# solve trigonometry equation

• May 26th 2008, 07:24 PM
Kuczaj
solve trigonometry equation
Hey, I'm just like... completely zoned out:

sin x = cos ( 2x + 15 )
• May 27th 2008, 05:34 AM
topsquark
Quote:

Originally Posted by Kuczaj
Hey, I'm just like... completely zoned out:

sin x = cos ( 2x + 15 )

Recall that
$sin(x) = cos \left ( x + \frac{\pi}{2} \right )$

Thus
$sin(x) = cos \left ( x + \frac{\pi}{2} \right ) = cos(2x + 15)$

-Dan
• May 27th 2008, 06:28 PM
Kuczaj
Wow, thanks for that that, even you didn't solve for X. Anyways, I found the answer and it's 25.
• May 27th 2008, 07:08 PM
Reckoner
Quote:

Originally Posted by topsquark
Recall that
$sin(x) = cos \left ( x + \frac{\pi}{2} \right )$

Thus
$sin(x) = cos \left ( x + \frac{\pi}{2} \right ) = cos(2x + 15)$

-Dan

This is incorrect. For example, let $x = \frac{\pi}2$. The actual identity should be $\sin x = \cos\left(\frac{\pi}2 - x\right)$. Then we have:

$\cos\left(\frac{\pi}2 - x\right) = \cos(2x + 15)$

Note, however, that there will be infinitely many solutions due to the periodicity of the cosine.

And Kuczaj, did you even try your answer? I get

$\begin{array}{rcr@{.}l}
\sin 25^\circ & \approx & 0&42261826174\\
\cos\left(2\left(25^\circ\right) + 15\right) & \approx & -0&986467363994
\end{array}$