tan(arcsin(5/13)) calculated without a calculator: how?

[DanteQ]

tan(arcsin(5/13))

You should be able to calculate this one without a calculator. But how?

PS. The answer should be 5/12

 

[Deleted member 4993]

Re: tan(arcsin(5/13))

You should be able to calculate this one without a calculator. But how?

PS. The answer should be 5/12

Let

\(\displaystyle \sin^{-1}(\frac{5}{13}) \, = \, \theta\) <<< edited the number

then

\(\displaystyle \tan(\theta) \, = \, \frac{sin(\theta)}{\sqrt{1-\sin^2(\theta)}}\)

 

[Loren]

Re: tan(arcsin(5/13))

Draw your right triangle in quadrant I. Make the angle at the origin resemble an angle whose sine is 5/13. That means the vertical side is about 5 and the hypotenuse is about 13. Use the Pythagorean Thm. to determine the length of the adjacent side. Now you have the necessary information to determine the tangent of the angle.

 

[soroban]

Re: tan(arcsin(5/13))

Hello, DanteQ!

This what Loren said . . .

\(\displaystyle \tan\left[\arcsin\left(\tfrac{5}{13}\right)\right]\)

\(\displaystyle \arcsin\left(\tfrac{5}{13}\right)\text{ is some angle, }\theta\:\hdots\:\text{ and we want: }\tan\theta\)

\(\displaystyle \text{So we have: }\:\theta \:=\:\arcsin\left(\tfrac{5}{13}\right) \quad\Rightarrow\quad \sin\theta \:=\:\frac{5}{13} \:=\:\frac{opp}{hyp}\)

\(\displaystyle \text{So }\theta\text{ is in a right triangle with: }\pp = 5,\;hyp = 13\)

                              *
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               13     *       |
                  *           | 5
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      * - - - - - - - - - - - * 
                  adj

. . \(\displaystyle \text{Using Pythagorus, we find that: }\:adj = 12\)


\(\displaystyle \text{Therefore: }\:\tan\theta \:=\:\frac{opp}{adj} \:=\:\frac{5}{12}\)

 

[DanteQ]

Thanx soroban.

It's completely clear now