# Weird basic question

• Jan 28th 2010, 03:04 AM
Punch
Weird basic question
$\displaystyle sin(x+50)=2cosx$

It's weird that this question doesn't have any special angle or any sort, I don't know how to solve this question with $\displaystyle 50^\circ$ as the angle!!
• Jan 28th 2010, 03:23 AM
CaptainBlack
Quote:

Originally Posted by Punch
$\displaystyle sin(x+50)=2cosx$

It's weird that this question doesn't have any special angle or any sort, I don't know how to solve this question with $\displaystyle 50^\circ$ as the angle!!

$\displaystyle \sin(x+50)=\sin(x)\cos(50)+\cos(x)\sin(50)=2\cos(x )$

so:

$\displaystyle \tan(x)=\frac{2-\sin(50)}{\cos(50)}$

CB
• Jan 28th 2010, 03:26 AM
Punch
Quote:

Originally Posted by CaptainBlack
$\displaystyle \sin(x+50)=\sin(x)\cos(50)+\cos(x)\sin(50)=2\cos(x )$

so:

$\displaystyle \tan(x)=\frac{2-\sin(50)}{\cos(50)}$

CB

Thanks but how do I solve for x with this? Also, how did you relate tan(x) with this?
• Jan 28th 2010, 03:28 AM
CaptainBlack
Quote:

Originally Posted by Punch
Thanks but how do I solve for x with this?

arctan

CB
• Jan 28th 2010, 03:29 AM
Punch
Quote:

Originally Posted by CaptainBlack
arctan

CB

SOrry but what is arctan?
• Jan 28th 2010, 04:03 AM
HallsofIvy
It is the inverse function to tangent. tan(arctan(x))= x and arctan(tan(x))= x as long as x is between $\displaystyle -pi/2$ and $\displaystyle \pi/2$. Also called $\displaystyle tan^{-1}(x)$ which must not be confused with $\displaystyle \frac{1}{tan(x)}= cot(x)$.