Evaluate the following inverse function?

Question

cos[ 2sin^-1(5/13)]

Answer 1

Call sin^-1(5/13) Angle A

this expression is therefore calling for cos (2A) for which you have a choice of three formulas. One of those formulas only uses the sine of A, which you are given as 5/13.

cos (2A)
= 1 - 2sin²A
= 1-2(5/13)²
= 1-50/169
= 119/169

Answer 2

cos(2x) = cos^2x - sin^2x = 1 - 2sin^2 x

x= arcsin(5/13)
sinx = 5/13
sin^2 x = 25/169

cos(2x) = 1 - 2*25/169
=(169-50)/169
= 119/169

Answer 3

cos (2 arcsin (5/13))

First, use the double angle formula:

cos² (arcsin (5/13)) - sin² (arcsin (5/13))

Using the Pythagorean theorem:

1 - 2 sin² (arcsin (5/13))

Simplifying:

1 - 2 (5/13)²
1-50/169
119/169

And we are done.

Answer 4

cos(2sin^-1(5/13)) = sqrt(1-sin^2(2sin^-1(5/13))) = sqrt(1-4sin^2(sin^-1(5/13))cos^2(sin^-1(5/13)) = sqrt(1-4*(5/13)^2 * (1-(5/13^2))) = sqrt( 1-14400/13^4) = sqrt(14161/13^4) = 119/169

so...there it is.