# Inverse Trig

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• Dec 7th 2007, 08:41 PM
shogunhd
Inverse Trig
The problem is

Determine the exact value of

sin (2 arccos (8/17))

The triangle for sin (arccos x) = sqroot(1-x^2). I have no idea how the answer is (240/289).
• Dec 7th 2007, 09:00 PM
Soroban
Hello, shogunhd!

Quote:

Determine the exact value of: . $\sin\left[2\arccos\left(\frac{8}{17}\right)\right]$

Let $\theta \:=\:\arccos\left(\frac{8}{17}\right)\quad\Rightar row\quad\cos\theta \:=\:\frac{8}{17} \:=\:\frac{adj}{hyp}$

So, $\theta$ is in a right triangle with: . $adj = 8,\;hyp = 17$
. . Using Pythagorus, we find that: . $opp = 15$
Hence: . $\sin\theta = \frac{15}{17},\;\cos\theta = \frac{8}{17}$

The problem becomes: . $\sin(2\theta) \;=\;2\sin\theta\cos\theta \:=\:2\left(\frac{15}{17}\right)\left(\frac{8}{17} \right) \;=\;\frac{240}{289}$
• Dec 8th 2007, 05:24 AM
Krizalid
Quote:

Originally Posted by shogunhd
Determine the exact value of

sin (2 arccos (8/17))

Similar to this problem.

Since $\sin (\arccos \varphi ) = \sqrt {1 - \varphi ^2 } \implies \sin (2\arccos \varphi ) = 2\varphi \sqrt {1 - \varphi ^2 } .$

Plug $\varphi=\frac8{17}$ and you're done.