How calculate cos((arcsin(12/13))/2) ?

Expert Answers

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You need to use the following substitution, such that:

`arcsin (12/13) = theta => sin theta = 12/13`

Hence, replacing `theta ` for `arcsin (12/13)` yields:

`cos ((arcsin(12/13))/2) = cos(theta/2)`

Using half angle identity, yields:

`cos(theta/2) = sqrt((1 + cos theta)/2)`

You need to evaluate `cos theta` , using the Pythagorean identity `sin^2 theta + cos^2 theta = 1` , such that:

`cos^2 theta = 1 - sin^2 theta`

Since `sin theta = 12/13, ` yields:

`cos^2 theta = 1 - (12/13)^2 => cos^2 theta = (169 - 144)/169`

`cos^2 theta = 25/169 => cos theta = 5/13`

`cos(theta/2) = sqrt((1 + 5/13)/2) => cos(theta/2) = sqrt(18/26)`

`cos(theta/2) = sqrt(9/13) => cos(theta/2) = 3/sqrt13 => cos(theta/2) = 3sqrt13/13`

Replacing back `arcsin (12/13)` for `theta,` yields:

`cos ((arcsin(12/13))/2) = 3sqrt13/13`

Hence, evaluating the given expression, yields cos `((arcsin(12/13))/2) = 3sqrt13/13` .

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