Results 1 to 10 of 10

Math Help - Trigo Identities....

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    8

    Trigo Identities....

    2tanx/1-tan^2x = 2/cotx-tanx


    csc^2x/cox^2x = sec^2x + csc^2


    (sinx+cosx)(tanx+cotx) = secx+cscx

    and


    1-sinx/1+sinx = (secx - tanx)^2

    Thanks alot
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2009
    Posts
    806
    Thanks
    4

    Trig.

    2tanx/1-tan^2x = 2/cotx-tanx
    2tanx/1-tan^2x = tan2x = sin2x/cos2x
    = 2sinx*cosx/ (cos^2x - sin^2x). Divide numerator and denomenator by sinx*cosx, you get
    = 2/(cotx - tanx)



    csc^2x/cox^2x = sec^2x + csc^2
    sec^2x + csc^2
    = 1/cos^2x + 1/sin^2x
    = (sin^2x + cos^2x)/ sin^2x*cos^2x
    = 1/(
    sin^2x*cos^2x)
    = csc^2x/cos^2x
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2009
    Posts
    8
    thanks, how about the 3rd problem?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Sep 2009
    Posts
    8
    Quote Originally Posted by sa-ri-ga-ma View Post
    2tanx/1-tan^2x = 2/cotx-tanx
    2tanx/1-tan^2x = tan2x = sin2x/cos2x
    = 2sinx*cosx/ (cos^2x - sin^2x). Divide numerator and denomenator
    what happened in the 2sinx*cosx? why did it became like did? i thought tan= sin/cos and not sin*cos? please clarify
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,751
    Thanks
    651
    Hello, cannon!

    \frac{1-\sin x}{1+\sin x} \:=\: (\sec x - \tan x)^2
    Multiply the left side by: . \frac{1-\sin x}{1-\sin x}

    . . \frac{1-\sin x}{1 + \sin x}\cdot{\color{blue}\frac{1-\sin x}{1-\sin x}} \;\;=\;\;\frac{(1-\sin x)^2}{1-\sin^2\!x} \;\;=\;\; \frac{(1-\sin x)^2}{\cos^2\!x} . = \;\left(\frac{1-\sin x}{\cos x}\right)^2 \;=\;\left(\frac{1}{\cos x} - \frac{\sin x}{\cos x}\right)^2 \;=\;(\sec x - \tan x)^2

    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello cannon
    Quote Originally Posted by cannon View Post
    2tanx/1-tan^2x = 2/cotx-tanx
    Quote Originally Posted by cannon View Post
    what happened in the 2sinx*cosx? why did it became like did? i thought tan= sin/cos and not sin*cos? please clarify
    \frac{2\sin x\cos x}{\cos^2x-\sin^2x}=\frac{\dfrac{2\sin x\cos x}{\color{red}\sin x \cos x}}{\dfrac{\cos^2x-\sin^2x}{\color{red}\sin x\cos x}}

    =\frac{2}{\dfrac{\cos^2x}{\sin x\cos x}-\dfrac{\sin^2x}{\sin x\cos x}}=\frac{2}{\dfrac{\cos x}{\sin x}-\dfrac{\sin x}{\cos x}}=\frac{2}{\cot x - \tan x}
    Quote Originally Posted by cannon View Post
    thanks, how about the 3rd problem?
    (\sin x +\cos x)(\tan x + \cot x) = (\sin x +\cos x)\Big(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\Big)

    = (\sin x +\cos x)\Big(\frac{\sin^2 x+\cos^2 x}{\cos x\sin x}\Big)

    =\frac{\sin x+\cos x} {\cos x\sin x}

    =\sec x + \csc x

    Grandad
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Sep 2009
    Posts
    8
    how did you get cos^2x-sin^2x?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Sep 2009
    Posts
    8
    and for the last 2 problems hehe...

    cosx/1+sinx = 1-sinx/cosx



    sinx/1-cotx + cosx/1-tanxx = sinx +cosx
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello cannon
    Quote Originally Posted by cannon View Post
    how did you get cos^2x-sin^2x?
    I was picking up on the reply by sa-ri-ga-ma:
    Quote Originally Posted by sa-ri-ga-ma View Post
    2tanx/1-tan^2x = 2/cotx-tanx
    2tanx/1-tan^2x = tan2x = sin2x/cos2x
    = 2sinx*cosx/ (cos^2x - sin^2x). Divide numerator and denomenator by sinx*cosx, you get
    = 2/(cotx - tanx)
    and your query
    Quote Originally Posted by cannon View Post
    what happened in the 2sinx*cosx? why did it became like did? i thought tan= sin/cos and not sin*cos? please clarify
    As far as your next post is concerned:
    Quote Originally Posted by cannon View Post
    and for the last 2 problems hehe...

    cosx/1+sinx = 1-sinx/cosx



    sinx/1-cotx + cosx/1-tanxx = sinx +cosx
    this thread is now too complicated! Please have another attempt at these questions - they are very similar to the ones we've shown you already. If you still can't do them, re-post them and show us your attempts in a new thread. (See Rule #14).

    Grandad
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Sep 2009
    Posts
    8
    I didn't get what you mean in your first 2 statements. Anyways, I already solved the last problem, sorry for being lazy lol and thanks a lot!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trigo Identities!
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: February 19th 2010, 03:06 AM
  2. Trigo identities
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: March 23rd 2009, 09:15 AM
  3. Trigo identities--Pls Help!!
    Posted in the Math Topics Forum
    Replies: 14
    Last Post: November 29th 2008, 02:56 AM
  4. trigo identities
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: July 17th 2007, 04:38 AM
  5. proving trigo identities
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: June 28th 2007, 12:38 PM

Search Tags


/mathhelpforum @mathhelpforum