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How do you evaluate #cos(pi/3)cos((5pi)/12)-sin(pi/3)sin((5pi)12)#?

1 Answer

Jul 11, 2016

#-1/sqrt2.#

Explanation:

Recall that #cos(A+B)=cosAcosB-sinAsinB#

So, given Exp. #=cos(pi/3+5pi/12)=cos(9pi/12)=cos(3pi/4)=cos (pi-pi/4)=-cos(pi/4)=-1/sqrt2.#

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