Question

Решите пожалуйста ( с объяснением, если можно):
$sin( frac{2 pi }{3} - frac{x}{4} )*cos( frac{ pi }{6} + frac{x}{4} )*sin frac{x}{4}$
( Ответ: $frac{1}{4}sin frac{3x}{4}$)

Answer 1

sin(2π/3 -x/4)*cos(π/6+x/4)*sinx/4 = 1/2 *(sin(2π/3 - x/4 +π/6+x/4) +
+sin(2π/3 - x/4 -π/6-x/4) ) *sinx/4 = 1/2 * (sin5π/6 + sin(π/2 -x/2)) *sinx/4  =
 1/2 * (1/2 + cosx/2) *sinx/4  =(1/4)*sinx/4  +1/2*sinx/4*cosx/2 = 
(1/4)*sinx/4  +(1/4)*( sin(x/4 -x/2) +cos(x/4 +x/2) ) =
(1/4)*sinx/4  +(1/4)* sin( -x/4) +(1/4)*cos3x/4 =(1/4)*sinx/4  -(1/4)* sinx/4 +
+(1/4)*cos3x/4= (1/4)*cos3x/4.

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использованы формулы : sin(π/2 -α) =cosα и
sinα*cosβ =(sin(α+β) + sin(α-β))/2 .