Contents · 2020-06-25 · Grade 11 Trigonometry Notes 2020 7 | P a g e Reduction Formulae The trig...

Grade 11 Trigonometry Notes 2020

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Contents Caps References .................................................................................................................... 2

Trig Ratios in 90 Triangles .................................................................................................. 3

Ratios and Co-Ratios ......................................................................................................... 3

Application of These Ratios .............................................................................................. 3

The Trig Ratios on The Cartesian Plane............................................................................. 4

Application: Using Diagrams to Solve Problems ............................................................... 5

Special Angles and Reduction Formulae ............................................................................... 6

Special Angles .................................................................................................................. 6

Reduction Formulae .......................................................................................................... 7

Very Big Positive or Negative Angles ............................................................................. 10

Trigonometry equations................................................................................................... 11

Applying Special Angles and the Reduction Formulae .................................................... 12

Trigonometric Identities ...................................................................................................... 14

How to prove identities ................................................................................................... 14

Mixed Exercise 1 ................................................................................................................ 16

Trigonometric Equations and the General Solution .............................................................. 16

tan θ = k .......................................................................................................................... 16

cos θ = k .......................................................................................................................... 18

sin θ = k .......................................................................................................................... 21

Further Examples ............................................................................................................ 23

Trigonometric Equations General Solutions .................................................................... 24

Trigonometric equations in the form: a sin θ ± b cos θ = 0.............................................. 25

Trigonometric equations for which factorising is necessary ............................................. 26

Trigonometric equations: when identities are invalid ....................................................... 26

Solving Trigonometric Equations without the use of a Calculator .................................... 28

Mixed exercise 2 ................................................................................................................. 29

Trigonometric Functions ..................................................................................................... 31

Difference between Grade 10 and Grade 11 Functions .................................................... 31

Solving Problems in Two Dimensions ................................................................................. 39

The Sine, Cosine and Area Rules..................................................................................... 39

Tips for Solving 2D Problems ......................................................................................... 41

Using the Sine, Cosine and Area Rules ............................................................................ 41


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Caps References Progression in terms of concepts and skills

GRADE 10 GRADE 11 GRADE 12

1. Define the trigonometric

ratios sin θ, cos θ and tan

θ, using right-angled

triangles.

2. Extend the definitions of

sin θ, cos θ and tan θ for

0 ≤ θ ≤ 360.

3. Define the reciprocals of

the trigonometric ratios

cosec θ, sec θ and cot θ,

using right-angled

triangles (these three

reciprocals should be

examined in Grade 10

only).

4. Derive values of the

trigonometric ratios for

the special cases

(without using a

calculator)

θ ∈ {0; 30; 45; 60; 90}.

5. Solve two-dimensional

problems involving

right-angled triangles.

6. Solve simple

trigonometric equations

for angles between 0

and 90.

7. Use diagrams to

determine the numerical

values of ratios for

angles from 0 to 360.

1. Derive and use the

identities

, k an

odd integer; and

2. Derive and use reduction

formulae to simplify the

following expressions:

( ); ( );

( );

( );

( );

( );

( );

( );

( );

( );

( );

3. Determine for which values

of a variable an identity

holds.

4. Determine the general

solutions of trigonometric

equations. Also, determine

solutions in specific

intervals.

5. Prove and apply the sine,

cosine and area rules.

6. Solve problems in two

dimensions using the sine,

cosine and area rules.

1. Proof and use of the

compound angle

identities:

( )

( )

2. Solve problems in two

and three dimensions.


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Trig Ratios in 90 Triangles

Ratios and Co-Ratios

When we start with trig ratios, we use right-angled triangles to start off. The ratios are sin ,

cos and tan while the reciprocals are cot , sec and cosec .

When dealing with the trig ratios in a right-angled triangle, the sides of the triangle are

named as follows:

The side opposite the angle we are working with is y

The side opposite the 90 angle (hypotenuse) is r

The remaining side is x

OR

For the triangles above:

Trig Ratio Reciprocal

Application of These Ratios

Solve KLM:

C

A

B 𝜽

r

x

y

A

B C

r

x

y

𝜽


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The Trig Ratios on The Cartesian Plane

(NOTE: In the diagram below, is in the standard position)

Trig Ratio Reciprocal

Please note – that reciprocal ratios is not

important after Grade 10

You can use the theorem of Pythagoras:

To calculate if you have x and y.

To calculate if you have x and r.

To calculate if you have y and r.

NOTE:

r is always positive, since r represents the distance from the origin to the ordered

number pair on the terminal arm of the angle .

Since x and y are coordinates and represent the position of a point, they can either be

negative or positive.

The signs of coordinates x and y depend on the position of the angle.

Does the angle lie in quadrant I, II, III or IV?

Therefore, the signs of the trig ratios depend on the signs of x and y.

Since r is always positive, it has no influence on the sign of the trig ratio.

The CAST diagram is a diagram of positives – it indicates which trig ratio is

positive in which quadrant:

All ratios are positive in quadrant I (A)

Only (and ) is positive in quadrant II (S)

Only tan (and ) is positive in quadrant III (T)

Only cos (and ) is positive in quadrant IV (C)

𝑥 ( ; )

𝑃(𝑥 ; 𝑦)

𝑦

𝑥

𝑦 r

A

C

S

T

( - ; - ) ( + ; - )

( + ; + ) ( - ; + )


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Application: Using Diagrams to Solve Problems

Worked Example 1

Given: and .

Make use of a diagram and determine the value of:

a) tan b) sin c) cos

SOLUTION

a)

b)

=

c) cos

=

We will revisit questions like this once reduction formula and co-ratios have been addressed.

Do the Revision exercise on page 137 to make sure your Grade 10 knowledge is up to date!

Given information:

𝜽

𝑦

𝑥 ? 𝑥

𝑦

To determine the quadrant:

𝑠𝑖𝑛𝜃

(sin = negative)

𝑐𝑜𝑠𝜃 (cos = positive)

Because of quadrant 4 x is +12


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Special Angles and Reduction Formulae

Special Angles

The trig ratios of the ‘special angles’ are used frequently, especially in problems where a

calculator may not be used. The special angles are: 0; 30; 45; 60; 90; 180; 270;

360

The ratios of the special angles must be memorised by making use of

one of the methods below. The method chosen depends on the preference of the teacher.

ON THE CARTESIAN PLANE

r = 2

USING THE CASIO CALCULATOR

The CASIO Calculator will automatically give you the answers to the special angles. The

question in the exam will say “without using a calculator”. So even if we use the calculator

we need to ensure that we show all steps

Example

Without using a calculator find the value of sin 30ᵒ×tan 60ᵒ + cos 30°. Show all steps.

3

2

32

2

3

2

3

2

33

2

1

30cos60tan30sin

.

We will revisit questions like this once reduction formula and co-ratios have been addressed.

𝟏 ; 𝟑

𝟐 ; 𝟐

𝟑 ; 𝟏

𝟔𝟎

𝟒𝟓

𝟑𝟎


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Reduction Formulae

The trig ratio of every single obtuse or reflex angle can be rewritten as the trig ratio of an

acute angle.

(180 – ); (180 + ) and (360 – )

Use your calculator and determine the value of sin 20°, sin 160°, sin 200° and sin 340°.

What do you notice?

Use your calculator and determine the value of cos 40°, cos 140°, cos 220° and cos 320°.

What do you notice?

Use your calculator and determine the value of tan 88°, tan 92°, tan 268° and tan 272°

What do you notice?

You should have notice a similarity in the answer of the four trigonometry ratios asked in

each question.

Think back to the trigonometric graphs you drew in Grade 10. Remember that the shape

stayed the same? Once we worked through the first 90° the curve is just transformed (by

reflections in either a vertical line of the x – axis. The means that the values on the y-axis is

repeated over and over.

So what is the link between 20°, 160°, 200° and 340°? Or 40°, 140°, 220° and 320°? Or

88°, 92°, 268° and 272°? Or between all 12 angles?

The obvious and easy link to find is that 20°, 40° and 88° are all in the first quadrant; 160°,

140° and 92° are all in the second quadrant; 200°, 220° and 268° are all in the third

quadrant and 340°, 320° and 272° are all in the fourth quadrant.

Looking closer we can see that:

160° = 180° – 20° 200° = 180° + 20° 340° = 360° – 20°

140° = 180° – 40° 220° = 180° + 40° 320° = 360° – 40°

92° = 180° – 88° 268° = 180° + 88° 272° = 360° – 88°

When the original (first) angle is deviated from the x – axis – although the angles become

bigger, the trigonometry ratios stay the same except for, in two case each, the sign changes.

You already know that the x and y values repeat themselves BUT in different signs. Refer

to our work in the Cartesian plane.

You should remind yourself that the y value of the Cartesian Plane is positive in the first

and second quadrant; hence sin α is positive in the first and second quadrant (sin 20° and

sin 160°), but the y value of the Cartesian Plane is negative in the third and fourth

quadrant; hence sin α is negative in the third and fourth quadrant (sin 200° and sin 340°).

You should also remember r

ysin when we work in the Cartesian Plane; r is always

positive. Hence whether sin α is positive depends on the y – value.

Similar: You should remind yourself that the x value of the Cartesian Plane is positive in

the first and fourth quadrant; hence cos α is positive in the first and fourth quadrant (cos40°

and cos 320°), but the x value of the Cartesian Plane is negative in the second and third

quadrant; hence cos α is negative in the second and third quadrant (cos 140° and cos 220°).

You should also remember r

xcos when we work in the Cartesian Plane; r is always


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positive. Hence whether sin α is positive depends on the x – value.

When we work with tan α we have to start with the fact that x

ytan . Consider the

Cartesian Plane. Both x and y can be positive or negative. In the first quadrant, x and y are

both positive and hence tan α is positive (tan 88°). In the third quadrant, x and y are both

negative and hence tan α is positive (tan 278°). In the second quadrant, x is negative and y

is positive so tan α will be negative (tan 92°) and finally in the fourth quadrant, x is

positive and y is negative so tan α will be negative (tan 272°).

We also have to address (360° + α). If α is an acute angle – this places the angle in the first

quadrant; which means all three the trigonometric ratios will be positive.

What about (–α)? Where will we find this angle on the Cartesian Plane? Is we rewrite this

as (0° – α) it is easy to see that we are in the fourth quadrant

How can we generalise all of this?

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

The CAST diagram plays a key role when using the reduction formulae. It is important

that you understand the ‘logic’ of the CAST diagram to get to answers ‘logically’.

However, you will have to memorise the diagram itself plus a few useful TIPS that will

make this section of trig easy.

Examples

Rewrite the following as the trigonometry ratio of a positive, acute angle:

𝟎 / 𝟑𝟔𝟎

𝟗𝟎

𝟏𝟖𝟎

𝟐𝟕𝟎

A

𝜽

I

T

(𝟏𝟖𝟎 𝜽)

III

C

(𝟑𝟔𝟎 𝜽)

IV

S

(𝟏𝟖𝟎 𝜽)

II


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a) ( ) b) ( ) c) ( ) d) ( ) e) ( ) see this as ( )

Do Exercise 1 on page 141

(90 – ), (90 + ), (270 – ) and (270 + )

According to the CAPS document only ( ) should be covered.

( ) works the same as ( ), just in different quadrants.

You might find ( ) useful and time-saving when solving problems. (This a

nice to know and you should please not worry about this)

Use your calculator again.

Compare sin 60° and cos 30°. Both answers in 2

3. We can say:

sin 60° = sin(90° – 30°) (because 90° – 30° = 60°)

= cos 30°

Compare cos 74° and sin 16°. Both answers, rounded off to 4 decimals, is 0,2756. We can

say:

cos 74° = cos(90° – 16°) (because 90° – 16° = 74°)

= sin 16°

I II

III IV

A S

T C

𝟎 / 𝟑𝟔𝟎

𝟗𝟎

𝟏𝟖𝟎

𝟐𝟕𝟎

(𝟗𝟎 𝜽) (𝟗𝟎 𝜽)

(𝟐𝟕𝟎 𝜽) (𝟐𝟕𝟎 𝜽)

𝜽


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TIP: The name of the trig ratio changes to the name of the co-function ONLY WHEN WE

WORK with ( ) [or with ( )], i.e. the VERTICAL axis.

NOTE: It is only when the question is asking (forcing) you to work with a co-ratio that you

do that.

( )

( )

( )

( ) ( ) ( )

( ) ( )

THE SECRET IS …

When working with reduction formulae and co-ratios, you must always consider three

things in your answer……

The sign (+ or –)

The trig ratio

An acute angle


Very Big Positive or Negative Angles

You may add or subtract as many 360o’s as you like. The angles are named ‘co terminal’

angles. The value of the trig ratios of co terminal angles is equal.

Consider: cos 1 456 o

Repeatedly subtract 360o until you get an

angle less than 360

1 456 o – 360

o = 1 096

o

1 096 o – 360

o = 736

o

736 o – 360

o = 376

o

376 o – 360

o = 16

o

Thus: cos 1456 o = cos 16

o

Consider: cos (– 568)o

Repeatedly add 360o until you get an

angle that is greater than 0 but less than

360

– 568o+ 360

o = –208

o

– 208o + 360

o = 152

o

So, cos 152 o

= cos (180 o – 28

o)

= -cos 28 o

Thus: cos (-568o) = cos 152

o

= – cos 28 o




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Trigonometry equations

Worked Example 1

If ( ) determine the value(s) of the following in terms of t:

a) ( ) b) ( ) c) ( ) d) ( )

a) ( ) ( ) ( )

b) ( ) ( ) ( ( ) ( ))

( ) = ( )

Make use of a diagram and determine the value of:

c) ( )

and

Using Pythagoras, you can calculate

( )

=

d) ( ) ( ) ( )



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Applying Special Angles and the Reduction Formulae

Worked Example 1

a) Show that: ( )

( ) ( )

b) Hence, calculate without the use of a calculator, the value of:

( )

(Leave your answer in surd form)

Please note: The word “Hence” is often used in a follow up question. It implies that

you have to apply the first part in answering the second part. If this was the exam

and you could not do part (a) you can still use the identity to answer part (b).

SOLUTION

a) ( )

( ) ( )

( ( )) ( )

( ) ( )

( ) ( )

( ) ( )

= RHS

b)

( )

( )

( ) ( )

(using (a) when )


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Worked Example 2

Evaluate without a calculator and showing all steps: ( )

( )

SOLUTION: ( )

( )

( )

( )

(

) (

)

Worked Example 3

If 08sin10 and 0tan calculate without the use of a calculator and with the aid of

a diagram the value of:

a) (sin + cos )2

b) 180tan

c) 90cos

5

4sin

10

8sin

8sin10

3

9

1625

5)4(

2

2

222

222

x

x

x

x

ryx

870° – 2(360°) = 150°

–1020° + 3(360°) = 60°


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a) (sin + cos )2

25

49

25

241

5

4

5

321

coscos.sin2sin 22

b) 180tan

3

4

3

4

tan

c) 90cos

5

4

5

4

sin


Trigonometric Identities

NOTE:

There are two main identities taught in Grade 11:

In Grade 12 you will be introduced to the compound angle identities. Double angle

identities are derived from the compound angle identities.

We can use these identities to prove other identities.

How to prove identities

There are always two sides to prove equal when dealing with identities.

Start with simplifying the least complicated side. (And at least get a mark)

Look for squared terms and see if you can use substitution to simplify.

Where possible, change everything to sine and cosine.

Trigonometric expressions can often need to be simplified like in algebraic

expressions, so check to see if the expression has like terms, fractions or factors

before substituting by an identity.


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Worked Examples

Prove the following identities:

a)

= 1 + sin x

b)

22

23

sin1

cos

270sin

810cos

cos.tan

cos.sinsin

SOLUTION (a)

Note: The RHS cannot be simplified so I move to the LHS

x

xxxLHS

sin1

cossincos 422

RHS = 1 + sin x

x

xxx

sin1

cossincos 222

= x

x

sin1

)1(cos2

= x

x

sin1

sin1 2

=

x

xx

sin1

sin1sin1

= xsin1

RHSLHS

SOLUTION (b)

Note: The RHS can be simplified so I start there before I move to the LHS:

RHS:

cos

1

cos

cos

sin1

cos

2

2

LHS:

RHS

cos

1

1

0

cos

1

1

90cos

cossin

1sin

)1(

720810cos

1

cos

cos

sin

cossinsin

270sin

810cos

cos.tan

cos.sinsin

2

22

2

23





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Mixed Exercise 1

1. ; and ( ; ) are points on the

Cartesian plane, as shown in the diagram

below. and .

Determine, without the use of a calculator,

the value of:

1.1

1.2 ( ) 1.3

2. If and , determine the value of where

and ∈ [ ; ].

3. Simplify the following:

)cos()720cos()90cos()180sin(

)360cos()180tan(

xxxx

xx

4. Simplify: ( ) ( ) ( ) ( )

5. Prove, without the use of a calculator, that:

Trigonometric Equations and the General Solution

Before dealing with this topic it is absolutely essential for you to ensure that your calculator

is on “degrees”.

tan θ = k

We are going to do a short investigation:

Find the numerical value of the following:

tan 45º, tan (–135º), tan 225º, tan (–315º)

The function value for the four different angles gives the same numerical value.

The inverse operation – finding the angle – only gives one answer.

Find the angle if the numerical value is given:

tan θ = 1 451tan 1 but what about the other values (–135º,

225º or –315º)?


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We need to look at the graph of y = tan θ to understand this phenomenon.

Points A, B, C and D have a y-value of 1 – this corresponds with tan 45º = 1, tan (–135º) = 1,

tan 225º = 1 and tan (–315º) = 1. When the calculator has to get the inverse answer, it seems

as though it uses the answer closes to the y – axis, i.e. 45º. It is important to note at this point

that each point differs 180º from the next point.


tan 150º, tan (–30º), tan 330º, tan (–210º)


tan θ = 3

3

30

3

3tan 1 but what about the other values

(150º, 330º or –210º)?


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Points E, F, G and H have a y-value of 3

3 – this corresponds with tan 150º =

3

3 ,

tan (–30º) = 3

3 , tan 330º =

3

3 and tan (–210º) =

3

3 . When the calculator has to get

the inverse answer, it again uses the answer closes to the y – axis, i.e. -30º. It is important to

note, again, that each point differs 180º from the next point.

The formula to solve a “tan θ” – equation is

If tan θ = k Rk

then Znnk ;180.)(tan 1

EXAMPLE 12 (of textbook done easier)

(c) Solve for θ if 2 tan θ = –9, 360;0 . Round answers off to two decimal places.

180..........47111,77

;180.2

9tan

2

9tan

9tan2

1

n

Znn

To find the applicable answers in the range, substitute n with integers:

n = 0 θ = –77,47º

n = 1 θ = 102,53º ✓

n = 2 θ = 282,53º ✓

n = 3 θ = 362,53º ×

Hence the applicable answers is θ = 102,53º or θ = 282,53º

cos θ = k


cos 55º, cos (–55º), cos 305º, cos (–305º)




cos θ = 0,5769 555769,0cos 1 but what about the other values (–55º,

305º or –305º)?


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We need to look at the graph of y = cos θ to understand this phenomenon.

Points A, B, C and D have a y-value of 0,5736 – this corresponds with cos 55º = 0,5736,

cos (–55º) = 0,5736, cos 225º = 0,5736 and cos (–315º) = 0,5736. When the calculator has to

get the inverse answer, it seems as though it uses the positive answer closes to the y – axis,

i.e. 55º. It is important to note at this point that

There are always a positive and a negative value of angles with the same function

value(B and C; A and D).

From A to C and from B to D there is a 360º difference.


cos 120º, cos (–120º), cos 240º, cos (–240º)


cos θ = 2

1

120

2

1cos 1 but what about the other values

(–120º, 240º or –240º)?


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1 – this corresponds with cos 120º =

2

1 ,

cos (–120º) =2

1 , cos 240º =

2

1 and cos (–240º) =

2

1 . When the calculator has to get the

inverse answer, it again uses the positive answer closes to the y – axis, i.e. 120º. It is

important to note, again, that

There are always a positive and a negative value of angles with the same function

value (E and H; F and G).

From E to G and from F to H there is a 360º difference.

The formula to solve a “cos θ” – equation is

If cos θ = k 1;1k

then Znnk ;360.)(cos 1

EXAMPLE 12

(b) Solve for θ if cos θ = –7

5, 360;0 . Round answers off to two decimal places.

...5846914,135

;360.7

5cos

7

5cos

1

Znn


θ = 135,58º

n = 0 θ = 135,58º ✓


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n = 1 θ = 495,58º ×

θ = –135,58º

n = 0 θ = –135,58º ×

n = 1 θ = 224,42º ✓

n = 2 θ = 584,42 ×

Hence the applicable answers is θ = 135,58º or θ = 224,42º

sin θ = k


sin 55º, sin 125º, sin (–235º), sin (–305º)




sin θ = 0,8192 558192,0sin 1 but what about the other values (125º,

–235º or –305º)?

We need to look at the graph of y = sin θ to understand this phenomenon.

Points A, B, C and D have a y-value of 0,8192 – this corresponds with sin 55º = 0,8192,

sin 125º = 0,8192, sin (–235º) = 0,8192 and sin (–315º) = 0,8192. When the calculator has to

get the inverse answer, it seems as though it uses the positive answer closes to the y – axis,

i.e. 55º. It is important to note at this point that

There is always an acute angle (C), the angle given by the calculator.

The next angle (D) has to be calculated by using the formula (180º – the acute angle).

From A to C and from B to D there is a 360º difference.


sin (–30º), sin (–150º), sin 210º, sin 330º


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sin θ = 2

1

30

2

1sin 1 but what about the other values

(–150º, 210º or 330º)?


1 – this corresponds with sin (–150º) =

2

1 ,

sin (–30º) =2

1 , sin 210º =

2

1 and sin 330º =

2

1 . When the calculator has to get the

inverse answer, it again uses the answer closes to the y – axis, i.e. –30º. It is important to

note, again, that

There is always a negative acute angle (F), the angle given by the calculator.

The next angle (G) has to be calculated by using the formula (180º – the acute angle).

From F to H and from G to E there is a 360º difference.

The formula to solve a “sin θ” – equation is

If sin θ = k 1;1k

then Znnk

k

;360.sin180

sin

1

1

PLEASE NOTE the two answers.


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EXAMPLE 12

(a) Solve for θ if sin θ = 2

1; 360;0

360.150

30

;360.

2

1sin180

2

1sin

2

1sin

1

1

n

Znn


θ = 30º

n = 0 θ = 30º ✓

n = 1 θ = 390º ×

θ = 150º

n = 0 θ = 150º ✓

n = 1 θ = 410 º ×

Hence the applicable answers is θ = 30º or θ = 150º


Further Examples

EXAMPLE 13

(a) Solve for A is sin (A – 24º) = –0,7 and A 360;0 rounded off to two decimals.

360.43,248

43,20

360.427004,224

427004,4424

;360.)7,0(sin180

)7,0(sin24

7,024sin

1

1

nA

nA

ZnnA

A

A = 339,57º or A = 248,43º


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(b) Solve for A if cos 2A = –0,867; 360;0A , rounded off to two decimals.

180.....0555,75

360...1118697,1502

;360.)867,0(cos2

867,02cos

1

nA

nA

ZnnA

A

A = 75,06º or A = 255,06º or A = 104,94º or A = 284,94º


Trigonometric Equations General Solutions

EXAMPLE 14

(a) Determine the general solution of sin θ = 2

1

(b) Find θ if 360;360

Solutions

(a) sin θ = 2

1

360.150

30

;360.

2

1sin180

2

1sin

1

1

n

Znn

(b) θ = 30º or θ = – 330º or θ = 150º or θ = –210º

EXAMPLE 15

(a) Determine the general solution of tan 2θ = –4

(b) Find θ if 180;90 , rounded off to two decimals.

(a) tan 2θ = –4

90...98187,37

180....9637,752

;180.)4(tan2 1

n

n

Znn

(b) θ = –37,98º or θ = 52,02º or θ = 142,02º


25 | P a g e

Summarised notes on finding the general solution

If tan θ = k, Znnk ;180.)(tan 1

If cos θ = k, Znnk ;360.)(cos 1

If sin θ = k, Znnk

k

;360.sin180

sin

1

1


Trigonometric equations in the form: a sin θ ± b cos θ = 0

EXAMPLE 16

(a) Find the general solution of the equation 3 sin θ = 2 cos θ.

(b) Solve for θ if sin θ + cos θ = 0; 360;0

Solutions

(a) The first step is to divide both sides by cos θ because you would like to work with a

single trigonometric ratio and

tan

cos

sin .

180.69,33

;180.3

2tan

3

2tan

2tan3

cos

cos2

cos

sin3

cos2sin3

1

n

Znn

(b) sin θ + cos θ = 0

180.45

;180.1tan

1tan

cos

cos

cos

sin

cossin

1

n

Znn

θ = 135º or θ = 315º


26 | P a g e

Trigonometric equations for which factorising is necessary

EXAMPLE 17

(a) Find the general solution for 1 + sin θ = cos2 θ

(b) Solve for θ, if 2cos2 θ + cos θ = 3

Solutions

(a) As you can see there is a sine and a cosine ratio in this equation. In this regard,

dividing by cos θ will not simplify the equation. The 1 as well as the cos2 θ creates

a problem.

We will choose to work with only sine in this example because it is easier to

write cos2 θ in terms of sin θ than writing sin θ in terms of cos θ.

1 + sin θ = cos2 θ

1 + sin θ = 1 – sin2θ

sin2θ + sin θ = 0

sin θ(sin θ + 1) = 0

sin θ = 0 or sin θ + 1 = 0

sin θ = –1

360.180

0

;360.0sin180

0sin

1

1

n

Znn

360.270

90

;360.1sin180

1sin

1

1

n

Znn

(b) 2cos2 θ + cos θ = 3

2cos2 θ + cos θ – 3 = 0

(2cos θ + 3)(cos θ – 1) = 0

2 cos θ + 3 = 0 cos θ – 1 = 0

solution No

2

3cos

3cos2

360.0

;360.1cos

1cos

1

n

Znn


Trigonometric equations: when identities are invalid

An identity is a statement of equality that is true for all values (except those values for which

the identity is not defined). Previously, equations containing fractions were solved. When

solving these types of equations, we always stated the restrictions to prevent division by zero.

We will now determine the restrictions for a trigonometric identity. In other words, the

value(s) for which the identity will be undefined.


27 | P a g e

EXAMPLE 18

(a) Consider the identity 1sin

cos.tan

x

xx.

For which values of x will the identity above be invalid? State the general

solution.

(b) Consider the identity

cos

1

sin1

costan

.

For which values of θ will the identity above be invalid for the interval

180;180 ?

Solutions

(a) First you consider the tangent ratio:

You will recall the graph of y = tan x has asymptotes at x = ±90º; ±270º……

We can summarise this as:

x = 90º + n.180º; Zn

Secondly consider division by zero:

The identity will be invalid if any denominator equals zero. In this example it

means if sin x = 0.

360.180

0

;360.)0(sin180

)0(sin

0sin

1

1

nx

Znnx

x

(b) Tangent ratio:

θ = 90º + n.180º; Zn

Division by zero:

1 + sin θ = 0 cos θ = 0

sin θ = –1

360.270

90

;360.)1(sin180

)1(sin

1

1

n

Znn

360.90

;360.)0(cos 1

n

Znn

Hence the three solutions are giving us similar answers and the identity will be

undefined for θ = 90º and θ = –90º



28 | P a g e

Solving Trigonometric Equations without the use of a Calculator

In this section a knowledge of special angles will be useful. The special angles are 0 º; 30 º;

45 º; 60 º; 90 º; 180 º; 270 º; 360 º.

The diagram for special angles is provided below.

EXAMPLE 19

Solve for θ if 2

2cos and 360;0

Solution

Znn

;360.135

2

2cos

θ = 135º or θ = 225º


Up until this point all the work, we have done will be ONE questions in the exam of about

25 marks.


29 | P a g e

Mixed exercise 2

Answer the following questions:

1. If determine the value of each of the following in terms of WITHOUT USING A CALCULATOR.

1.1

1.2 cos 56°

1.3 cos 34 °

1.4 tan 34 °

2. Determine the value of the following expression: )720cos().180(sin

sin).90cos(2

3. If 900 < A < 360

0 and tan A = , determine without the use of a calculator.

3.1 sin A

3.2 cos2A – sin

2A

4. Given that sin x = t, express the following in terms of t, without the use of calculator.

4.1 cos (x – 900)

4.2 tan x

5. Calculate without the use of a calculator:

332cos.28sin.118tan

208cos2

6. If and [ ; ] determine by using a sketch, the value of

, without the use of a calculator.

7. Simplify to a single trigonometric ratio of x: ( ) ( )

( )

8. Simplify: xx

xxx 22

sin)180sin(

)90(sin.)360tan(.)180cos(

9. Simplify:

)450cos().sin(

)360tan().180cos().180sin(

xx

xxx

3

2


30 | P a g e

10. In the diagram below, P (–15 ; m) is a point in the third quadrant and 17cos β + 15=0.

WITHOUT USING A CALCULATOR, determine the value of the following:

10.1 m

10.2 sin β + tan β

10.3 34.cos(180° – β) + 17.cos(90° + β)

11. Evaluate, without using a calculator:

β

.

P (–15; –m)

x

y

O


31 | P a g e

Trigonometric Functions

Difference between Grade 10 and Grade 11 Functions

In Grade 10, you plotted the basic graphs of xy sin ; xy cos ; xy tan where

∈ [ ; ]. In Grade 11, you plot graphs within the interval [ ; ]

qpxkay

qpxkay

qpxkay

)(tan

)(cos

)(sin

In Grade 10 you need to know the effect of a and q

In Grade 11 you need to know the effect of k and p

The parameters a; p; q and k affect cos x and sin x in the same way. The tan x graph

behaves differently to both sin x and cos x, because of the asymptotes and the difference

in range.

The function for sin x and cos x are wave- like shapes whereas tan x is a repeated curve

shape.

Because of the wave-shape of the graphs of sin x and cos x, these two graphs have an

amplitude (a). The amplitude is the height from the rest value q to the maximum or the

minimum.

All the three functions have a period which depends on the value of k. The period is the

length required for the graph to make one complete shape.

Knowing the features and the characteristics of the function will help in finding the

equation and interpreting the graph. NB

xy tan has asymptotes, and they should not be part of the domain i.e. kx 18090

Make sure you can use the calculator to draw the functions as it will help you to save

time during the exams.

First make sure that your table mode works only with f(x) and not f(x) and g(x)

qwR41

Above is for the CASIO fx-82ZA

For the CASIO fx-991ZA we use

qwR51

When you work in table mode, we always work in steps of 15 when we are drawing trig

graphs

Worked Example 1

Draw the following sets of graphs on the same set of axes and investigate the effect of parameter a;

p; q and k on the graphs.

a) ( ) , ( ) , ( ) ; for ∈ [ ; ] b) ( ) ; ( ) ; ( ) ; for ∈ [ ; ] c) ( ) ; ( ) ( ); for ∈ [ ; ] d) ( ) ; ( ) ( ); for ∈ [ ; ]


32 | P a g e

SOLUTIONS

a)

The effect of k on the graph:

Value of k Period

( ) , so 1 complete shape within 360

( ) , so 2 complete shapes within 360

( ) , so 3 complete shapes within 360

b)

The effect of a on the graphs:

( )

( )

( ) The effect of q on the graphs: Vertical shift

The graph of ( ) is the shift of f(x) by 2 units up,

If q is + the basic graph will move upward


33 | P a g e

If q is - the basic graph will move downward

c)

d)

The effect of p on the graph: Horizontal Shift

The graph of m(x) is the shift of the basic graph f(x) by to the left.

The graph of j(x) is the shift of the basic graph f(x) by 45 to the right with the amplitude of 2.

– p shifts the graph to the right.

+ p shifts the graph to the left.


34 | P a g e

Worked Example 2

Draw the graphs of ( ) ; ( ) and ( ) for

∈ [ ; ]on the same set of axes.

Hence investigate the effects of the parameters a and q on the graphs of f(x) and g(x) and then

on h(x).

a widens or narrows the graph y = tan x. When a becomes bigger the graph narrows

closer to the line , and when it becomes smaller the graph widens moving away

from The graph of y = tan x has asymptotes which sin x and cos x graphs do not have. The

asymptote is a value where the graph is undefined. It is usually indicated by a dotted

line on the graph.

The effect of the parameters a and q on the graph ( ) is the same as on

the graph of sin x as discussed in the previous Worked Example.

The effect of q on the graph y = tan x is the same as on the graphs of y = cos x and y = sin

x.

Worked Example 3

On the same set of axes, draw sketch graphs of sin 1f x x and cos 2g x x for

90 ;270x . Then answer the questions that follow:

a. Give the range of f.

b. For which values if x, 90 ;270x , is:

i. 0xg

ii. f x g x

iii. . 0f x g x


35 | P a g e

The drawing of the graphs are simply done by using your calculator to develop the table and

carefully plotting the graphs. There are letter values on the graphs to assist in the explanation

of question (b). In the exam you will not really do that – unless specifically asked.

a. Range: 2;0y

b. There are three types of questions asked here. Question (i) involves only the graph of

g(x), whereas question (ii) and (iii) involves both graph.

i. The question is asking for which x – values is the graph of g(x) > 0. We need to

remember that g(x) represents the y –values. The y – axis is positive above the x – axis,

so basically this question is asking “where is the graph of g(x) above the x – axis. The

points involved are E, F, G and H.

So in English we will answer “between E and F and again between G and H”.

In Mathematics we will write:

-45° < x < 45° or 135° < x < 225°.

ii. In this question we have to compare the two graphs and find out where the y – values of

f(x) are smaller than the y – values of g(x).

We can see that the y – values of the two graphs are equal at A, B, C and D. The y –

values of f(x) is lying lower than those of g(x) between A and B as well as between C

and D. If the graph is lower, the y – values will be smaller. The Mathematical answer

will be:

-30° ≤ x ≤ 0° or 180° ≤ x ≤ 210°

iii. Although this question involves both graphs, the question is not asking you to compare

the two graphs, but when will the product of the two graphs’ y – values be negative.

The product of two values is negative of one is positive and the other negative. So in

general this means that the graphs must NOT be on the same side of the x – axis. If both

graphs are below the x – axis, the y – values are both negative and the product of the

two will be positive.

For this specific question, the values of f(x) are always positive, so the answer depends

on the y – value of g(x).

-90° ≤ x ≤ -45° or 45° ≤ x ≤ 135° or x ≥ 225°.


36 | P a g e

Exercise on Trigonometry Functions

1. xxf 2sin)( for 90180 x is shown in the sketch below:

1.1 Write down the range of f. (2)

1.2 Determine the period of

xf

2

3. (2)

1.3 Draw the graph of )30cos()( xxg for 90180 x on the same set of

axes. Clearly label ALL x- intercepts and turning points. (4)

1.4 Determine the general solution of )30cos(2sin xx (5)


37 | P a g e

2. The graphs of the functions xaxf tan)( and xbxg cos)( for 2700 x are

shown in the diagram below. The point ( )2;225 lies on f. The graphs intersect at points

P and Q.

2.1 Determine the numerical values of a and b. (4)

2.2 Determine the minimum value of g(x) +2. (2)

2.3 Show that if the x-coordinate of P is , then the x-coordinate of Q is ( )

3

3.1 Write down the solutions for 13sin x on the interval ]180;90[ (2)


38 | P a g e

3.2 Give the maximum value of h of h(x) = f(x) – 1 (2)

3.3 Draw the graph of g(x) = 3cos x for ∈ [ ; ] on the same set of axis. (3)

3.4 Use the graphs to determine how many solutions there are to the equation

on the interval [ ; ]

(2)

3.5 Use the graphs to solve: f(x).g(x) < 0 (4)




Do Revision Exercise on page 195


39 | P a g e

Solving Problems in Two Dimensions

Any triangle can be solved, if THREE properties of the triangle are given/known, of which

one must be a side length, by using:

The trig ratios in RIGHT-ANGLED triangles

The sine or cosine rule

The Sine, Cosine and Area Rules

TYPES OF QUESTIONS:

Numeric (calculations) problems

Non-numeric (prove type) problems

According to the CAPS document, you must be able to:

Establish (prove) the rules

Apply the rules in solving 2D problems

When to use what?

If the triangle is

RIGHT-ANGLED

If the triangle is NOT

right-angled

Use the trig RATIOS

Sine or Cosine Rule:

COSINE RULE if ……

3 sides of the triangle are given

2 sides and an included angle of the

triangle is given

SINE RULE if ……

Any condition that does NOT satisfy

the cosine rule

AREA RULE if ……

…. “area” is mentioned (only)

Remember:

3 properties of a triangle, of

which at least one is a side, must

be given in a triangle in order to

work in that triangle


40 | P a g e

In any ABC the rules are applied as follow:

A

B C

b

A

b

B C

c

a

c

a

Sine rule

If an ANGLE is asked

𝑠𝑖𝑛𝐴

𝑎

𝑠𝑖𝑛𝐵

𝑏

𝑠𝑖𝑛𝐶

𝑐

If a SIDE is asked

𝑎

𝑠𝑖𝑛𝐴

𝑏

𝑠𝑖𝑛𝐵

𝑐

𝑠𝑖𝑛𝐶

Cosine rule

If an ANGLE is asked

𝑐𝑜𝑠𝐴 𝑏 𝑐 𝑎

𝑏𝑐

𝑐𝑜𝑠𝐵 𝑎 𝑐 𝑏

𝑎𝑐

𝑐𝑜𝑠𝐶 𝑎 𝑏 𝑐

𝑎𝑏

If a SIDE is asked

𝑎 𝑏 𝑐 𝑏𝑐 𝑐𝑜𝑠𝐴 𝑏 𝑎 𝑐 𝑎𝑐 𝑐𝑜𝑠𝐵 𝑐 𝑎 𝑏 𝑎𝑏 𝑐𝑜𝑠𝐶

Area rule 𝐀𝐫𝐞𝐚 𝐨𝐟 𝐀𝐁𝐂

𝑎𝑏 𝑠𝑖𝑛𝐶 or

𝑏𝑐 𝑠𝑖𝑛𝐴 or

𝑎𝑐 𝑠𝑖𝑛𝐵


41 | P a g e

Tips for Solving 2D Problems

1. The diagram usually consists of 2 or more triangles with COMMON sides.

2. One of the triangles is sometimes right-angled, so you can use the trig ratios to solve it.

(In triangles without right angles, the Sine, Cosine and Area rules must be applied.)

You can also apply the Sine rule in a right-angled triangle.

3. Make use of basic Geometry to obtain additional information, such as vertical opposite

angles, interior angles of a triangle, etc.

4. Start in the triangle that contains the most information, then move along to the triangle

in which the required line/angle is.

5. In applications, we often use angles of DEPRESSION and ELEVATION. Both are

measured from the horisontal.

Using the Sine, Cosine and Area Rules

Worked Example 1

A soccer player (S) is 15 m from the back line of a soccer field (CH). She aims towards the

goal (GH). The angle from the left goal post (G) to the soccer player is 116°. The goal posts

are 7,32 metres apart. The diagram below represents the situation.

a) Calculate the size of SGC ˆ .

b) Calculate SG, the distance between the soccer player and the left goal post FG.

c) Calculate the size of HSG ˆ , the angle within which the soccer player could possibly score

a goal.

SOLUTION:

a) 64ˆSGC (angles on straight line CGH)

b) In CGS , right-angled at C

Angle of elevation

Angle of depression

HORISONTAL

H

15

G C 7,3

2 116°

S

𝐶𝐺𝑆 is right-angled

use the trig ratios


42 | P a g e

= 16,689 02911 … metres

16,69 metres

c) In SGH we know the lengths of two sides (SG and GH and the size of the included

angle ) so we use the cosine rule

116cos69,1632,7269,1632,7222SH

√( ) ( ) ( ) ( ) SH = 20,957 389 36

20,96 metres

Enough information in SGH is known to use either the sine or cosine rule to calculate .

Using the sine rule:

(

)

= 18,293 926 57

18,3

Using the cosine rule:

( ) ( ) ( )

( )( )

(( ) ( ) ( )

( )( )

= 18,296 286 19

18,3

Worked Example 2

In the figure below ˆSPQ = , ˆPQS and PQ = h. PQ and SR are perpendicular on RQ.

Prove that

sin .cos

sin

hRS

Solution:


43 | P a g e

In ΔPQS:

180ˆQSP

sin

sin.SQ

sinsin

SQ

180sinsin

SQ

SsinPsin

h

h

h

sp

In ΔSRQ:

90ˆRQS

sin

cos.sin.

cossin

sin.

cosSQRS

90sinSQ

RS

h

hRS






Do Revision Exercise on page 185


44 | P a g e

Extra Exercise

1. Triangle PQS represents a certain area of a park. R is a point on line PS such that QR

divides the area of the park into two triangular parts, as shown below. PQ

, RS

and RQ x units.

1.1 Calculate the size of .

1.2 Determine the area of Δ QRS in terms of x.

2. In the diagram below, ABCD is a cyclic quadrilateral with DC = 6 units, AD = 10

units, and .

Calculate the following, correct to ONE decimal place:

2.1 The length of BC

2.2 The area of ∆ABC

P

Q

R

S

𝑥

𝑥

𝑥

𝑥

6

D

C

B

A

100°

10

40°

Contents · 2020-06-25 · Grade 11 Trigonometry Notes 2020 7 | P a g e Reduction Formulae The trig...

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