Results 1 to 3 of 3

Math Help - Trigonometry Identities

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    10

    Trigonometry Identities

    Hey Im having trouble with these identities i was wondering if anyone else could figure them out??


    cosX/cosX-sinX = 1/1-tanX


    2tanX/tan(squared) -1 = 2/tanX-cotX


    tanXsinX/tanX-sinX = sinX/1-cosX


    (tanX + cotX)(secX + cosX) = secXtanX
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,749
    Thanks
    649
    Hello, Giggly2!

    There is a typo in the last one . . .


    \frac{\cos x}{\cos x-\sin x} \:=\:\frac{1}{1-\tan x}
    On the left side, divide top and bottom by \cos x

    . . \frac{\dfrac{\cos x}{\cos x}}{\dfrac{\cos x}{\cos x} - \dfrac{\sin x}{\cos x}} \;=\;\frac{1}{1 - \tan x}




    \frac{2\tan x}{\tan^2\!x -1} \:=\: \frac{2}{\tan x-\cot x}
    On the left, divide top and bottom by \tan x

    . . \frac{\dfrac{2\tan x}{\tan x}} {\dfrac{\tan^2\!x}{\tan x} - \dfrac{1}{\tan x}} \;=\;\frac{2}{\tan x - \cot x}




    \frac{\tan x\sin x}{\tan x-\sin x} \:=\: \frac{\sin x}{1-\cos x}

    We have: . \frac{\dfrac{\sin x}{\cos x}\cdot\sin x}{\dfrac{\sin x}{\cos x} - \sin x}

    Multiply by \frac{\cos x}{\cos x}\!:\;\;\frac{\cos x\left(\dfrac{\sin^2\!x}{\cos x}\right)}{\cos x\left(\dfrac{\sin x}{\cos x} - \sin x\right)} \;=\;\frac{\sin^2\!x}{\sin x -\sin x\cos x}

    Factor and reduce: . \frac{\sin^2\!x}{\sin x(1 - \cos x)} \;=\;\frac{\sin x}{1 - \cos x}




    (\tan x + \cot x)(\sec x \;{\color{red}-}\;  \cos x) \:=\: \sec x\tan x

    The left side is: . \left(\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}\right)\cdot\left(\frac{1}{\cos x} - \cos x\right)

    . . = \;\frac{\overbrace{\sin^2\!x + \cos^2\!x}^{\text{This is 1}}}{\sin x\cos x} \cdot \frac{\overbrace{1-\cos^2\!x}^{\text{This is }\sin^2\!x}}{\cos x} \;=\;\frac{1}{\sin x\cos x}\cdot\frac{\sin^2\!x}{\cos x}

    . . = \;\frac{\sin x}{\cos^2\!x} \;=\; \frac{1}{\cos x}\cdot\frac{\sin x}{\cos x} \;=\;\sec x \tan x

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2009
    Posts
    10

    Thumbs up

    Thank you soooo much, that was so helpful
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trigonometry Identities? Please Help.
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: December 7th 2010, 08:35 AM
  2. Trigonometry help with identities
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: July 10th 2010, 12:40 PM
  3. Trigonometry Identities
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: April 30th 2009, 03:39 AM
  4. Trigonometry Help (Identities)
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: January 6th 2009, 05:46 AM
  5. Trigonometry Identities
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: June 18th 2008, 02:25 PM

Search Tags


/mathhelpforum @mathhelpforum