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Math Help - trig differentiating

  1. #1
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    [SOLVED] trig differentiating

    d/dx (cosx+sinx/cosx-sinx)

    I'm having a bit of trouble with trying to prove LHS = RHS (or the other way around for that matter)


    I have simplified the LHS to be 2/1-2cosxsinx but how do i get the RHS?
    Last edited by ipokeyou; April 10th 2010 at 02:30 AM.
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  2. #2
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    Quote Originally Posted by ipokeyou View Post
    d/dx (cosx+sinx/cosx-sinx)

    I'm having a bit of trouble with trying to prove LHS = RHS (or the other way around for that matter)


    I have simplified the LHS to be 2/1-2cosxsinx but how do i get the RHS?
    1. You don't have a right hand side, so how can you prove that LHS=RHS.

    2. Your expression is unclear.

    Is it \frac{d}{dx}\left(\frac{\cos{x} + \sin{x}}{\cos{x} - \sin{x}}\right)?
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  3. #3
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    ah yes sorry, i left out the RHS which is meant to be sec^2(pi/4 + x)

    and yes the LHS is as you quoted. sorry for the confusion
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    Quote Originally Posted by ipokeyou View Post
    ah yes sorry, i left out the RHS which is meant to be sec^2(pi/4 + x)

    and yes the LHS is as you quoted. sorry for the confusion
    Dear ipokeyou,

    So your expression is,

    \frac{d}{dx}\left(\frac{\cos{x} + \sin{x}}{\cos{x} - \sin{x}}\right)=sec^2\left(\frac{\pi}{4} + x\right)

    Am I correct??
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  5. #5
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    Quote Originally Posted by Sudharaka View Post
    Dear ipokeyou,

    So your expression is,

    \frac{d}{dx}\left(\frac{\cos{x} + \sin{x}}{\cos{x} - \sin{x}}\right)=sec^2\left(\frac{\pi}{4} + x\right)

    Am I correct??
    yes you are correct
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  6. #6
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    Dear

    Yep. I must be correct, since using the above expression I managed to show that the RHS=LHS.

    First consider,

    \frac{\cos{x} + \sin{x}}{\cos{x} - \sin{x}}=\frac{sin\left(\frac{\pi}{2}-x\right)+sinx}{sin\left(\frac{\pi}{2}-x\right)-sinx}=\frac{2sin\frac{\pi}{4}cos\left(\frac{\pi}{4  }-x\right)}{2cos\frac{\pi}{4}sin\left(\frac{\pi}{4}-x\right)}=\frac{cos\left(\frac{\pi}{4}-x\right)}{sin\left(\frac{\pi}{4}-x\right)}


    =\tan\left(\frac{\pi}{2}-\frac{\pi}{4}+x\right)

    I hope you could continue from here!!
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  7. #7
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    Quote Originally Posted by Sudharaka View Post
    Dear

    Yep. I must be correct, since using the above expression I managed to show that the RHS=LHS.

    First consider,

    \frac{\cos{x} + \sin{x}}{\cos{x} - \sin{x}}=\frac{sin\left(\frac{\pi}{2}-x\right)+sinx}{sin\left(\frac{\pi}{2}-x\right)-sinx}=\frac{2sin\frac{\pi}{4}cos\left(\frac{\pi}{4  }-x\right)}{2cos\frac{\pi}{4}sin\left(\frac{\pi}{4}-x\right)}=\frac{cos\left(\frac{\pi}{4}-x\right)}{sin\left(\frac{\pi}{4}-x\right)}


    =\tan\left(\frac{\pi}{2}-\frac{\pi}{4}+x\right)

    I hope you could continue from here!!
    Could i ask, how did you get:

    \frac{2sin\frac{\pi}{4}cos\left(\frac{\pi}{4}-x\right)}{2cos\frac{\pi}{4}sin\left(\frac{\pi}{4}-x\right)}

    and also i think you meant:

    =\cot\left(\frac{\pi}{2}-\frac{\pi}{4}+x\right)

    instead of tan? please correct me if i'm mistaken.
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  8. #8
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    Quote Originally Posted by ipokeyou View Post
    Could i ask, how did you get:

    \frac{2sin\frac{\pi}{4}cos\left(\frac{\pi}{4}-x\right)}{2cos\frac{\pi}{4}sin\left(\frac{\pi}{4}-x\right)}

    and also i think you meant:

    =\cot\left(\frac{\pi}{2}-\frac{\pi}{4}+x\right)

    instead of tan? please correct me if i'm mistaken.
    Dear ipokeyou,

    1) Do you know the sum-to-product rule?? If not better refer: Sum-to-Product formulas (Proof of these formulas could be found in many basic trignometric books.)

    2) \frac{cos\left(\frac{\pi}{4}-x\right)}{sin\left(\frac{\pi}{4}-x\right)}=\cot\left(\frac{\pi}{4}-x\right)=\tan\left(\frac{\pi}{2}-\frac{\pi}{4}+x\right)

    Does this solve your problems?? If not please don't hesitate to reply.
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  9. #9
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    Quote Originally Posted by Sudharaka View Post
    Dear ipokeyou,

    1) Do you know the sum-to-product rule?? If not better refer: Sum-to-Product formulas (Proof of these formulas could be found in many basic trignometric books.)
    ah yes, its just i havent learnt the sum-to-product rules yet but i get it. would there be a way to simplify it without it? just wondering.

    and

    i still dont get how you get from \cot\left(\frac{\pi}{4}-x\right)=\tan\left(\frac{\pi}{2}-\frac{\pi}{4}+x\right)

    sorry for these silly questions
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  10. #10
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    Quote Originally Posted by ipokeyou View Post
    ah yes, its just i havent learnt the sum-to-product rules yet but i get it. would there be a way to simplify it without it? just wondering.

    and

    i still dont get how you get from \cot\left(\frac{\pi}{4}-x\right)=\tan\left(\frac{\pi}{2}-\frac{\pi}{4}+x\right)

    sorry for these silly questions
    See Complementary Angle Identities: Proofs of trigonometric identities - Wikipedia, the free encyclopedia
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  11. #11
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    Quote Originally Posted by mr fantastic View Post
    ah thanks for the reminder!! i got it now thanks
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