Prove that 2sin150*cos315*cos(90-x)/sin(360-x)cos(-240)sin135=2

Answer 1

sin 150 = sin 30 = 1/2
cos 315 = cos 45 = 1/√2
cos(90-x) = sin x
sin (360-x) = -sin x
cos -240 = cos 240 = -cos 60 = -1/2
sin 135 = 1/√2

So you have

2(1/2)(1/√2)(sin x)
-------------------------
(-sin x)(-1/2)(1/√2)

Think you can handle it now?
As I mentioned on your previous post, just reduce everything to reference angles in Quadrant I.

Answer 2

Thank you Is the answer complete?

Answer 3

Of course the answer is complete. I have evaluated every trig function given.

What else can there be? What makes you think that there might be something missing?