Extra Practice Solutions Trigonometry

Extra Practice Solutions: Trigonometry 1. For this first set of problems, you should be able to use our two triangles (π/6, π/3, π/2 and π/4, π/4, π/2) and the unit circle. Evaluate the following (exactly, without a calculator): (a) sin(3π/4) SOLUTION: By the unit circle, sin(3π/4) = sin(π/4) and √ this can be read off of √ a standard triangle- The value is 1/ 2 (or equivalently, 2/2). (b) cos(−5π/4) SOLUTION: In the unit circle, −5π/4 is in quadrant II, and is the same as 3π/4. The cosine in this quadrant is negative, so the answer is: 1 cos(−5π/4) = − cos(π/4) = − √ 2 (c) tan(2π/3) SOLUTION: In the unit circle, 2π/3 is in the second quadrant. In this quadrant, the sine is positive and the cosine is negative, so the tangent is negative. Use this together with the 30-60-90 triangle: √ √ 3 =− 3 tan(2π/3) = − tan(π/3) = − 1 (d) sec(7π/6) SOLUTION: In the unit circle, we see that 7π/6 is π/6 into the third quadrant, so sine and cosine is negative. Use this, together with the 30-60-90 triangle: sec(7π/6) = 1 2 1 =− −√ cos(7π/6) cos(π/6) 3 (e) csc(29π/6) SOLUTION: The angle is π/6 shy of 30π/6, which is 5π. On the unit circle, this is equivalent to π. Overall, the angle is the same as 5π/6 on the unit circle. Since this is in quadrant II, the sine is positive. This, together with the 30-60-90 triangle gives us: csc(29π/6) = 1 1 = =2 sin(29π/6) sin(π/6) (f) tan(π/4) SOLUTION: At π/4, sine is equal to cosine, to the tangent is 1. For the next two questions, we want to use the general ideas: 1 • For the “regular” trig functions, sine and cosine are periodic with period 2π. The tangent is periodic with period π. • For A cos(ωt) or A sin(ωt), the amplitude is A, the period is 2π/ω and the frequency is the reciprocal of the period, or ω/(2π). For the tangent, the period of tan(ωt) would be π/ω. • Adding a constant would just shift the graph up/down, and would not change the amplitude or period. 2. What is the amplitude, period and frequency for f (x) = 1 + 2 cos(3x) SOLUTION: Amplitude is 2, period is 2π/3, frequency is 3/(2π). 3. What is the period of f (x) = tan(πx)? f (x) = cos(x/π)? SOLUTION: The period of tan(πx) is 1. The period of cos(x/π) is 2π 2 4. Solve for x: (a) 2 cos(x) + 1 = 0 SOLUTION: First, cos(x) = − 21 . On the unit circle, we can have angles either in quadrant II or quadrant III. By the 30-60-90 triangle, the solution to cos(x) = 1/2 is x = π/3. Putting this in the second and third quadrants: x= 2π 4π , or we can add any multiple of 2π 3 3 (b) 3 cot2 (x) = 1 SOLUTION: First the algebra to get √ 1 cot(x) = ± √ ⇒ tan(x) = ± 3 3 The tangent function is positive in quadrants I and III, and negative in quadrants II and IV. Furthermore, using a 30-60-90 triangle, x = π/3. Therefore, the solutions to: √ π 4π tan(x) = 3 ⇒ x = , 3 3 And √ 2π 5π tan(x) = − 3 ⇒ x = , 3 3 And we can add any multiple of π to these. In fact, with that simplification, we could say: √ π tan(x) = 3 ⇒ x = + kπ for k = 0, ±1, ±2, ... 3 and √ 2π tan(x) = − 3 ⇒ x = + kπ for k = 0, ±1, ±2, ... 3 2 (c) sin(x) > cos(x) To solve this, we first look to see where sin(x) = cos(x). On the unit circle, this is at π/4 and 5π/4. Also on the unit circle, this set of points correspond to where the y coordinate is larger than the x coordinate. π 5π SOLUTION: < x < 4 4 5. For the following, recall that, to compute sin−1 (x) means to find an angle θ so that sin(θ) = x. Normally, this would mean that there are an infinite number of possible angles (because the sine is not one-to-one); in order to make the inverse sine a function, we restricted the possible angles to be between −π/2 and π/2. Similarly, we defined the other inverse functions. Here we go: (a) sin−1 (0) SOLUTION: 0 = sin−1 (0) either by unit circle or the graph of sine. (b) sin−1 (1) SOLUTION: π/2 is the angle (the only one in [−π/2, π/2]). (c) arcsin(1/2) SOLUTION: Use a 30-60-90 triangle to see that the angle is π/6 (and this is the only angle). (d) sin−1 (2) SOLUTION: The domain of the inverse sine is [−1, 1], so we cannot compute this quantity (it does not exist). √ (e) tan−1 (− 3) SOLUTION: The angles used in the inverse tangent are almost the same as the inverse sine (delete the endpoints). The tangent is negative in quadrant IV, and using the 30-60-90 triangle we see that the angle is −π/3. (f) tan−1 (0) = 0 (g) tan−1 (1) = π/4 (h) sec−1 (−2) SOLUTION: This is the angle whose secant is −2, which means that the cosine of the angle is −1/2. Since we typically restrict the domain of the cosine to be [0, π] (to get its inverse), we’ll do the same with the secant, except for the secant there is a vertical asymptote at π/2 (since cos(π/2) = 0). With that restriction, we need the angle to be in quadrant II. Using a 30-60-90 triangle, we see that the cosine is 1/2 means the angle is π/3. Putting this in quadrant II gives us our answer: sec−1 (−2) = 2π 3 √ (i) sec−1 (2/ 3) SOLUTION: Similar to the √ previous reasoning, we look for an angle in [0, pi/2) ∪ (π/2, π] for which cos(θ) = 32, which is π/6. 3 6. The trick on these is that, while the inverse function undoes the original function: f −1 (f (x)) = x ⇒ sin(sin−1 (x)) and sin−1 (sin(x)) = x we have to use the restricted domain and range (so that the function is one-to-one). Same idea applies to the tangent and inverse tangent. Here we go: (a) sin−1 (sin(π)) SOLUTION: Since π is not in our restricted domain of [−π/2, π/2], you should note that the answer is NOT π. Rather, sin(π) = 0 and sin−1 (0) = 0, so the answer is 0. (b) sin(sin−1 (3/5) SOLUTION: Since 3/5 is in the restricted domain of the inverse sine (which is [−1, 1]), the answer is 3/5 (that is, we apply the rule sin(sin−1 (x)) = x). (c) tan−1 (tan(π/4)) SOLUTION: π/4 (d) tan−1 (tan(π)) = 0 by reasoning similar to the first problem. 7. Simplify the following expressions (using a triangle). Also think about the value(s) of x for which the simplification is valid. (a) tan(sin−1 (x)) SOLUTION: Let θ = sin−1 (x), or sin(θ) = x1 . Build a right triangle with this relationship (opposite√ θ has length x, hypotenuse has length 1). Then the length of the side adjacent is 1 − x2 by the Pythagorean Theorem. Finally, we compute the tan: x tan(θ) = tan(sin−1 (x)) = √ 1 − x2 (b) cos(tan−1 (x)) SOLUTION: Let θ = tan−1 (x), or tan(θ) = x1 . Build a right triangle with this relationship (opposite√θ has length x, adjacent has length 1). Then the length of the hypotenuse is 1 + x2 by the Pythagorean Theorem. Now compute the cosine: 1 cos(θ) = cos(tan−1 (x)) = √ 1 + x2 (c) sec(sin−1 (x)) SOLUTION: This triangle is the same as in part (a)- From that: cos(θ) = √ 1 1 − x2 ⇒ 4 sec(sin−1 (x)) = √ 1 − x2 (d) tan(sec−1 (x)) SOLUTION: Make this triangle consistent with the relation: cos(θ) = 1 x so that the length of the side adjacent is 1,√the length of the hypotenuse is x, which makes the length of the side opposite x2 − 1. Therefore, √ tan(sec−1 (x)) = x2 − 1 (e) cos(2 sin−1 (x)) SOLUTION: Using the hint, if we let θ = sin−1 (x), then we need to find cos2 (θ) and sin2 (θ). Build a triangle consistent with the relation: sin(θ) = x 1 so that the length of the side adjacent to θ is cos(2θ) = cos2 (θ) − sin2 (θ) = √ √ 1 − x2 . Then: 1 − x2 2 − x2 = 1 − 2x2 (f) sin(2 tan−1 (x)) Using the hint, if θ = tan−1 (x), the we want to compute sin(θ) and cos(θ).√The tri√ 2 angle has the same lengths as (b), so: sin(θ) = x/ 1 − x and cos(θ) = 1/ 1 − x2 : sin(2θ) = 2 sin(θ) cos(θ) = 5 2x 1 − x2