Question

Question #17729

Answer 1

See the proof below

Explanation:

We need

`sin2alpha=2sinalphacosalpha`

`cos2alpha=1-2sin^2alpha=2cos^2alpha-1`

Therefore,

`(sin2alpha*cosalpha)/((1+cos2alpha)(1+cosalpha))`

`=(2sinalpha*cosalpha*cosalpha)/((1+1-2sin^2alpha)(1+cosalpha))`

`=(2sinalpha(1-sin^2alpha))/(2(1-sin^2alpha)(1+cosalpha))`

`=sinalpha/(1+cosalpha)`

`=(2sin(alpha/2)cos(alpha/2))/(1+2cos^2alpha-1)`

`=(2sin(alpha/2)cos(alpha/2))/(2cos^2(alpha/2))`

`=sin(alpha/2)/cos(alpha/2)`

`=tan(alpha/2)`

`QED`

Answer 2

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

Explanation:

We know `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,

`(sin 2a)/(1+cos 2a). (cos a)/(1 + cos a)`

= `(2 sin a cos a)/[1+1 - 2sin^2a] . cosa/[1+cosa]`

= `[2sina cos^2a]/[(2-2sin^2a)(1+cosa)]`

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

= `[sinacos^2a]/[cos^2a.(1+cosa)]`

= `sina/[1+cosa]`

= `[2sin(a/2)cos(a/2)]/[1+2cos^2(a/2)-1]`

= `[2sin(a/2)cos(a/2)]/[2cos^2(a/2)]`

= `sin(a/2)/cos(a/2)`

= `tan(a/2)`

Answer 3

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:

`(sin(2a)cos(a))/(1 + cos(2a) + cos(a) + cos(2a)cos(a)) = tan(a/2)`

Substitute `cos^2(a) - sin^2(a)` for `cos(2a)`:

`(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 `2sin(a)cos(a)` for `sin(2a)`

`((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)` for `1 - sin^2(a)`:

`((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)` in common:

`(2sin(a)cos(a))/(cos(a) + cos(a) + 1 + cos^2(a) - sin^2(a)) = tan(a/2)`

Substitute `cos^2(a)` for `1 - sin^2(a)`:

`(2sin(a)cos(a))/(cos(a) + cos(a) + cos^2(a) + cos^2(a)) = tan(a/2)`

Combine like terms in the denominator:

`(2sin(a)cos(a))/(2cos(a) + 2cos^2(a)) = tan(a/2)`

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

`sin(a)/(1 + cos(a)) = tan(a/2)`

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