Results 1 to 4 of 4

Math Help - Very Difficult Trig Identity Cot(2X) - Cot(3X) = 1/2SecXCsc(3X)

  1. #1
    Newbie
    Joined
    Apr 2008
    Posts
    1

    Very Difficult Trig Identity Cot(2X) - Cot(3X) = 1/2SecXCsc(3X)

    Cot(2X) - Cot(3X) = 1/2SecXCsc(3X)

    Here are the Identities I (read: You) have to work with

    cscX = 1/sinX
    secX = 1/cosX
    cotX = 1/tanX

    tanX = sinX/cosX
    cotX = cosX/sinX

    sin(x+y) = sinXcosY + cosXsinY
    sin(x-y) = sinXcosY - cosXsinY

    cos(x+y) = cosXcosY - sinXsinY
    cos(x-y) = cosXcosY + sinXsinY

    tan(x+y) = tanX+tanY/1-tanXtanY
    tan(x-y) = tanX-tanY/1+tanXtanY

    cos2X = (cos(X))^2 - (sin(x))^2
    cos2X = 1-2(sin(x))^2
    cos2X = 2(cos(x))^2 - 1

    sin2X = 2sinXcosX

    tan2X = 2tanX/1- (tan(x))^2
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Mar 2008
    Posts
    14
    Hi, I tried this one for a while with no luck. But just from plugging it into a calculator, I'm skeptical that it is in fact an identity.

    cot(2x) - cot(3x) = \frac{1}{2sec(x)csc(3x)}

    Are not equal when put into a calculator. Maybe I plugged the numbers in incorrectly, or maybe I misinterpreted your formula, but as far as I know, the above is not an identity. Could I get verification on this?
    Last edited by mrbuttersworth; April 19th 2008 at 01:52 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,914
    Thanks
    779
    Hello, D4mien!

    \cot(2x) - \cot(3x) \:= \:\frac{1}{2}\sec x\csc 3x

    \cot(2x) - \cot(3x) \;=\; \frac{\cos(2x)}{\sin(2x)} - \frac{\cos(3x)}{\sin(3x)} \;=\;\frac{\overbrace{\sin(3x)\cos(2x) - \sin(2x)\cos(3x)}^{\text{This is }\sin(3x-2x)}}{\sin(2x)\sin(3x)}

    . . . . . = \;\frac{\sin(x)}{\sin(2x)\sin(3x)} \;=\;\frac{\sin(x)}{2\sin(x)\cos(x)\cdot\sin(3x)} \;=\;\frac{1}{2\cos(x)\sin(3x)}

    . . . . . = \;\frac{1}{2}\cdot\frac{1}{\cos(x)}\cdot\frac{1}{\  sin(3x)} \;=\;\frac{1}{2}\sec(x)\csc(3x)

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2008
    Posts
    14
    Oh I see now, I misread the identity.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Difficult-ish Trig Identity
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: December 5th 2011, 09:54 PM
  2. Difficult Trig Identity Question
    Posted in the Trigonometry Forum
    Replies: 11
    Last Post: December 5th 2010, 08:11 AM
  3. Difficult Trig Proof
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: December 9th 2008, 07:38 PM
  4. difficult identity
    Posted in the Trigonometry Forum
    Replies: 8
    Last Post: March 23rd 2008, 02:59 PM
  5. difficult trig identity help plz!
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: November 3rd 2007, 12:21 PM

Search Tags


/mathhelpforum @mathhelpforum