How do you evaluate $Sin^2 (pi/6) - 2sin (pi/6) cos (pi/6) + cos^2 (-pi/6)$ ?

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How do you evaluate $Sin^2 (pi/6) - 2sin (pi/6) cos (pi/6) + cos^2 (-pi/6)$ ?

2 Answers

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Answer 1 · 2016-08-24T04:13:03

$1/2(2-sqrt3)$

Explanation:

Let us note that, because, $cos(-x)=cosx......(star)$ , the given expression

$=sin^2(pi/6)-2sin(pi/6)cos(pi/6)+cos^2(pi/6)$

$=(sin(pi/6)-cos(pi/6))^2$

$=(1/2-sqrt3/2)^2={(1-sqrt3)/2}^2$

$=1/4(1-2sqrt3+3)$

$=1/4(4-2sqrt3)$

$=1/2(2-sqrt3)$

Alternatively, we can directly plug in the respective values, keeping

$(star)$ in mind, to get,

The Reqd. Value $=(1/2)^2-2(1/2)(sqrt3/2)+(sqrt3/2)^2$

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

$=1-sqrt3/2=1/2(2-sqrt3)$ , as before!

Enjoy Maths.!

Answer 2 · 2016-08-24T04:22:21

$1 - sqrt3/2$

Explanation:

Since
$sin^2 (pi/6) + cos^2 (- pi/6) = 1$
$2sin (pi/6).cos (pi/6) = sin (pi/3)$ (identity: sin 2a = 2sin a.cos a)
There for, the given expression is equal to:
$S = 1 - sin (pi/3) = 1 - sqrt3/2$

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How do you evaluate $Sin^2 (pi/6) - 2sin (pi/6) cos (pi/6) + cos^2 (-pi/6)$ ?

2 Answers

Explanation:

Explanation:

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How do you evaluate Sin^2 (pi/6) - 2sin (pi/6) cos (pi/6) + cos^2 (-pi/6)Sin^2 (pi/6) - 2sin (pi/6) cos (pi/6) + cos^2 (-pi/6)?

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Explanation:

Explanation:

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How do you evaluate $Sin^2 (pi/6) - 2sin (pi/6) cos (pi/6) + cos^2 (-pi/6)$ ?