Russell Gillen

2022-01-14

How to prove $\mathrm{sin}\left(x-a\right)+\mathrm{sin}\left(x+a\right)=2\mathrm{sin}\left(x\right)\mathrm{cos}\left(a\right)$?

Mary Herrera

Expert

HINT
$\mathrm{sin}\left(x+y\right)=\mathrm{sin}x\mathrm{cos}y+\mathrm{cos}x\mathrm{sin}y$

sirpsta3u

Expert

Using $\mathrm{sin}\left(x+y\right)=\mathrm{sin}\left(x\right)\mathrm{cos}\left(y\right)+\mathrm{cos}\left(x\right)\mathrm{sin}\left(y\right)$ as suggested
This means that:
$\mathrm{sin}\left(x+a\right)=\mathrm{sin}\left(x\right)\mathrm{cos}\left(a\right)+\mathrm{cos}\left(x\right)\mathrm{sin}\left(a\right)$
$\mathrm{sin}\left(x-a\right)=\mathrm{sin}\left(x\right)\mathrm{cos}\left(-a\right)+\mathrm{cos}\left(x\right)\mathrm{sin}\left(-a\right)=\mathrm{sin}\left(x\right)\mathrm{cos}\left(a\right)-\mathrm{cos}\left(x\right)\mathrm{sin}\left(-a\right),$
since $\mathrm{cos}\left(-x\right)=\mathrm{cos}\left(x\right)$ and $\mathrm{sin}\left(-x\right)=-\mathrm{sin}\left(x\right)$
Adding $\mathrm{sin}\left(x-a\right)$ and $\mathrm{sin}\left(x+a\right)$ together gives,
$\mathrm{sin}\left(x+a\right)+\mathrm{sin}\left(x-a\right)=2\mathrm{sin}\left(x\right)\mathrm{cos}\left(a\right)$

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