# I cannot figure out how which identities to use to solve this problem. `cos[(1/2)sin^(-1) 8/17] ` - I know the answer is `(4sqrt(17))/17 ` but I can't figure out how to get there. Help please!

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The identity needed here is the half-angle identity. For cosine, this identity is

` cos^2 (x/2) = (1 + cosx)/2 `

In your example, angle x is the arcsine of 8/17 (`sin^(-1) (8/17) ` ). So, the first step is to calculate the ` cos(sin^(-1) 8/17) ` .

To do this, we need to use the Pythagorean identity:

`cos^2x + sin^2x = 1 `

The sine of arcsine, by definition, is the angle itself, so

`sin(sin^(-1) 8/17) = 8/17 `

Then, `sin^2(sin^(-1) 8/17) = 8^2/17^2 = 64/289 `

From Pythagorean identity,

`cos^2(sin^(-1) 8/17) = 1 - sin^2 (sin^(-1) 8/17) = 1 - 64/289 = (289 - 64)/289 = 225/289 `

Taking square root, we find the cosine:

`cos(sin^(-1) 8/17) = sqrt(225/289) = 15/17 `

Now we can plug this cosine into the half-angle identity and find the square of the cosine of half-arcsine of 8/17:

` cos^2[(1/2)sin^(-1) 8/17] = (1 + cos(sin^(-1) 8/17))/2 = (1 + 15/17)/2`

This fraction can be simplified by multiplying both numerator and denominator by 17:

`cos^2[(1/2)sin^(-1) 8/17] = (17 + 15)/34 = 32/34 = 16/17 `

Taking square root results in the cosine we are looking for:

`cos[(1/2)sin^(-1) 8/17] = sqrt(16/17) = 4/sqrt(17) = (4sqrt(17))/17 ` (The last step is to multiply both numerator and denominator by ` sqrt(17) ` , in order to rationalize the denominator.)

Hope this helps!

In this case there is a simpler method to evaluate `cos(sin^(-1) (8/17)) ` :

Note that 8-15-17 represents a primitive pythagorean triple. Thus a simpler "textbook" solution is to see that `sin^(-1)(8/17)=alpha ==> sin alpha = 8/17 `

So `cos(sin^(-1)(8/17))=cos alpha =15/17 `