Question

3(sin 2x − cos 2x) write expression only in sine.

I got sin=-(√3)/3 cos= (√3)/3

Does anyone know what to do next?

Answer 1

Is it:

\[\sin^2x\]

Or

\[\sin2x?\]

Answer 2

sin2x

Answer 3

And how do you get that values for sin and cos??

Answer 4

using the Pythagorean theorem

Answer 5

Can you show me how you used here?

Answer 6

wait i just figured it out =-= i did it wrong lol

Answer 7

\[3(\sin2x - (1 - 2\sin^2x)) \implies 3(\sin2x - 1 + 2\sin^2x) \implies 6\sin^2x +3\sin2x - 3\]

Answer 8

I have used this:

\[\huge \color{green}{\cos2x = 1 - 2\sin^2x}\]

Answer 9

oh ok thanks :)