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

see below

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

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