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

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hala718 | High School Teacher | (Level 1) Educator Emeritus

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`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`

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