Trigonometry - Solutions to Problem Set

1. Find the acute angle that has the same sin-, cos-, and tan-values as 380○,−450○, 1082○, 5π radians, 11π

2radians, and −26π.

Solution: The acute angles with the same trig ratios as 380○ is 20○. 90○

has the same cos-value as −450○, but there is no acute angle with the samesin- and tan-values. The acute angle with the same trig ratio values as1082○ is 2○. 90○ has the same cos-value as 11π

2, but there is no acute angle

with the same sin- and tan-values. An angle of 0 has the same trig ratiovalues as −26π.

2. Convert the following angle to radians: 45○, −27○, and 120○.

Solution: 45○ = π4, −27○ = −3π

20, 120○ = 2π

3.

3. Convert the following angles (given in radians) to degrees: 3π, π12

, and−2π.

Solution: 3π = 540○,π

12= 15○ − 2π = −360○.

4. A triangle has a 90○ angle. The side facing that angle has length 5. Oneof the other angles is 30○. Find the third angle, the cos-value of the thirdangle (using special triangles), and the length of the side adjacent to thethird angle (using the previous two answers).

Solution: The third angle is 180○ − 90○ − 30○ = 60○. Then cos(60○) =cos (π

3) = 1

2. Let the length of the adjacent side be x. Using the triangle

that was given in the question, cos(60○) = x5. But that is supposed to

equal 12. Therefore, the adjacent side has length x = 10.

5. Find all angles θ between 0 and 2π for which tan θ is not defined. Explainyour answer.

Solution: tan θ = sin θcos θ

. Therefore, tan θ is not defined when cos θ = 0. When0 ≤ x ≤ 2π, this occurs at θ = π

2and θ = 3π

2.

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6. Are the sin θ and cos θ functions defined for all angles θ? Or are therevalues of θ where they do not exist? Explain.Solution: We get the sin θ and cos θ functions by defining them on an-gles in a right angle triangle and then extending them to all angles, byparametrizing the unit circle. As a result, we can consider the set {cos θ}to be the set of x coordinates of points on the unit circle (with repetition)and {sin θ} to be the set of y coordinates (with repetition). Therefore,since the unit circle has a point corresponding to every angle θ for allθ ∈ R, both functions are defined for angles θ.

7. Find the angle θ between 0 and 2π such that cot θ = cot π4

but θ ≠ π4

.

Solution: cot π4= 1tanπ

4Since π

4is in the first quadrant, the opposite and

adjacent sides of the π2− π

4π4

have positive lengths. In order for angle tohave the same tan-value as π

4, both its opposite and adjacent sides would

have to have negative length, putting the angle in the third quadrant, atan angle of π

4from the x-axis. This angle is 5π

4.

8. If a triangle has two angles (30○ and 50○ respectively) separated by a sideof length 8, is it possible to find the lengths of the other two sides usingSine Law or Cosine Law? If not, why not? If one of those Laws makes itpossible, which one and why?

Solution: To use Cosine Law, we would need at least two side lengths,which we do not have. However, we can find the third angle θ = 180○ −50○ − 30○ = 100○. We now have an angle (100○) and the length of the sideacross from it (8). To find the side x across from the 50○ angle, we canuse Sine Law: sin 100○

8= sin 50○

x. By replacing the 50○ with 30○, we can find

the third side length.

9. Graph y = sin(x − π), y = cosx + 1, and y = tanx.

Solution: The first graph is the graph of y = sinx shifted to the right byπ radians. Likewise, the second graph is the graph of y = cosx shifted upby one. All three of y = sinx, y = cosx, and y = tanx can be easily foundon the internet or in any calculus textbook.

10. A kite flies over a fence and starts to descend. The person holding thestring attached to the kite is holding it one meter above the ground. Thefence is two meters high at a distance of 4 meters from the person. At theinstant the kite touches the fence, there is 3 meters of string between thekite and the fence. At that instant, how high off the ground is the kite?Solution: Using similar triangles, we can see that the hand holding thekite, the string, and the fence creates a triangle with sides of length 4, 1,and√

17 respectively. (12 + 42 = 17) Using similar triangles, we see that√17+3√17

= x1

where x is the height of the kite.

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