Question

sin pi/12 × cos^3 pi/12 - sin^3 pi/12 × cos pi/12 - 1/4 sin pi/3

Answer 1

$sin\frac{\pi}{12}*cos^3\frac{\pi}{12}-sin^3\frac{\pi}{12}*cos\frac{\pi}{12}-\frac{1}4sin\frac{\pi}{3}\\\\sin\frac{\pi}{12}*cos\frac{\pi}{12}(cos^2\frac{\pi}{12}-sin^2\frac{\pi}{12})-\frac{1}4sin\frac{\pi}{3}\\\\\frac{1}4*(4sin\frac{\pi}{12}*cos\frac{\pi}{12}*cos\frac{\pi}{6}-sin\frac{\pi}{3})\\\\\frac{1}4(2sin\frac{\pi}{6}*cos\frac{\pi}{6}-sin\frac{\pi}{3})\\\\\frac{1}4(sin\frac{\pi}{3}-sin\frac{\pi}{3})\\\\\frac{1}4*0\\\\0$

$(cos^2a-sin^2a=cos2a)\\(2sina*cosa=sin2a)$