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Math Help - A Couple Trig Identities

  1. #1
    Member ~berserk's Avatar
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    A Couple Trig Identities

    This is the beginning of a couple of identities that i will probably need assistance on, all help given is greatly appreciated.

    \displaystyle{\frac{\tan x - \cot x}{\tan x + \cot x}=2\sin^2x-1}
    first i switched them all to sin and cos
    then i found like denominators and combined
    then i multiplied by the reciprocal leading me to here:

    \displaystyle{\frac{\sin x - \cos x}{\cos x \sin x} \bullet \frac{\cos x \sin x}{\sin x + \cos x}}

    then got \displaystyle{\frac{\sin x - \cos x}{\sin x + \cos x} = 2\sin^2x-1}
    I'm unsure where to go from here
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  2. #2
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    Quote Originally Posted by ~berserk View Post
    This is the beginning of a couple of identities that i will probably need assistance on, all help given is greatly appreciated.

    \displaystyle{\frac{\tan x - \cot x}{\tan x + \cot x}=2\sin^2x-1}
    first i switched them all to sin and cos
    then i found like denominators and combined
    then i multiplied by the reciprocal leading me to here:

    \displaystyle{\frac{\sin x - \cos x}{\cos x \sin x} \bullet \frac{\cos x \sin x}{\sin x + \cos x}}

    then got \displaystyle{\frac{\sin x - \cos x}{\sin x + \cos x} = 2\sin^2x-1}
    I'm unsure where to go from here
    If you did the 1st one that way, your squares have gone missing...

    \displaystyle\frac{\left[\frac{sinx}{cosx}-\frac{cosx}{sinx}\right]}{\left[\frac{sinx}{cosx}+\frac{cosx}{sinx}\right]}=\frac{\left[\frac{1}{sinxcosx}\right]}{\left[\frac{1}{sinxcosx}\right]}\;\frac{sin^2x-cos^2x}{sin^2x+cos^2x}

    =sin^2x-cos^2x

    since sin^2x+cos^2x=1

    and now use cos^2x=1-sin^2x to finish
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  3. #3
    Member ~berserk's Avatar
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    also there is this one where I am unsure where to start: (\sin x + \cos x)^2 = 1 +\sin 2x
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    Quote Originally Posted by ~berserk View Post
    also there is this one where I am unsure where to start: (\sin x + \cos x)^2 = 1 +\sin 2x
    (\sin x + \cos x)^2 = \left(\sin^2{x}+\cos^2{x}\right)+2\sin{x}\cos{x} = ...
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  5. #5
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    Quote Originally Posted by ~berserk View Post
    also there is this one where I am unsure where to start: (\sin x + \cos x)^2 = 1 +\sin 2x
    That's another sin^2x+cos^2x=1 identity in disguise.

    That's a crucial one to remember!

    (sinx+cosx)(sinx+cosx)=sinx(sinx+cosx)+cosx(sinx+c  osx)

    =sin^2x+sinxcosx+cosxsinx+cos^2x=sin^2x+cos^2x+2si  nxcosx

    Finish up with the identity sin2x=2sinxcosx
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