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.

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- September 20th 2006, 04:31 PMdnlstffrdIdentities
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. - September 20th 2006, 04:51 PMtopsquark
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