Решите пожалуйста ( с объяснением, если можно):
[tex]sin( \frac{2 \pi }{3} - \frac{x}{4} )*cos( \frac{ \pi }{6} + \frac{x}{4} )*sin \frac{x}{4} [/tex]
( Ответ: [tex] \frac{1}{4}sin \frac{3x}{4} [/tex])
[tex]sin( \frac{2 \pi }{3} - \frac{x}{4} )*cos( \frac{ \pi }{6} + \frac{x}{4} )*sin \frac{x}{4} [/tex]
( Ответ: [tex] \frac{1}{4}sin \frac{3x}{4} [/tex])
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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.
* * * * * * * * * * * * * * * * * * * * * * * * * *
использованы формулы : sin(π/2 -α) =cosα и
sinα*cosβ =(sin(α+β) + sin(α-β))/2 .