Question

What is the exact value of `Sin^2 (pi/6) - 2sin (pi/6) cos (pi/6) + cos^2 (-pi/6)`?

Answer 1

Value is `(1-sqrt3/2)`

Explanation:

To solve this we need value of `sin(pi/6)`, `cos(pi/6)` and `cos(-pi/6)`. As `sin(pi/6)=1/2` and `cos(pi/6)=cos(-pi/6)=sqrt3/2`, putting these

`sin^2(pi/6)−2sin(pi/6)cos(pi/6)+cos^2(−pi/6)`

=`(1/2)^2-2*1/2*sqrt3/2+(sqrt3/2)^2`

=`1/4-sqrt3/2+3/4`

= `1-sqrt3/2`