187=(1/2)*(2cos2θcos(θ/2)-2cos3θcos(9θ/2))
=(1/2)*(cos(5θ/2)+cos(3θ/2)-cos(15θ/2)-cos(3θ/2))
=1/2*(cos5θ/2)-cos(15θ/2))
=1/2*(2sin5θsin(5θ/2))
=sin5θsin5θ/2
prove..
1.cos2θcosθ/2-cos3θcos9θ/2=sin5θsin5θ/2
2.sinAsin(A+2B)-sinBSin(B+2A)=sin(A-B)sin(A+B)
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2 Answers
187187=1/2*(2sinAsin(A+2B)-2sinBsin(B+2A))
=1/2*(cos2B-cos(2A+2B)-cos2A+cos(2A+2B))
=1/2*(cos2B-cos2A)
=1/2*(2sin(A+B)sin(A-B))
=sin(A+B)sin(A-B)
Shreyas Venkatesan thanks man !!! :)Upvote·0· Reply ·2013-08-12 07:44:44