# prove identity help!

• Dec 2nd 2010, 07:23 AM
maths99
prove identity help!
can someone help in prove the identity:

2sin2 θ
1+cos2
is equal to tan2 θ

in english 2sin sqaured theta divided by 1 + cos2 eqauls to tan squared theta

any help by great
• Dec 2nd 2010, 08:56 AM
elemental
Do you mean:

$\displaystyle {tan}^2 x = \frac{2{sin}^2 x}{1+cos 2x}$ ?

If so, then here's a proof:

$\displaystyle {tan}^2 x = \frac{1 - cos 2x}{1 + cos 2x}$

$\displaystyle 1 - cos 2x = 1 - (1 - 2{sin}^2 x) = 2{sin}^2 x$

Substitute the new numerator.

$\displaystyle {tan}^2 x = \frac{2{sin}^2 x}{1 + cos 2x}$

QED.
• Dec 2nd 2010, 09:31 AM
Quote:

Originally Posted by maths99
can someone help in prove the identity:

$\displaystyle \displaystyle\frac{2sin^2\theta}{1+cos2\theta}=tan ^2\theta$

In english.... 2sin squared theta divided by 1 + cos2theta equals tan squared theta.

Any help would be great.

The left-hand-side is expressed in terms of sine and cosine.
We can also express the right-hand side in terms of sine and cosine...

$\displaystyle \displaystyle\frac{2sin^2\theta}{1+cos2\theta}=\le ft(\frac{sin\theta}{cos\theta}\right)^2=\frac{sin^ 2\theta}{cos^2\theta}$ ?

$\displaystyle sin^2\theta$ is common to both sides, so divide both sides by $\displaystyle sin^2\theta$

$\displaystyle \displaystyle\frac{2}{1+cos2\theta}=\frac{1}{cos^2 \theta}$ ?

Divide both sides by 2.

$\displaystyle \displaystyle\frac{1}{1+cos2\theta}=\frac{1}{2cos^ 2\theta}$ ?

Now you only need determine if the denominators are equal...
• Dec 2nd 2010, 09:42 AM
VonNemo19
Why do people just go and solve the problem for the newbie? This is his first post and who knows...Maybe this problem is part of a quiz for one of his online assignments....

Don't be too eager to please guys. That's all I'm saying.

Peace
• Dec 2nd 2010, 09:44 AM
maths99
thanks for the help
• Dec 2nd 2010, 09:44 AM
Quote:

Originally Posted by VonNemo19
Why do people just go and solve the problem for the newbie? This is his first post and who knows...Maybe this problem is part of a quiz for one of his online assignments....

Don't be too eager to please guys. That's all I'm saying.

Peace

To whom?
• Dec 2nd 2010, 09:54 AM
VonNemo19
Quote:

To whom?

Archie Meade. I've watched you for a while and you are genuinely here as a helper. Your reply worked as a guide for the newb. So, rest assured, that I wasn't directing my post to you. Let's have a logic session, shall we?

$\displaystyle \text{ArchieMeade}\wedge\text{elemental}$

$\displaystyle \neg\text{ArchieMeade}$

$\displaystyle \rightarrow{elemental}$
• Dec 2nd 2010, 10:00 AM
maths99
just to let you know it wasnt part of an assignment or on line test. just im doing lots to gain practise in my civil engineer course, and was stuck on this. thanks for your help peace