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Math Help - proving a trig identity

  1. #1
    Senior Member
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    proving a trig identity

    This is my work so far for proving this trig identity. What should I do next? Is there anything I did wrong thus far?

    cotx=tan(-(x-pi/2))

    Right side:

    tan(-x+pi/2)
    [-tanx+tanpi/2]/[1-(-tanx)(tanpi/2)]
    [-sinx/cosx+tanpi/2]/[1+(sinx/cosx)(tanpi/2)]

    also, tanpi/2 is undefined... what does that do to the question?

    thank you in advance!
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  2. #2
    Junior Member JoanF's Avatar
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    Quote Originally Posted by skeske1234 View Post
    This is my work so far for proving this trig identity. What should I do next? Is there anything I did wrong thus far?

    cotx=tan(-(x-pi/2))

    Right side:

    tan(-x+pi/2)
    [-tanx+tanpi/2]/[1-(-tanx)(tanpi/2)]
    [-sinx/cosx+tanpi/2]/[1+(sinx/cosx)(tanpi/2)]

    also, tanpi/2 is undefined... what does that do to the question?

    thank you in advance!
    Hi!



    tan(pi/2) is undefined...well, I think you don't need that to solve this problem. If there is something you don't understand, just say it.
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  3. #3
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    Hello, skeske1234!

    You didn't do anything wrong . . .


    Prove: . \cot x\:=\:\tan\left[\text{-}\left(x-\tfrac{\pi}{2}\right)\right]

    The right side is: . \tan\left(\tfrac{\pi}{2} - x\right) \;=\;\frac{\sin\left(\frac{\pi}{2}-x\right)}{\cos\left(\frac{\pi}{2}-x\right)} \;= \;\frac{\sin\frac{\pi}{2}\cos x - \cos\frac{\pi}{2}\sin x} {\cos\frac{\pi}{2}\cos x + \sin\frac{\pi}{2}\sin x }

    . . =\;\frac{1\cdot\cos x - 0\cdot\sin x}{0\cdot\cos x + 1\cdot\sin x} \;=\;\frac{\cos x}{\sin x} \;=\;\cot x

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  4. #4
    Junior Member JoanF's Avatar
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    So, forget what my resolution. I didn't know that you can use these formulas ( sin(a-b) and cos(a-b). Here, in Portugal, you can't use these formulas to solve the problem.
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