos3 x sin 3
x + 0.375 = 0
sin
13. If tan θ + sin φ = 3
2 & tan² θ + cos² φ =
7
4 then find the general value of θ & φ .
14. Solve for x , the equation 13 18− tanx = 6 tan x − 3, where − 2 π < x < 2 π .
15. Find the general solution of sec 4 θ − sec 2 θ = 2 .
16. Solve the equation 3
2 sin x − cos x = cos² x .
17. Solve for x: 2 3 4
1 8 2 22sin sin . cosx x x+
= +π .

x . tan 4 x = tan2
x − tan2 3 x + tan 4
x .
20. Find the values of x, between 0 & 2 π, satisfying the equation; cos 3x + cos 2x = sin 3
2
P
21. Solve: cos 2
x cos 6 x = − 1 .
22. Solve the equation, sin2 4 x + cos2 x = 2 sin 4
x cos4 x .
EXERCISE # 1EXERCISE # 1EXERCISE # 1EXERCISE # 1
1. D 2. B 3. D 4. C 5. B 6. C 7. B
8. B 9. D 10. A 11. D 12. D 13. D 14. C
15. D 16. AD 17. ABC 18. ABCD 19. CD
20. BC
1.
6

, n ∈ Ι
3
m
− π
π 2
, n ∈ Ι
, n ∈ Ι
, n ∈ I
14. α − 2 π; α − π, α, α + π, where tan α = 2
3
π , n ∈ Ι
16. x = (2 n + 1)π, , n ∈ Ι or 2 n π ±
π 3
, n ∈ Ι
12
20. π π
π π π
π 2
, n∈I
Triangle
Index:
3. Answer Key
1
Properties & Solution of Triangle 1. Sine Rule:
In any triangle ABC, the sines of the angles are proportional to the opposite sides i.e.
Csin
c
Bsin
b
Asin
a == .
c
ba + =
2
ba + =
2
Asin
a =
Bsin
b =
Csin
⇒ a = k sinA, b = k sinB and c = k sinC
L.H.S. = c
= R.H.S. Hence L.H.S. = R.H.S. Proved
Example : In any ABC, prove that (b2 – c2) cot A + (c2 – a2) cot B + (a2 – b2) cot C = 0
Solution. We have to prove that (b2 – c2) cot A + (c2 – a2) cot B + (a2 – b2) cot C = 0
from sine rule, we know that
a = k sinA, b = k sinB and c = k sinC∴ (b2 – c2) cot A = k2 (sin2B – sin2C) cot A sin2B – sin2C = sin (B + C) sin (B – C) ∴ (b2 – c2) cot A = k2 sin (B + C) sin (B – C) cotA B + C = π – A
∴ (b2 – c2) cot A = k2 sin A sin (B – C) Asin
Acos cosA = – cos(B + C)
= – 2
k2
⇒ (b2 – c2) cot A = – 2
k2
Similarly (c2 – a2) cot B = – 2
k2
and (a2 – b2) cot C = – 2
k2
[sin 2A – sin 2B] ..........(iii)
adding equations (i), (ii) and (iii ), we get (b2 – c2) cot A + (c2 – a2) cot B + (a2 – b2) cot C = 0 Hence Proved
Self Practice Problems In any ABC, prove that
1. a sin
b c
2
+ − or a² = b² + c² − 2bc cos A = b2 + c2 + 2bc cos (B + C)
(ii) cos B = c a b
c a
+ −
Example : In a triangle ABC if a = 13, b = 8 and c = 7, then find sin A.
Solution. cosA = bc2
3 Ans.
*Example : In a ABC, prove that a(b cos C – c cos B) = b2 – c2
Solution. We have to prove a (b cosC – c cosB) = b2 – c2. from cosine rule we know that
cosC = ab2
Hence L.H.S. = R.H.S. Proved

++
1
1. The sides of a triangle ABC are a, b, 22 baba ++ , then prove that the greatest angle is 120°.
2. In a triangle ABC prove that a(cosB + cosC) = 2(b + c) sin2
2
A .
3. Projection Formula:(i) a = b cosC + c cosB (ii) b = c cosA + a cosC (iii) c = a cosB + b cosA Example : In a triangle ABC prove that a(b cosC – c cosB) = b2 – c2
Solution. L.H.S. = a (b cosC – c cosB) = b (a cosC) – c (a cosB) .. .... ... ...(i)
From projection rule, we know that b = a cosC + c cosA ⇒ a cosC = b – c cosA
& c = a cosB + b cosA ⇒ a cosB = c – b cosA Put values of a cosC and a cosB in equation (i), we get
L.H.S. = b (b – ccos A) – c(c – b cos A) = b2 – bc cos A – c2 + bc cos A = b2 – c2
= R.H.S. Hence L.H.S. = R.H.S. Proved
Note: We have also proved a (b cosC – ccosB) = b2 – c2 by using cosine – rule in solved *Example. Example : In a ABC prove that (b + c) cos A + (c + a) cos B + (a + b) cos C = a + b + c.
Solution. L.H.S. = (b + c) cos A (c + a) cos B + (a + B) cos C 3
= b cos A + c cos A + c cos B + a cos B + a cos C + b cos C = (b cos A + a cos B) + (c cos A + a cos C) + (c cos B + b cos C) = a + b + c = R.H.S.
Hence L.H.S. = R.H.S. Proved
1. 2
(i) tan 2
+ cot
C
2
Example : Find the unknown elements of the ABC in which a = 3 + 1, b = 3 – 1, C = 60°.
Solution. a = 3 + 1, b = 3 – 1, C = 60°
A + B + C = 180°
∴ A + B = 120° .......(i)
tan
π = 45°
⇒ A – B = 90° .......(ii) From equation (i) and (ii ), we get
A = 105° and B = 15° Now,
From sine-rule, we know that Asin
a =
Bsin
b =
Csin
c
1. In a ABC if b = 3, c = 5 and cos (B – C) = 25
7 , then find the value of tan
2
A .

−
5. Trigonometric Functions of Half Angles:
(i) sin A
a b c+ +
(iv) sin A = )cs)(bs)(as(s bc
2 −−− =
bc
2
2
1 ca sin B = s s a s b s c( ) ( ) ( )− − −
Example : In a ABC if a, b, c are in A.P. then find the value of tan 2
A . tan
∴ tan 2
A . tan
b ........(i)
it is giv en that a, b, c are in A.P. ⇒ 2b = a + c
s = 2
∴ tan 2
1 Ans.
Example : In a ABC if b sinC(b cosC + c cosB) = 42, then find the area of the ABC.
Solution. b sinC (b cosC + c cosB) = 42 ........(i) given From projection rule, we know that
a = b cosC + c cosB put in (i), we get ab sinC = 42 ........(ii)
= 2
+

+
∴ 2s – b – a = c
cot cot
cot cot
α β
Example : If the median AD of a triangle ABC is perpendicular to AB, prove that tan A + 2tan B = 0.
Solution. From the figure, we see that θ = 90° + B (as θ is external angle of ABD)
Now if we apply m-n rule in ABC, we get (1 + 1) cot (90 + B) = 1. cot 90° – 1.cot (A – 90°) ⇒ – 2 tan B = cot (90° – A) ⇒ – 2 tan B = tan A ⇒ tan A + 2 tan B = 0 Hence proved.
Example : The base of a triangle is divided into three equal parts. If t 1 , t
2 , t
subtended by these parts at the opposite vertex, prove that
4
.
Solution. Let point D and E divides the base BC into three equal parts i.e. BD = DE = DC = d (Let) and let α, β and γ be the angles subtended by BD, DE and EC respectively at their opposite vertex. ⇒ t
1 = tanα, t
Now in ABC
BE : EC = 2d : d = 2 : 1 ∴ from m-n rule, we get
(2 + 1) cotθ = 2 cot (α + β) – cotγ ⇒ 3cotθ = 2 cot (α + β) – cotγ .........(i) again in ADC DE : EC = x : x = 1 : 1 ∴ if we apply m-n rule in ADC, we get
(1 + 1) cotθ = 1. cotβ – 1 cotγ
2cotθ = cotβ – cotγ .........(ii) from (i) and (ii), we get
θ
θ
cot2
cot3 =
cotcot
cot)cot(2
⇒ 3cotβ – 3cotγ = 4cot (α + β) – 2 cotγ ⇒ 3cotβ – cotγ = 4 cot (α + β)
⇒ 3cotβ – cotγ = 4
−βα
cotcot
1cot.cot
⇒ 3cot2β + 3cotα cotβ – cotβ cotγ – cotα cotγ = 4 cotα cotβ – 4 ⇒ 4 + 3cot2β = cotα cotβ + cotβ cotγ + cotα cotγ ⇒ 4 + 4cot2β = cotα cotβ + cotα cotγ + cotβ cotγ + cot2β ⇒ 4(1 + cot2β) = (cotα + cotβ) (cotβ + cotγ )
⇒ 4
1 Hence proved
1. In a ABC, the median to the side BC is of length 3611
1
− and it div ides angle A into the angles of
30° and 45°. Prove that the side BC is of length 2 units.
8. Radius of Circumcirlce :
Example : In a ABC prove that sinA + sinB + sinC = R
s
Asin
a =
Bsin
b =
Csin
= R2
R
s .
Example : In a ABC if a = 13 cm, b = 14 cm and c = 15 cm, then find its circumradius.
Solution. R = 4
∴ R = 84.4
Example : In a ABC prove that s = 4R cos 2
A
1
1. a cot A + b cotB + cos C = 2(R + r).
2. 4
−1
c
s =
R
r .
3. If α, β, γ are the distances of the vertices of a triangle from the corresponding points of contact with the
incircle, then prove that y
y
(i) r =
A
A
2 sin
(i) r 1 =
A
2 + r
2 + r
3 – r
=
= R.H.S. Hence L.H.S. = R.H.S. proved
Example : If the area of a ABC is 96 sq. unit and the radius of the escribed circles are resp

XI Mathematics IIT JEE Advanced Study Package 2014 159

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