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

1 Answer
Jul 3, 2015

#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#