Solve the problem.

$\mathrm{sin}2x-\mathrm{sin}4x$

$\mathrm{sin}2x-\mathrm{sin}4x$

Macioccujx
2022-07-30
Answered

Solve the problem.

$\mathrm{sin}2x-\mathrm{sin}4x$

$\mathrm{sin}2x-\mathrm{sin}4x$

You can still ask an expert for help

minotaurafe

Answered 2022-07-31
Author has **22** answers

$\mathrm{sin}2x-\mathrm{sin}4x=\mathrm{sin}2x-2\mathrm{sin}2x.\mathrm{cos}2x=\mathrm{sin}2x(1-2\mathrm{cos}2x)$

$=\mathrm{sin}2x(1-\mathrm{cos}2x+1)=2\mathrm{sin}x\mathrm{cos}x(1-2\mathrm{cos}2x)$

$=\mathrm{sin}2x(1-\mathrm{cos}2x+1)=2\mathrm{sin}x\mathrm{cos}x(1-2\mathrm{cos}2x)$

Ismael Molina

Answered 2022-08-01
Author has **2** answers

$\mathrm{sin}(\alpha +\beta )=\mathrm{sin}\alpha \mathrm{cos}\beta +\mathrm{cos}\alpha \mathrm{sin}\beta $

$\mathrm{sin}(\alpha -\beta )=\mathrm{sin}\alpha \mathrm{cos}\beta -\mathrm{cos}\alpha \mathrm{sin}\beta -$

$\mathrm{sin}(\alpha +\beta )-\mathrm{sin}(\alpha -\beta )=2\mathrm{cos}\alpha \mathrm{sin}\beta $ , assume $\alpha +\beta =p$ and $\alpha -\beta =q$

$\alpha +\beta =p$

$\alpha -\beta =q+2\alpha =p+q=>\alpha =1/2(p+q)$

$\alpha +\beta =p$

$\alpha -\beta =q-2\beta =p-q=>\beta =1/2(p-q)$

$\therefore \mathrm{sin}p-\mathrm{sin}q=2\mathrm{cos}1/2(p+q)\mathrm{sin}1/2(p-q)$

$\mathrm{sin}2x-\mathrm{sin}4x=2\mathrm{cos}1/2(2x+4x)\mathrm{sin}1/2(2x-4x)$

$=2\mathrm{cos}1/2(6x)\mathrm{sin}1/2(-2x)$

$=2\mathrm{cos}3x\ast \mathrm{sin}-x=>\mathrm{sin}(-x)=-\mathrm{sin}x$

$=-2\mathrm{cos}3x\mathrm{sin}x$

$\mathrm{sin}(\alpha -\beta )=\mathrm{sin}\alpha \mathrm{cos}\beta -\mathrm{cos}\alpha \mathrm{sin}\beta -$

$\mathrm{sin}(\alpha +\beta )-\mathrm{sin}(\alpha -\beta )=2\mathrm{cos}\alpha \mathrm{sin}\beta $ , assume $\alpha +\beta =p$ and $\alpha -\beta =q$

$\alpha +\beta =p$

$\alpha -\beta =q+2\alpha =p+q=>\alpha =1/2(p+q)$

$\alpha +\beta =p$

$\alpha -\beta =q-2\beta =p-q=>\beta =1/2(p-q)$

$\therefore \mathrm{sin}p-\mathrm{sin}q=2\mathrm{cos}1/2(p+q)\mathrm{sin}1/2(p-q)$

$\mathrm{sin}2x-\mathrm{sin}4x=2\mathrm{cos}1/2(2x+4x)\mathrm{sin}1/2(2x-4x)$

$=2\mathrm{cos}1/2(6x)\mathrm{sin}1/2(-2x)$

$=2\mathrm{cos}3x\ast \mathrm{sin}-x=>\mathrm{sin}(-x)=-\mathrm{sin}x$

$=-2\mathrm{cos}3x\mathrm{sin}x$

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