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what is the exact value of `sin ^2 (pi/6) -2 sin( pi/6) cos (pi/6) + cos ^2...
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High School Teacher
`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`
Posted by hala718 on March 30, 2012 at 4:20 AM (Answer #1)
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