Evaluating a Composition of Functions

[nycmathguy]

Section 4.7

Question 52

cos (arcsin(4/5))

Let u = arcsin(4/5)

This leads to sin u = 4/5.

The range of the arcsin function is [-pi/2 to pi/2] and sin u is positive. We are in quadrant 1.

I know that sin u = opp/hyp.

I drew a right triangle in quadrant 1 to find the adjacent side. According to a^2 + b^2 = c^2, the adj side = 3.

The original problem, after the u-substitution became cos u = adj/hyp. In this case, cos u = 3/5.

So, cos (arcsin(4/5)) = 3/5.

You say?

 

[HallsofIvy]

sine is 4/5 and sine is "opposite side over hypotenuse" so we can take the opposite side to be 4 and the hypotenuse to be 5. Taking the near side to be x, x^2+ 16= 25. x^2= 25- 16= 9 so, yes, x= 3 and the cos(arcsin(4/5))= 3/5. Finally!!

 

[nycmathguy]

sine is 4/5 and sine is "opposite side over hypotenuse" so we can take the opposite side to be 4 and the hypotenuse to be 5. Taking the near side to be x, x^2+ 16= 25. x^2= 25- 16= 9 so, yes, x= 3 and the cos(arcsin(4/5))= 3/5. Finally!!

I got it right.