help solving trig equations for x

[jetguy35]

My problem is tan^-1 x = sin^-1 24/25

It says to solve for x algebraically.

I am pretty lost but here is what I start doing and then get lost.

tan(tan^-1 x) = tan(sin^-1 24/25)
x= tan(sin(sin^-1 24/25)
x=?

Is this a completely wrong start or attempt? I am not quite sure where to go from here. Please help.

Thanks!

 

[soroban]

Hello, jetguy35!

\(\displaystyle \text{Solve: }\;\tan^{-1} x \:=\: \sin^{-1}\left(\frac{24}{25}\right)\)

\(\displaystyle \text{Let }\,\theta \:=\:\sin^{-1}\left(\frac{24}{25}\right) \quad\Rightarrow\quad \sin\theta \:=\:\frac{24}{25} \:=\:\frac{opp}{hyp}\)


\(\displaystyle \theta\text{ is in a right triangle with: }\,opp = 24,\;hyp = 25\)

\(\displaystyle \text{Pythagorus tells us that: }\:adj = 7\)

\(\displaystyle \text{Hence: }\:\tan\theta \:=\:\frac{opp}{adj} \:=\:\frac{24}{7} \quad\Rightarrow\quad \theta \:=\:\tan^{-1}\left(\frac{24}{7}\right)\)

. . \(\displaystyle \text{That is: }\:\sin^{-1}\left(\frac{24}{25}\right) \;=\;\tan^{-1}\left(\frac{24}{7}\right)\)


\(\displaystyle \text{The equation becomes: }\;\tan^{-1} x \;=\;\tan^{-1}\left(\frac{24}{7}\right)\)

. . \(\displaystyle \text{Therefore: }\;x \;=\;\frac{24}{7}\)

 

[jetguy35]

Thank you. It makes more sense now. I will give it a shot on some more problems!