Crystal Wheeler

2022-07-03

Solving $\frac{\mathrm{sin}\left(2v\right)+\mathrm{sin}\left({40}^{\circ }\right)}{\mathrm{sin}\left({40}^{\circ }-2v\right)}=3$

jugf5

Expert

HINT:
Using Prosthaphaeresis Formula $\left(\mathrm{sin}C+\mathrm{sin}D\right)$,
$2\mathrm{sin}\left(v+{20}^{\circ }\right)\mathrm{cos}\left(v-{20}^{\circ }\right)=-3\cdot 2\mathrm{sin}\left(v-{20}^{\circ }\right)\mathrm{cos}\left(v-{20}^{\circ }\right)$
$2\mathrm{cos}\left(v-{20}^{\circ }\right)\left\{\mathrm{sin}\left(v+{20}^{\circ }\right)+3\mathrm{sin}\left(v-{20}^{\circ }\right)\right\}=0$
So, either $\mathrm{cos}\left(v-{20}^{\circ }\right)=0$ or $\mathrm{sin}\left(v+{20}^{\circ }\right)+3\mathrm{sin}\left(v-{20}^{\circ }\right)=0$
But $\mathrm{cos}\left(v-{20}^{\circ }\right)=0$ will make the left hand side of the given expression
$\frac{0}{0}$
Finally use
$\mathrm{sin}\left(A±B\right)$
formula

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