For some reason, i have a problem with these, and only these.
[sin(x)+cos(2x)-1]/[cos(x)-sin(2x)]=tan(x)
and
tan(x/2)=(sin(x))/(1+cos(x)), for which I can get it to tan(x/2)=sin(x)tan(x), but don't know the identity to go further.
I already answered the first one in the other thread.
I would think of the second one with the replacement y = x/2. (I hate 1/2 angle problems!) That transforms the problem to showing:
tan(y) = sin(2y) / [1 + cos(2y)]
The RHS is
sin(2y) = 2sin(y)cos(y)
cos(2y) = 2[cos(y)]^2 - 1
So 1 + cos(2y) = 1 + 2[cos(y)]^2 - 1 = 2[cos(y)]^2
Dividing:
sin(2y) / [1 + cos(2y)] = [2sin(y)cos(y)]/[2{cos(y)}^2] = sin(y)/cos(y) = tan(y).
- Dan