Question #17729

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Question #17729

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Answer 1 · 2017-02-11T16:01:08

See the proof below

Explanation:

We need

$sin 2 α = 2 sin α cos α$ $sin2alpha=2sinalphacosalpha$

$cos 2 α = 1 - 2 {sin}^{2} α = 2 {cos}^{2} α - 1$ $cos2alpha=1-2sin^2alpha=2cos^2alpha-1$

Therefore,

$\frac{sin 2 α \cdot cos α}{(1 + cos 2 α) (1 + cos α)}$ $(sin2alpha*cosalpha)/((1+cos2alpha)(1+cosalpha))$

$= \frac{2 sin α \cdot cos α \cdot cos α}{(1 + 1 - 2 {sin}^{2} α) (1 + cos α)}$ $=(2sinalpha*cosalpha*cosalpha)/((1+1-2sin^2alpha)(1+cosalpha))$

$= \frac{2 sin α (1 - {sin}^{2} α)}{2 (1 - {sin}^{2} α) (1 + cos α)}$ $=(2sinalpha(1-sin^2alpha))/(2(1-sin^2alpha)(1+cosalpha))$

$= \frac{sin α}{1 + cos α}$ $=sinalpha/(1+cosalpha)$

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

$= \frac{2 sin (\frac{α}{2}) cos (\frac{α}{2})}{2 {cos}^{2} (\frac{α}{2})}$ $=(2sin(alpha/2)cos(alpha/2))/(2cos^2(alpha/2))$

$= \frac{sin (\frac{α}{2})}{cos (\frac{α}{2})}$ $=sin(alpha/2)/cos(alpha/2)$

$= tan (\frac{α}{2})$ $=tan(alpha/2)$

$Q E D$ $QED$

Answer 2 · 2017-02-11T16:15:33

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

Explanation:

We know $sin 2 a = 2 sin a cos a and cos 2 a = 1 - 2 {sin}^{2} a$ $sin 2a = 2 sina cosa and cos2a = 1 - 2 sin^2a$ and

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

Now put all values in the problem,

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

= $\frac{2 sin a cos a}{1 + 1 - 2 {sin}^{2} a} . \frac{cos a}{1 + cos a}$ $(2 sin a cos a)/[1+1 - 2sin^2a] . cosa/[1+cosa]$

= $\frac{2 sin a {cos}^{2} a}{(2 - 2 {sin}^{2} a) (1 + cos a)}$ $[2sina cos^2a]/[(2-2sin^2a)(1+cosa)]$

= $\frac{2 sin a {cos}^{2} a}{2 (1 - {sin}^{2} a) (1 + cos a)}$ $[2sinacos^2a]/[2(1-sin^2a)(1+cosa)$

= $\frac{sin a {cos}^{2} a}{{cos}^{2} a . (1 + cos a)}$ $[sinacos^2a]/[cos^2a.(1+cosa)]$

= $\frac{sin a}{1 + cos a}$ $sina/[1+cosa]$

= $\frac{2 sin (\frac{a}{2}) cos (\frac{a}{2})}{1 + 2 {cos}^{2} (\frac{a}{2}) - 1}$ $[2sin(a/2)cos(a/2)]/[1+2cos^2(a/2)-1]$

= $\frac{2 sin (\frac{a}{2}) cos (\frac{a}{2})}{2 {cos}^{2} (\frac{a}{2})}$ $[2sin(a/2)cos(a/2)]/[2cos^2(a/2)]$

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

= $tan (\frac{a}{2})$ $tan(a/2)$

Answer 3 · 2017-02-11T16:19:54

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:

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

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

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

Substitute $2 sin (a) cos (a)$ $2sin(a)cos(a)$ for $sin (2 a)$ $sin(2a)$

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

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

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

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

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

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

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

Combine like terms in the denominator:

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

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

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

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

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Question #17729

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