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How do you evaluate #cos^2(pi/12)-sin^2(pi/12)#?

1 Answer

Shwetank Mauria

May 2, 2016

#cos^2(pi/12)-sin^2(pi/12)=sqrt3/2#

Explanation:

To evaluate #cos^2(pi/12)-sin^2(pi/12)#, recall the identity

#cos2theta=cos^2theta-sin^2theta#

Hence, #cos^2(pi/12)-sin^2(pi/12)=cos2xx(pi/12)=cos(pi/6)=sqrt3/2#

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