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

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 $\tan(x) = \dfrac{\sin(x)}{\cos(x)}$.

Phaptitude
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    Generally, any question that involves a trig function of an inverse trig function can be answered by drawing a right triangle, marking an angle, assigning the sides in the proper ratio, using Pythagoras to get the third side, and reading off the trig function. – Ross Millikan Oct 16 '13 at 16:00

5 Answers5

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Hint: 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

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

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

  • That's quite smart, I would have not thought of that method. – Phaptitude Oct 16 '13 at 15:57
  • @Phaptitude That is the only method you should ever need to use for this type of problem, even when you have abstract terms in the function, like if I were trying to solve $\sin^{-1}(\sec(4x/6y))$. – AJMansfield Oct 17 '13 at 00:54
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Here is an approach. Let

$$ y=13 \sin \left( \tan ^{-1} (\dfrac{12}{5}) \right) \implies \sin \left( \tan ^{-1} (\dfrac{12}{5}) \right)=\frac{y}{13} \longrightarrow (1)$$

$$ \implies \cos \left( \tan ^{-1} (\dfrac{12}{5}) \right)=\sqrt{1-\left(\frac{y}{13}\right)^2} \longrightarrow (2). $$

Now, dividing $(1)$ by $(2)$ and simplifying gives

$$ 12^213^3-13^2y^2=0\implies y=\pm 12 \implies y=12. $$

Note: We used the identity

$$ \tan(\tan^{-1}(x))=x. $$

Mhenni Benghorbal
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Have the Pytagoric triangle of sides $5$, $12$ and $13$, and use the definitions of the trigonometric functions in a rectangle triangle.

Carlos Eugenio Thompson Pinzón
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Let $\displaystyle\tan^{-1}\frac{12}5=\theta$

$\displaystyle\implies(i)\tan\theta=\frac{12}5$

and $\displaystyle(ii)-\frac\pi2\le\theta\le\frac\pi2 $ (using the definition of principal value)

$\displaystyle\implies\cos\theta\ge0$

$\displaystyle\implies\cos\theta=\frac1{\sec\theta}=+\frac1{\sqrt{1+\tan^2\theta}}=\cdots$

$\displaystyle\implies \sin\theta=\tan\theta\cdot\cos\theta=\cdots$

lab bhattacharjee
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$\tan^{-1} \left({\frac{12}{5}}\right)\in (0,\frac{\pi}{2})$ since $\frac{12}{5}>0$ now we see $\tan^{-1}\left({\frac{12}{5}}\right)=\sin^{-1}\left({+\frac{12}{13}}\right)$ .So the given expression is $13.\frac{12}{13}=12$ and proved.

shadow10
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