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Math Help - Proving trigonometric equations:

  1. #1
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    Can someone help me prove the following identities:

    (1+cos^2x/1+tan^2x) tanx=cotx

    e) secx-cosx=sinxtanx f)sinxcosx/sinx=1+tanx\tanx g)cosx/sinx-sinx/secx=cos^2xcotx



    h)cotx-tanx/cotx+tanx=2cos^2x-1 i)3sin^2x+4cos2x=3+cos^2x j)9sec^2x-5tan^2x=5+4sec^2x



    k) (2cos^2x-1)^2/sin^4x-cos^4x=2sin^2x-1 L)(sin^2x/1-cosx)-1=cosx



    m)1+cos2x/sin2x=cotx



    Okay, here are these ones for now... the rest are not the same so I want to have these down first Thank you so much for helping by the way! I wish I was a math wizz too!
    Last edited by mr fantastic; April 21st 2009 at 06:13 PM. Reason: Removed the 100 pt sized text, Merged posts
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  2. #2
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    Trig Identities

    Hello etheen
    Quote Originally Posted by etheen View Post
    Can someone help me prove the following identities:

    (1+cos^2x/1+tan^2x) tanx=cotx
    This is not true. Try it when x = \tfrac{\pi}{4} for example.

    e) secx-cosx=sinxtanx
    \sec x -\cos x = \frac{1}{\cos x}-\cos x= \frac{1-\cos^2x}{\cos x}

    Can you do it from here? (Use \sin^2x + \cos^2x =1)

    f)sinxcosx/sinx=1+tanx\tanx
    Again, this is not true. The LHS is \cos x and the RHS is \cot x + 1, and they are not identical.

    g)cosx/sinx-sinx/secx=cos^2xcotx
    \frac{\cos x}{\sin x}-\frac{\sin x}{\sec x}= \frac{\cos x}{\sin x}-\sin x\cos x

    =\frac{\cos x - \sin^2 x \cos x}{\sin x}

    Factorise the numerator and then use \sin^2x + \cos^2x =1.

    Hopefully, you've got the idea now, and can do some of the remaining questions yourself. Show us your attempts if you want more help.

    Grandad
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  3. #3
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    okay for e) um....1-cos^2x =sin^2x and cosx=cottsint ...okay I'm stuck now:|



    by the way what do you mean that f) is not true, grandad?
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  4. #4
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    Quote Originally Posted by etheen View Post
    okay for e) um....1-cos^2x =sin^2x and cosx=cottsint ...okay I'm stuck now:|


    Hi etheen,

    Here's the finish to (e). Work with the LHS

    \sec x -\cos x={\color{red}\sin x \tan x}

    \frac{1}{\cos x}-\cos x=

    \frac{1-\cos^2 x}{\cos x}=

    \frac{\sin^2 x}{\cos x}=

    \sin x \left(\frac{\sin x}{\cos x}\right)=

    {\color{red}\sin x \tan x}
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  5. #5
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    Hm, I see alittle!

    can youplease exaplin exactly what explains when we do? 1/cosx-cosx?

    you do LCD so we had to do 1 times cosx to get the denominator to be the same and once it was the same, we did cosx times cosx because we did that to the bototm right?
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  6. #6
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    Trig Identities

    Hello etheen
    Quote Originally Posted by etheen View Post
    by the way what do you mean that f) is not true, grandad?
    You are asking us to prove that
    f)sinxcosx/sinx=1+tanx\tanx
    is true for all values of x - and it simply isn't!

    Are you sure you have written the question down correctly?

    Grandad

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  7. #7
    A riddle wrapped in an enigma
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    Quote Originally Posted by masters View Post
    Hi etheen,

    Here's the finish to (e). Work with the LHS

    \sec x -\cos x={\color{red}\sin x \tan x}

    \frac{1}{\cos x}-\cos x=

    \frac{1-\cos^2 x}{\cos x}=

    \frac{\sin^2 x}{\cos x}=

    \sin x \left(\frac{\sin x}{\cos x}\right)=

    {\color{red}\sin x \tan x}
    Quote Originally Posted by etheen View Post
    Hm, I see alittle!

    can youplease exaplin exactly what explains when we do? 1/cosx-cosx?

    you do LCD so we had to do 1 times cosx to get the denominator to be the same and once it was the same, we did cosx times cosx because we did that to the bototm right?
    Does this help?

    \frac{1}{\cos x}-\cos x=\frac{1}{\cos x}-\frac{\cos^2 x}{\cos x}=\frac{1-\cos^2 x}{\cos x}
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