can someone help in prove the identity:
is equal to tan2 θ2sin2 θ1+cos2
in english 2sin sqaured theta divided by 1 + cos2 eqauls to tan squared theta
any help by great
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.
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...
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?
$\displaystyle \text{ArchieMeade}\wedge\text{elemental}$
$\displaystyle \neg\text{ArchieMeade}$
$\displaystyle \rightarrow{elemental}$