How do you prove #sin(x+[pi/6])*sin(x-[pi/6]) = (sin x)^2 - 1/4#?

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Answer 1 · 2016-05-04T19:13:43

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Use Properties: #sin(A+B)=sinAcosB+cosAsinB#

#sin(A-B)=sin A cos B-cos A sin B#

Left Side:#=(sinxcos(pi/6) + cosx sin (pi/6)) xx (sinxcos(pi/6) - cosx sin (pi/6))#

#=sin^2xcos^2(pi/6)-cos^2xsin^2(pi/6)#

#=sin^2x * (sqrt3/2)^2 - cos^2x * (1/2)^2#

#=3/4 sin^2x-1/4 cos^2 x#

#=3/4 sin^2 x-1/4(1-sin^2x)#

#=3/4 sin^2 x-1/4+1/4 sin^2 x#

#=(3/4 sin^2 x+1/4 sin^2 x)-1/4#

#=sin^2x-1/4#

#=# Right Side