1. ## Trig equation

Hi

$\displaystyle cos(\frac{\pi}{4}+\theta)=\frac{1}{2}\cdot cos(\frac{\pi}{4}-\theta)$

How do I solve these now again?

Thx!

2. Originally Posted by Twig
Hi

$\displaystyle cos(\frac{\pi}{4}+\theta)=\frac{1}{2}\cdot cos(\frac{\pi}{4}-\theta)$

How do I solve these now again?

Thx!
use this

$\displaystyle cos(A+B)=cosAcosB-sinAsinB$
and

$\displaystyle cos(A-B)=cosAcosB+sinAsinB$

3. $\displaystyle cos(\frac{\pi}{4}+\theta)=\frac{1}{2}\cdot cos(\frac{\pi}{4}-\theta)$

Compound Angle Formula

$\displaystyle cos\frac{\pi}{4}cos\theta-sin\frac{\pi}{4}sin\theta=\frac{1}{2}(cos\frac{\pi }{4}cos\theta+sin\frac{\pi}{4}sin\theta$

$\displaystyle \frac{cos\theta}{\sqrt{2}}-\frac{sin\theta}{\sqrt{2}}=\frac{1}{2}( \frac{cos\theta}{\sqrt{2}}+\frac{sin\theta}{\sqrt{ 2}})$

$\displaystyle *\sqrt{2}: cos\theta-sin\theta=\frac{1}{2}(cos\theta+sin\theta)$
$\displaystyle *2: 2cos\theta-2sin\theta=cos\theta+sin\theta$
$\displaystyle cos\theta-3sin\theta=0$
$\displaystyle cos\theta=3sin\theta$
$\displaystyle cos^2\theta=9sin^2\theta$
$\displaystyle 1=10sin^2\theta$
$\displaystyle \sqrt{\frac{1}{10}}=sin\theta$

Solve from there. I'm sleepy

4. ## thx

oh of course!

Thanks alot!

5. Hi I-Think

I belive the easiest approach after writing out the compound angle formulas are: $\displaystyle tan(\theta)=\frac{1}{3}$

6. Originally Posted by I-Think
$\displaystyle cos(\frac{\pi}{4}+\theta)=\frac{1}{2}\cdot cos(\frac{\pi}{4}-\theta)$

Compound Angle Formula

$\displaystyle cos\frac{\pi}{4}cos\theta-sin\frac{\pi}{4}sin\theta=\frac{1}{2}(cos\frac{\pi }{4}cos\theta+sin\frac{\pi}{4}sin\theta$

$\displaystyle \frac{cos\theta}{\sqrt{2}}-\frac{sin\theta}{\sqrt{2}}=\frac{1}{2}( \frac{cos\theta}{\sqrt{2}}+\frac{sin\theta}{\sqrt{ 2}})$

$\displaystyle *\sqrt{2}: cos\theta-sin\theta=\frac{1}{2}(cos\theta+sin\theta)$

$\displaystyle *2: 2cos\theta-2sin\theta=cos\theta+sin\theta$

$\displaystyle cos\theta-3sin\theta=0$

$\displaystyle cos\theta=3sin\theta$

$\displaystyle cos^2\theta=9sin^2\theta$

$\displaystyle 1=10sin^2\theta$

$\displaystyle \sqrt{\frac{1}{10}}=sin\theta$

Solve from there. I'm sleepy
make spaces between [tex][/ math] and the [tex][/ math] that makes it easy to read

7. A slightly different approach :
Originally Posted by Twig
$\displaystyle cos(\frac{\pi}{4}+\theta)=\frac{1}{2}\cdot cos(\frac{\pi}{4}-\theta)$

How do I solve these now again?
As $\displaystyle \cos\left(\frac{\pi}{4}-\theta\right)=\sin\left(\frac{\pi}{4}+\theta\right )$, the above equation can be written $\displaystyle \tan\left(\frac{\pi}{4}+\theta\right)=2$. Expanding the left hand side gives us
$\displaystyle \frac{\tan\frac{\pi}{4}+\tan\theta}{1-\tan\frac{\pi}{4}\cdot \tan\theta}=2\Longleftrightarrow \frac{1+\tan\theta}{1-1\cdot \tan\theta}=2\Longleftrightarrow 1+\tan \theta=2-2\tan \theta.$
Thus $\displaystyle \tan \theta=1/3$...