SOLUTION: Prove the following identity. cos2x/sinx = cscx-2sinx How do I go about this problem? I'm stuck on where to start and any help would be greatly appreciated.

Algebra ->  Trigonometry-basics -> SOLUTION: Prove the following identity. cos2x/sinx = cscx-2sinx How do I go about this problem? I'm stuck on where to start and any help would be greatly appreciated.       Log On


   

Question 1126914: Prove the following identity.
cos2x/sinx = cscx-2sinx
How do I go about this problem? I'm stuck on where to start and any help would be greatly appreciated.
Found 2 solutions by MathLover1, Boreal:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

cos%282x%29%2Fsin%28x%29+=+csc%28x%29-2sin%28x%29

start with left side and arrive to right side

cos%282x%29%2Fsin%28x%29 .....since cos%282x%29=cos%5E2%28x%29+-+sin%5E2%28x%29, we have

=%28cos%5E2%28x%29+-+sin%5E2%28x%29%29%2Fsin%28x%29

=cos%5E2%28x%29%2Fsin%28x%29+-+sin%5E2%28x%29%2Fsin%28x%29

=cos%5E2%28x%29%2Fsin%28x%29+-+sin%28x%29...............since cos%5E2%28x%29=1+-+sin%5E2%28x%29, we have

=%281+-+sin%5E2%28x%29%29%2Fsin%28x%29+-+sin%28x%29

=1%2Fsin%28x%29++-+sin%5E2%28x%29%2Fsin%28x%29+-+sin%28x%29

=1%2Fsin%28x%29++-+sin%28x%29+-+sin%28x%29.............since 1%2Fsin%28x%29+=csc%28x%29, we have
=csc%28x%29-2sin%28x%29 -> we have arrived to right side



Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
use cos^2 x- sin^2 x for cos2x
and cos^2 x=(1-sin^2 x)
so the left side becomes (1/sin x) (1-2 sin^2 x)
this is (1/sin x)(-2 sin x), dividing both terms by sin x.
or csc x-2 sin x