АЛГЕБРА !!! ДАЮ 20 БАЛЛОВ! ОЧЕНЬ НАДО. Created by elisawetta2. algebra-ru

АЛГЕБРА !!! ДАЮ 20 БАЛЛОВ! ОЧЕНЬ НАДО...

elisawetta2 @elisawetta2

July 2022 1 4 Report

АЛГЕБРА !!! ДАЮ 20 БАЛЛОВ! ОЧЕНЬ НАДО

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NNNLLL54

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$1)\; \; sinx\, cos6x-sin3x\, cos4x=\\\\=\frac{1}{2}\cdot (sin7x+sin(-5x))-\frac{1}{2}\cdot (sin7x+sin(-x)) =\\\\=\frac{1}{2}\cdot (sin7x-sin5x-sin7x+sinx)=\frac{1}{2}\cdot ( sinx-sin5x )=\\\\=\frac{1}{2}\cdot 2\, sin\frac{x-5x}{2}\cdot cos\frac{x+5x}{2}=sin(-2x)\cdot cos3x=-sin2x\cdot cos3x$

$2)\; \; sina\cdot sin( \beta - \alpha )+sin^2(\frac{\beta}{2}-a)=\\\\\star\; \; sinx\cdot siny=\frac{1}{2}\cdot (cos(x-y)-cos(x+y))\; \; \star \\\\\star \; \; sin^2x=\frac{1-cos2x}{2} \; \; \star \\\\=\frac{1}{2}\cdot (cos(2a- \beta )-cos \beta )+\frac{1-cos( \beta -2a)}{2}=\\\\\star \; \; cos( \beta -2a)=cos(2a- \beta )\; \; \star \\\\=\frac{1}{2}\cdot (cos(2a- \beta )-cos \beta +1-cos(2a-\beta ))=\\\\=\frac{1}{2}\cdot (1-cos\beta )= \frac{1-cos \beta }{2} =sin^2 \frac{ \beta }{2}$

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