Question

How do you evaluate `sin(arccos(1/2))`?

Answer 1

`sin(arccos(1/2))=sqrt(3)/2`

Explanation:

To calculate this, you need to know two identities:
`sin^2(x)+cos^2(x)=1 <=> sin^2(x)=1-cos^2(x)` `\color(white)(.............)` `(1)`
`cos(arccos(x))=x` `\color(white)(.............)` `(2)`

From `(2)`, we know that we need to have a cosine instead of a sine. We can convert a sine into a cosine by using `(1)`. If we want to use `(1)`, we need `sin^2`, so let's square the expression.
To allow us to square, we'll also immediately take the square root:
`sqrt((sin(arccos(1/2)))^2)=sqrt(sin^2(arccos(1/2)))`
Now, we can use `(1)`
`sqrt(1-cos^2(arccos(1/2)))=sqrt(1-cos(arccos(1/2))*cos(arccos(1/2)))`
Now, we can use `(2)`
`sqrt(1-1/2*1/2)=sqrt(1-1/4)=sqrt(3/4)=sqrt(3)/2`