# Proving Questions leading to 'Hence' solve for x

• Aug 20th 2011, 07:21 AM
sierratwo1
Proving Questions leading to 'Hence' solve for x
I have a question to ask, would anyone of you explain to me how to solve this?

1. Prove that (1+cosθ+sin2θ)/(1-cos2θ+sin2θ)=cotθ

I'm able to prove it but,,

Then the following part asks me solve the equation for 0≤θ≤2π

1+sin2θ=3cos2θ

• Aug 20th 2011, 08:20 AM
terrorsquid
Re: Proving Questions leading to 'Hence' solve for x
Is the first part supposed to be $(1 + cos(2\theta) +sin(2\theta))$ for the numerator?

If not, can you show me your proof :D

For the second part, wait till someone more experienced comes along; however, while you're waiting, here is my attempt at it...

Let $t = tan(\theta)$

$1 + sin(2\theta) = 3cos(2\theta)$

$\equiv~$ $~\frac{1+t^2}{1+t^2} + \frac{2t}{1+t^2} - \frac{3(1-t^2)}{1+t^2} = 0$

Solve for t...

$tan(\theta) = \frac{1}{2}$

$or$

$tan(\theta) = -1$

For $tan(\theta) = -1:$

$\theta = arctan(-1) = -\frac{\pi}{4}$ therefore $V=\{\frac{3\pi}{4}, \frac{7\pi}{4}\}$ are some solutions. Not 100% on how to solve for 1/2.
• Aug 20th 2011, 09:40 AM
Siron
Re: Proving Questions leading to 'Hence' solve for x
Probably you made a mistake in the numerator because I think the identity has to be $\frac{1+\cos(2x)+\sin(2x)}{1-\cos(2x)+\sin(2x)}=\cot(x)$ (like terrorsquid already noticed).

You can prove it this way (I use $x$ in stead of $\theta$):
$\frac{1+\cos(2x)+\sin(2x)}{1-\cos(2x)+\sin(2x)}=\frac{1+(2\cos^2(x)-1)+2\sin(x)\cos(x)}{1-(1-2\sin^2(x))+2\sin(x)\cos(x)}$
$=\frac{2\cos^2(x)+2\sin(x)\cos(x)}{2\sin^2(x)+2\si n(x) \cos(x)}$
$=\frac{2\cos(x)[\cos(x)+\sin(x)]}{2\sin(x)[\sin(x)+\cos(x)]}=\frac{\cos(x)}{\sin(x)}=\cot(x)$

The method Terrorsquid used is very useful.
For the other solution:
$\tan(\theta)=\frac{1}{2}$ that means:
$\theta=\arctan\left(\frac{1}{2}\right)+k\pi$