1. ## Trig equation

Hi

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

How do I solve these now again?

Thx!

2. Originally Posted by Twig
Hi

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

How do I solve these now again?

Thx!
use this

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

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

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

Compound Angle Formula

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

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

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

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: $tan(\theta)=\frac{1}{3}$

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

Compound Angle Formula

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

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

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

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

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

$cos\theta=3sin\theta$

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

$1=10sin^2\theta$

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

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

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