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Math Help - Proving trigonometric Identities

  1. #1
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    Proving trigonometric Identities

    Prove

    tan(pi/4 + x) - tan(pi/4 - x) = 2tan2x

    and

    tan^2A - tan^2B = (sin^2A - sin^2B) / (cos^2A cos^B)

    help on either would be very much appreciated. Thank you
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by BeckyDWood View Post
    Prove

    tan(pi/4 + x) - tan(pi/4 - x) = 2tan2x

    and

    tan^2A - tan^2B = (sin^2A - sin^2B) / (cos^2A cos^B)

    help on either would be very much appreciated. Thank you
    \frac{\sin(\frac{\pi}{4}+x)}{\cos ({\pi}{4} +x)} - \frac{\sin (\frac{\pi}{4}-x) }{\cos (\frac{\pi}{4} -x) }

    \frac{\sin(\frac{\pi}{4}+x)(\cos (\frac{\pi}{4} -x))-\sin (\frac{\pi}{4}-x)(\cos (\frac{\pi}{4} +x))}{\cos(\frac{\pi}{4}+x)(\cos (\frac{\pi}{4}-x)}

    \sin (A-B) = \sin A \cos B - \sin B \cos A

    \cos (A) \cos (B) = 1/2 (\cos(\frac{A+B}{2})+\cos (\frac{A-B}{2}))

    the numerator \sin\left( \frac{\pi}{4} +  x -(\frac{\pi}{4} -x )\right)

    the denominator

    1/2(\cos (\pi/4+x+\pi/4 -x)+\cos (\pi/4+x-(\pi/4-x)))
    Last edited by Amer; October 30th 2009 at 07:40 AM.
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  3. #3
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    Hello, Becky!

    For the first one, we can use these identities:

    . . \tan(A \pm B) \:=\:\frac{\tan A \pm \tan B}{1 \mp \tan A\tan B} \qquad\qquad \frac{2\tan A}{1 - \tan^2A} \;=\;\tan2A


    Prove: . \tan\left(\tfrac{\pi}{4} + x\right) - \tan\left(\tfrac{\pi}{4} - x\right) \;=\; 2\tan2x

    \tan\left(\tfrac{\pi}{4} + x\right) - \tan\left(\tfrac{\pi}{4} - x\right) \;\;=\;\;\frac{\tan\frac{\pi}{4} + \tan x}{1 - \tan\frac{\pi}{4}\tan x} - \frac{\tan\frac{\pi}{4} - \tan x}{1 + \tan\frac{\pi}{4}\tan x}\;= . \frac{1+\tan x}{1 - \tan x} - \frac{1 - \tan x}{1 + \tan x}

    . . = \;\;\frac{(1+\tan x)^2 - (1 - \tan x)^2}{(1-\tan x)(1 + \tan x)} \;\;=\;\;\frac{(1 + 2\tan x + \tan^2\!x) - (1 - 2\tan x + \tan^2\!x)}{1-\tan^2x}

    . . = \;\;\frac{4\tan x}{1 - \tan^2\!x} \;\;=\;\;2\left(\frac{2\tan x}{1-\tan^2\!x}\right) \;\;=\;\;2\tan2x

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