Rebecca Villa

2022-07-09

We have the expression $13\mathrm{sin}[{\mathrm{tan}}^{-1}({\displaystyle \frac{12}{5}})]$.

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 $\mathrm{tan}(x)={\displaystyle \frac{\mathrm{sin}(x)}{\mathrm{cos}(x)}}$

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 $\mathrm{tan}(x)={\displaystyle \frac{\mathrm{sin}(x)}{\mathrm{cos}(x)}}$

Sanaa Hinton

Beginner2022-07-10Added 15 answers

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.

$\mathrm{tan}t=\frac{12}{5}$

Recall that the tangent is the opposite side over the adjacent side.

Ximena Skinner

Beginner2022-07-11Added 7 answers

Let ${\mathrm{tan}}^{-1}\frac{12}{5}=\theta$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}(i)\mathrm{tan}\theta =\frac{12}{5}$

and $(ii)-\frac{\pi}{2}\le \theta \le \frac{\pi}{2}$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{cos}\theta \ge 0$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{cos}\theta =\frac{1}{\mathrm{sec}\theta}=+\frac{1}{\sqrt{1+{\mathrm{tan}}^{2}\theta}}=\cdots$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{sin}\theta =\mathrm{tan}\theta \cdot \mathrm{cos}\theta =\cdots$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}(i)\mathrm{tan}\theta =\frac{12}{5}$

and $(ii)-\frac{\pi}{2}\le \theta \le \frac{\pi}{2}$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{cos}\theta \ge 0$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{cos}\theta =\frac{1}{\mathrm{sec}\theta}=+\frac{1}{\sqrt{1+{\mathrm{tan}}^{2}\theta}}=\cdots$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{sin}\theta =\mathrm{tan}\theta \cdot \mathrm{cos}\theta =\cdots$

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