[jake beggs]

I solved the first one I posted but the second one still gets me I can get to

2sin(X)cos(X)-(sin(X))/(cos(X))=(sin(X))/(cos(X))cos(2X)

but I don't know which identity to use for cos(2X) in the left side.

the original identity is sin(2X)-tan(X)=tan(X)cos(2X)

[ebaines]

First, you shouldn't start a new thread when asking a follow-up question. Just hit "answer this question" to keep the conversation going.

You have 3 options to choose from when substituting for cos(2x):

1. cos (2x) = cos^2(x) - sin^2(x). This is the basic version which you should memorize.

2. Use the fact that sin^2(x) + cos^2(x) = 1 to either replace the sin^2(x) term with 1 - cos^2(x), or the cos^2(x) term with 1 - sin^2(x). So you end up with either:
2a: cos(2x) = 1 - 2sin^2(x), or
2b: cos(2x) = 2cos^2(x) -1

So now you have three versions of cos(2x). One of these makes for a nice simple substitution that immediately makes the right hand side look like the left and side.

[jake beggs]

I've tried to plug in all three and I still can't get it to match up.

[ebaines]

OK, here goes...

First, the left and side:
sin(2x)-tan(x) = 2sin(x)cos(x) - sin(x)/cos(x)

Now the right hand side:
tan(x)cos(2x) = sin(x)/cos(x) [2cos^2(x)-1]
= 2 sin(x)cos(x) - sin(x)/cos(x)
which is the same as the left hand side.

[Justwantfair]

I thought we didn't do homework problems on this forum.

[jake beggs]

Oh OK thanks I'm really not that good at math.