Section 4.4

#1. a) We need to find sin^{-1}(1/2). In other words, the value of t where sin(t) = 1/2. That means the y-coordinate of the terminal point must be 1/2, which means the value of t must be associated with π/6. Now the inverse sine function has principal values in the 1st & 4th quadrants. Since the y-coordinate is positive, we must be in the 1st quadrant. So a π/6 distance into quadrant 1 is, in fact, just π/6.

b) Same idea as in part (a), but this time we know the x-coordinate is 1/2. And the principal values for cosine are in quadrants 1 & 2.

c) This time we need to find t where cos(t) = 2. I hope that seems like a problem to you.

#2 - 8. Same ideas as the hint for #1.

#9 - 12. Make sure your calculator is in Radian mode, not Degree mode.

#13 - 28: Most of the problems in this set should be similar to the hints given for #13 & #19 below.

#13. We need to find sin(sin^{-1}(1/4)).

Step 1: Let's figure out the plus or minus sign for our final answer. Then we can ignore it along the way. Since the 1/4 is positive, that means the sine-inverse must give us an answer in the first quadrant. Then the sine of a 1st quadrant angle must be positive, so our final answer must be positive.

Step 2: Find the first quadrant solution to our problem. Let's replace the inside part, sin^{-1}(1/4), with A, so our final answer should be the value of sin(A). If the inside part equals A, then we have sin^{-1}(1/4) = A, which is equivalent (in the first quadrant) to sin(A) = 1/4. So we know that sin(A) = 1/4, but our final answer is the value of sin(A): So our answer to this step is 1/4.

Step 3: Mix the first two steps together. That is, take our answer of 1/4 and make it positive (which it is already). So our answer is 1/4.

#19. We need to find sin^{-1}(sin(-Pi/6)).

Step 1. Let's figure out which quadrant our final answer will be in. Then we can ignore the minus signs along the way. Since the -π/6 is in quadrant 4, and sine is negative in quadrant 4, the sin(-π/6) will be negative. Then we take a sine-inverse of a negative value, which means our final answer must be in quadrant 4.

Step 2: Find the first quadrant solution to our problem. That is, find sin^{-1}(sin(π/6)). But in the first quadrant, sine & sine-inverse cancel each other out, giving π/6.

Step 3: Mix the first two steps together. That is, take our answer of π/6 and move it into the 4th quadrant. Thus our answer is - π/6.

#29. We need to evaluate sin(cos^{-1}3/5). Let's call the inside part A, which means we are ultimately looking for sin(A). If A = cos^{-1}3/5, then that means cos(A) = 3/5. We set up a triangle that says cos(A) = 3/5, which means we put 3 on the adjacent side and 5 on the hypotenuse. Using the Pythagorean Theorem, we can calculate the opposite side to be 4 this time. Since we are ultimately looking for sin(A), and we know sine is opp/hyp, we get sin(A) = 4/5.

#33. We need to evaluate sec(sin^{-1}12/13). Let's call the inside part A, which means we are ultimately looking for sec(A). If A =sin^{-1}12/13, then that means sin(A) =12/13. We set up a triangle that says sin(A) = 12/13, which means we put 12 on the opposite side and 13 on the hypotenuse. Using the Pythagorean Theorem, we can calculate the adjacent side to be 5 this time. Since we are ultimately looking for sec(A), and we know secant is hyp/adj, we get sec(A) = 13/5.

#41. We want an expression for cos(sin^{-1}x). Let's think of sin^{-1}x as A, so we need to find cos(A). If sin^{-1}x = A, then we need to find an angle A whose sine is x. Think of a right-triangle with one angle A, and the opposite side=x, and the hypotenuse=1. That would cause the sine to be x/1. That would also cause the adjacent side to be sqrt(1-x^{2}). So finally cos(A) = adj/hyp = sqrt(1-x^{2}).

#44. We want an expression for cos(tan^{-1}x). Let's think of tan^{-1}x as A, so we need to find cos(A). If tan^{-1}x = A, then we need to find an angle A whose tangent is x. Think of a right-triangle with one angle A, and the opposite side=x, and the adjacent side=1. That would cause the tangent to be x/1. That would also cause the hypotenuse to be sqrt(x^{2}+1). So finally cos(A) = adj/hyp = 1/sqrt(x^{2}+1).

#49. I will use T in place of theta here.

a) Looking at the triangle on the picture, we can see that tan(T) = h/2. Since we want to express h as a function of T, we need to isolate h. Multiplying through by 2 gives us: h = 2tan(T).

b) This time we want T as a function of h. That is, we need to isolate T. Since tan(T) = h/2, we can take the inverse tangent of both sides and get: T =tan^{-1}(h/2).

#51. I will use T in place of theta here.

a) Looking at the triangle on the picture, we can see that sin(T) = h/680. Since we want to express T as a function of h, we need to isolate T. Taking the inverse sine of both sides we get: T = sin^{-1}(h/680).

b) If h = 500, then T = sin^{-1}(500/680). A calculator gives us T = 47.3^{o}.