Rebecca Villa

2022-07-09

We have the expression $13\mathrm{sin}\left[{\mathrm{tan}}^{-1}\left(\frac{12}{5}\right)\right]$.
Apparently the answer is 12, and I have to simplify it, and I'm assuming it means I have to show it's 12, without using a calculator.
Normally I show my own work in the questions, but in this case I have absolutely no clue how to. The only thing I know that might help is that $\mathrm{tan}\left(x\right)=\frac{\mathrm{sin}\left(x\right)}{\mathrm{cos}\left(x\right)}$

Sanaa Hinton

There is a right triangle with side lengths 5, 12 and 13. Draw this triangle, and choose one of the two non-right angles t for which
$\mathrm{tan}t=\frac{12}{5}$
Recall that the tangent is the opposite side over the adjacent side.

Ximena Skinner

Let ${\mathrm{tan}}^{-1}\frac{12}{5}=\theta$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\left(i\right)\mathrm{tan}\theta =\frac{12}{5}$
and $\left(ii\right)-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mathrm{cos}\theta \ge 0$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mathrm{cos}\theta =\frac{1}{\mathrm{sec}\theta }=+\frac{1}{\sqrt{1+{\mathrm{tan}}^{2}\theta }}=\cdots$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mathrm{sin}\theta =\mathrm{tan}\theta \cdot \mathrm{cos}\theta =\cdots$

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