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How do you evaluate # cos (pi/2) cos(pi/7) + sin(pi/2) sin(pi/7)#?

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Oct 16, 2016

Inserting #cos(pi/2)=0 and sin(pi/2)=1# in the given expression we get

#cos(pi/2)cos(pi/7)+sin(pi/2)sin(pi/7)#

#=0xxcos(pi/7)+1xxsin(pi/7)#

#=sin(pi/7)=0.434#

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