# Inverse Trig

• Dec 7th 2007, 07:41 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, 08:00 PM
Soroban
Hello, shogunhd!

Quote:

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

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

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

The problem becomes: .$\displaystyle \sin(2\theta) \;=\;2\sin\theta\cos\theta \:=\:2\left(\frac{15}{17}\right)\left(\frac{8}{17} \right) \;=\;\frac{240}{289}$
• Dec 8th 2007, 04:24 AM
Krizalid
Quote:

Originally Posted by shogunhd
Determine the exact value of

sin (2 arccos (8/17))

Similar to this problem.

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

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