math

sin2x-cotx = -cotxcos2x

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

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asked by sam
  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

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    posted by Reiny
  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

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    posted by Jacob

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