# prove identity help!

• Dec 2nd 2010, 08:23 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, 09:56 AM
elemental
Do you mean:

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

If so, then here's a proof:

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

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

Substitute the new numerator.

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

QED.
• Dec 2nd 2010, 10:31 AM
Quote:

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

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

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...

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

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

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

Divide both sides by 2.

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

Now you only need determine if the denominators are equal...
• Dec 2nd 2010, 10:42 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, 10:44 AM
maths99
thanks for the help
• Dec 2nd 2010, 10:44 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, 10:54 AM
VonNemo19
Quote:

$\text{ArchieMeade}\wedge\text{elemental}$
$\neg\text{ArchieMeade}$
$\rightarrow{elemental}$