Question

What is `cos(sin^(-1)(sqrt(2)/2))` ?

Answer 1

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

Explanation:

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

Hence:

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

So if `sin theta = sqrt(2)/2` then:

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

Note also that `sin^(-1) x` is always in the range `[-pi/2, pi/2]` and `cos theta >= 0` for `theta in [-pi/2, pi/2]`.

Hence we want the positive square root and we find:

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

Alternatively, just consider a right angled triangle formed by bisecting a unit square diagonally, with sides `1, 1, sqrt(2)`...

Remember that:

`sin theta = "opposite"/"hypotenuse"`

`cos theta = "adjacent"/"hypotenuse"`

So we can see from this triangle that:

`sin (pi/4) = cos (pi/4) = 1/sqrt(2) = sqrt(2)/2`

Since we are dealing with angles in Q1, we find:

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

Answer 2

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

Explanation:

We can solve this using that `sin(pi/4) = sqrt(2)/2` is well known, giving

`sin^(-1)(sqrt(2)/2) = pi/4`

`=> cos(sin^(-1)(sqrt(2)/2)) = cos(pi/4) = sqrt(2)/2`

However, let's look at a more general method for solving this kind of problem.

Suppose we are trying to find `cos(sin^(-1)(x))`. First, draw a right triangle with an angle `a` such that `sin(a) = x`, that is, `a = sin^(-1)(x)`.

Using the Pythagorean theorem, we can find the remaining side as `sqrt(1-x^2)`. With this, we have

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

If we set `x = sqrt(2)/2`, we find

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

`=sqrt(1-1/2)`

`=sqrt(1/2)`

`=1/sqrt(2)`

`=sqrt(2)/2`

giving us the same answer as above.