How do you express #cos(pi/ 2 ) * cos (( 7 pi) / 6 ) # without using products of trigonometric functions?

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Answer 1 · 2016-04-15T16:04:42

#cos (pi/2) cos ((7pi)/6) = 1/2 [cos ((10pi)/6)+cos(-(4pi)/6)] =0#

Sum and Product Properties:

#cos (A+B) + cos (A-B) = 2 cos A cos B#

#cosA cosB = 1/2 [cos(A+B)+cos(A-B)]#

#cos (pi/2) cos ((7pi)/6)=1/2 [cos(pi/2 +(7pi)/6)+cos(pi/2 -(7pi)/6)]#

#cos (pi/2) cos ((7pi)/6) = 1/2 [cos ((10pi)/6)+cos(-(4pi)/6)]#

#=1/2 [1/2+ -1/2]#

#=1/2 (0)#

#=0#