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

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Answer 1 · 2016-02-18T09:25:20

Value is $(1-sqrt3/2)$

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

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

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

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

= $1-sqrt3/2$

What is the exact value of 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)?