SOLUTION: prove 2cot2xcosx=cscx-2sinx I tried starting on the left. My work is 2(cos2x/sin2x)*cosx= 2(cos^2(x)-sin^2(x)/2sin(x)cos(x))*cos(x)= (cos^2(x)-sin^2(x)/sinxcosx)*cos(x)= an

Algebra ->  Trigonometry-basics -> SOLUTION: prove 2cot2xcosx=cscx-2sinx I tried starting on the left. My work is 2(cos2x/sin2x)*cosx= 2(cos^2(x)-sin^2(x)/2sin(x)cos(x))*cos(x)= (cos^2(x)-sin^2(x)/sinxcosx)*cos(x)= an      Log On


   

Question 997682: prove 2cot2xcosx=cscx-2sinx
I tried starting on the left. My work is
2(cos2x/sin2x)*cosx=
2(cos^2(x)-sin^2(x)/2sin(x)cos(x))*cos(x)=
(cos^2(x)-sin^2(x)/sinxcosx)*cos(x)=
and that's where I get stuck. If I multiply by that cos(x) then I come up with
cos^3(x)-sin^2(x)cos(x)/sinxcosx. Then cross out the cos(x)? but that doesn't help me either. Please help
Answer by addingup(3677) About Me  (Show Source):
You can put this solution on YOUR website!
Prove 2cot2xcosx=cscx-2sinx
Rewrite like this:
(cos(2x)/sin(2x))2cos(x)=? 1/sinx-2sin(x)
(2cos(x)cos(2x)/sin(2x))=? 1-2sin(x)^2/sin(x)
cross multiply and you get:
2cos(x)1-2sin(x)^2sin(x)=? sin(2x)(1-2sin(x)^2)
2cos(x)sin(x)-4cos(x)sin(x)^3=? sin(2x)(1-2sin(x)^2)
2cos(x)sin(x)-4cos(x)sin(x)^3=? 2cos(x)sin(x)-4cos(x)sin(x)^3
True, identity verified
NOTE: the question mark goes over the equal sign until the identity is proven true or false.