what is the exact value of

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

Show all the steps to this question

### 1 Answer | Add Yours

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

First, we know that:

`cos(-x) = cos(x) `

`==gt cos^2 (-pi/6) = cos^2 (pi/6)`

Also, we know that:

`2sinxcosx = sin2x`

`==> 2sin (pi/6) cos (pi/6) = sin2(pi/6)= sin (pi/3)`

==> Now we will substitute:

`==gt sin^2 (pi/6) - sin (pi/3) + cos^2 (pi/6)`

But `sin^2 pi/6 + cos^2 pi/6 = 1\`

` ==gt 1-sin pi/3 = 1- sqrt3/2 = (2-sqrt3)/2 ~~ 0.134`

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes