solve trigonometry equation

**Kuczaj** · May 26th 2008, 06:24 PM

Hey, I'm just like... completely zoned out:

sin x = cos ( 2x + 15 )

**topsquark** · May 27th 2008, 04:34 AM

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

**Kuczaj** · May 27th 2008, 05:28 PM

Wow, thanks for that that, even you didn't solve for X. Anyways, I found the answer and it's 25.

**Reckoner** · May 27th 2008, 06:08 PM

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

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

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