sin2x-cotx = -cotxcos2x

Question

Using the various trigonometric identities(i.e. double angle formulas, power reducing formulas, half angle formulas, quotient identities, etc.) verify the identity.

I first added cotx to both sides to get

sin2x = -cotxcos2x+cotx

then I tried dividing both sides by cos2x so i got

tan2x = -cotx + cotx/cos2x

now i don't know where to go from here or if i'm even on the right track

Answer 1

In proving an identity in trig equations,
you have to work one side independently from the other side.
(Using the typical rules of equations creates a paradox: you are using the rules of an equation to prove that it is an equation)

In this case the key thing is to remember the identities for
sin2x and cos2x, have them handy.

Left Side
= 2(sinx)(cosx) - cosx/sinx
= (2sin^2 x)(cosx) - cosx)/sinx
= cosx(2 sin^2 x - 1)/sinx
= (cosx/sinx)(-cos2x
= =cotx(cos2x)
= Right Side

Answer 2

I do not see how Reiny went from:

cosx(2sin^2x-1)/sinx
TO
(cosx/sinx)(-cos2x)

if you break it up you get
(cosx/sinx)((2sin^2x-1)/sinx)

help im doing a take home exam lol

Answer 3

sorry for being 13 years late, but the way that Reiny went from

cosx(2sin^2x-1)/sinx ---> (cosx/sinx)(-cos2x)

was by flipping the sign of the Pythagorean Identity: cos2x=1-2sin^2x.

flipping the equation around, we can see that
cos2x ---> -cos2x
1-2sin^2x ---> -1+2sin^2x

now we substitute in -cos2x into our equation :>, hope this helps anyone who is reading !