Question

How do you evaluate `cos(sin^-1((sqrt3/2))` without a calculator?

Answer 1

`cos(sin^(-1)(sqrt(3)/2)) = 1/2`

Explanation:

Consider an equilateral triangle with sides of length `2`, bisected to form two right angled triangles:

Remembering that `sin = "opposite"/"hypotenuse"`, we can see that:

`sin(pi/3) = sqrt(3)/2`

Since `sqrt(3)/2 > 0` and `pi/3` is in Q1, we can deduce:

`sin^(-1)(sqrt(3)/2) = pi/3`

From the same diagram, remembering `cos = "adjacent"/"hypotenuse"`, we can see that:

`cos(pi/3) = 1/2`

So:

`cos(sin^(-1)(sqrt(3)/2)) = cos(pi/3) = 1/2`

Answer 2

`cos(sin^(-1)(sqrt(3)/2)) = 1/2`

Explanation:

Starting from:

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

Subtract `sin^2 theta` from both sides to get:

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

Take the square root to find:

`cos theta = +-sqrt(1-sin^2 theta)`

If `theta = sin^(-1)(sqrt(3)/2)` then `sin(theta) = sqrt(3)/2` and we find:

`cos theta = +-sqrt(1-(sqrt(3)/2)^2) = +-sqrt(1-3/4) = +-sqrt(1/4) = +-1/2`

Further note that `sqrt(3)/2 > 0`, so `sin^(-1)(sqrt(3)/2)` must be in Q1. Hence `cos theta > 0` and we need to choose the positive square root.

So:

`cos(sin^(-1)(sqrt(3)/2)) = 1/2`