# Question 17729

Feb 11, 2017

See the proof below

#### Explanation:

We need

$\sin 2 \alpha = 2 \sin \alpha \cos \alpha$

$\cos 2 \alpha = 1 - 2 {\sin}^{2} \alpha = 2 {\cos}^{2} \alpha - 1$

Therefore,

$\frac{\sin 2 \alpha \cdot \cos \alpha}{\left(1 + \cos 2 \alpha\right) \left(1 + \cos \alpha\right)}$

$= \frac{2 \sin \alpha \cdot \cos \alpha \cdot \cos \alpha}{\left(1 + 1 - 2 {\sin}^{2} \alpha\right) \left(1 + \cos \alpha\right)}$

$= \frac{2 \sin \alpha \left(1 - {\sin}^{2} \alpha\right)}{2 \left(1 - {\sin}^{2} \alpha\right) \left(1 + \cos \alpha\right)}$

$= \sin \frac{\alpha}{1 + \cos \alpha}$

$= \frac{2 \sin \left(\frac{\alpha}{2}\right) \cos \left(\frac{\alpha}{2}\right)}{1 + 2 {\cos}^{2} \alpha - 1}$

$= \frac{2 \sin \left(\frac{\alpha}{2}\right) \cos \left(\frac{\alpha}{2}\right)}{2 {\cos}^{2} \left(\frac{\alpha}{2}\right)}$

$= \sin \frac{\frac{\alpha}{2}}{\cos} \left(\frac{\alpha}{2}\right)$

$= \tan \left(\frac{\alpha}{2}\right)$

$Q E D$

Feb 11, 2017

Proved R.H.S. = L.H.S.

#### Explanation:

We know $\sin 2 a = 2 \sin a \cos a \mathmr{and} \cos 2 a = 1 - 2 {\sin}^{2} a$ and

also cosa = 2cos^(a/2) -1 and sina = 2 sin(a/2)cos(a/2).

Now put all values in the problem,

$\frac{\sin 2 a}{1 + \cos 2 a} . \frac{\cos a}{1 + \cos a}$

= $\frac{2 \sin a \cos a}{1 + 1 - 2 {\sin}^{2} a} . \cos \frac{a}{1 + \cos a}$

= $\frac{2 \sin a {\cos}^{2} a}{\left(2 - 2 {\sin}^{2} a\right) \left(1 + \cos a\right)}$

= [2sinacos^2a]/[2(1-sin^2a)(1+cosa)#

= $\frac{\sin a {\cos}^{2} a}{{\cos}^{2} a . \left(1 + \cos a\right)}$

= $\sin \frac{a}{1 + \cos a}$

= $\frac{2 \sin \left(\frac{a}{2}\right) \cos \left(\frac{a}{2}\right)}{1 + 2 {\cos}^{2} \left(\frac{a}{2}\right) - 1}$

= $\frac{2 \sin \left(\frac{a}{2}\right) \cos \left(\frac{a}{2}\right)}{2 {\cos}^{2} \left(\frac{a}{2}\right)}$

= $\sin \frac{\frac{a}{2}}{\cos} \left(\frac{a}{2}\right)$

= $\tan \left(\frac{a}{2}\right)$

Feb 11, 2017

Change only one side and stop, when it looks like the other side.

#### Explanation:

I will change only the left side.

Multiply the numerator and the denominator:

$\frac{\sin \left(2 a\right) \cos \left(a\right)}{1 + \cos \left(2 a\right) + \cos \left(a\right) + \cos \left(2 a\right) \cos \left(a\right)} = \tan \left(\frac{a}{2}\right)$

Substitute ${\cos}^{2} \left(a\right) - {\sin}^{2} \left(a\right)$ for $\cos \left(2 a\right)$:

$\frac{\sin \left(2 a\right) \cos \left(a\right)}{1 + {\cos}^{2} \left(a\right) - {\sin}^{2} \left(a\right) + \cos \left(a\right) + \left({\cos}^{2} \left(a\right) - {\sin}^{2} \left(a\right)\right) \cos \left(a\right)} = \tan \left(\frac{a}{2}\right)$

Substitute $2 \sin \left(a\right) \cos \left(a\right)$ for $\sin \left(2 a\right)$

$\frac{\left(2 \sin \left(a\right) \cos \left(a\right)\right) \cos \left(a\right)}{1 + {\cos}^{2} \left(a\right) - {\sin}^{2} \left(a\right) + \cos \left(a\right) + \left({\cos}^{2} \left(a\right) - {\sin}^{2} \left(a\right)\right) \cos \left(a\right)} = \tan \left(\frac{a}{2}\right)$

Substitute ${\cos}^{2} \left(a\right)$ for $1 - {\sin}^{2} \left(a\right)$:

$\frac{\left(2 \sin \left(a\right) \cos \left(a\right)\right) \cos \left(a\right)}{{\cos}^{2} \left(a\right) + {\cos}^{2} \left(a\right) + \cos \left(a\right) + \left({\cos}^{2} \left(a\right) - {\sin}^{2} \left(a\right)\right) \cos \left(a\right)} = \tan \left(\frac{a}{2}\right)$

The numerator and denominator have a factor of $\cos \left(a\right)$ in common:

$\frac{2 \sin \left(a\right) \cos \left(a\right)}{\cos \left(a\right) + \cos \left(a\right) + 1 + {\cos}^{2} \left(a\right) - {\sin}^{2} \left(a\right)} = \tan \left(\frac{a}{2}\right)$

Substitute ${\cos}^{2} \left(a\right)$ for $1 - {\sin}^{2} \left(a\right)$:

$\frac{2 \sin \left(a\right) \cos \left(a\right)}{\cos \left(a\right) + \cos \left(a\right) + {\cos}^{2} \left(a\right) + {\cos}^{2} \left(a\right)} = \tan \left(\frac{a}{2}\right)$

Combine like terms in the denominator:

$\frac{2 \sin \left(a\right) \cos \left(a\right)}{2 \cos \left(a\right) + 2 {\cos}^{2} \left(a\right)} = \tan \left(\frac{a}{2}\right)$

There is a common factor of $2 \cos \left(a\right)$ in the numerator and denominator:

$\sin \frac{a}{1 + \cos \left(a\right)} = \tan \left(\frac{a}{2}\right)$

The above is a well know identity; you can substitute $\tan \left(\frac{a}{2}\right)$ into left side, if you like. Q.E.D.