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Math Help - Identity Proof!

  1. #1
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    Identity Proof!

    tan( pi/4 + x/2 ) = sec x + tan x

    I have worked on this for over two hours now.
    Help would be greatly appreciated

    Thanks!
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by dxscorp View Post
    tan( pi/4 + x/2 ) = sec x + tan x

    I have worked on this for over two hours now.
    Help would be greatly appreciated

    Thanks!
    using this
    \tan (a+b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}

    \tan \left(\frac{\pi}{4} + \frac{x}{2}\right)=\frac{1+\tan x/2}{1 - \tan x/2 }

    \tan \left(\frac{\pi}{4} + \frac{x}{2}\right)=\frac{1+\tan x/2}{1 - \tan x/2 } \left(\frac{1+\tan x/2}{1+\tan x/2} \right)

    can you continue?, you must know

    \cos 2x = \cos ^2 x - \sin ^2 x \Longrightarrow \cos x = \cos ^2 \left(\frac{x}{2}\right) - \sin ^2 \left(\frac{x}{2}\right)

    \sin 2x = 2\sin x \cos x \Longrightarrow \sin x = 2 \sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)
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  3. #3
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    Hello, dxscorp!

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

    \tan\left(\frac{\pi}{4} + \frac{x}{2}\right) \;=\;\frac{\tan\frac{\pi}{4} + \tan\frac{x}{2}}{1 - \tan\frac{\pi}{4}\tan\frac{x}{2}} \;=\;\frac{1 + \tan\frac{x}{2}}{1 - \tan\frac{x}{2}} . =\;\frac{1 + \dfrac{\sin\frac{x}{2}}{\cos\frac{x}{2}}} {1 - \dfrac{\sin\frac{x}{2}}{\cos\frac{x}{2}}} \;=\;\frac{\cos\frac{x}{2} + \sin\frac{x}{2}} {\cos\frac{x}{2} - \sin\frac{x}{2}}


    Multiply by \frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} + \sin\frac{x}{2}}\!:

    . . \frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}} \cdot {\color{blue}\frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} + \sin\frac{x}{2}}}

    . . . . . . =\;\;\frac{\overbrace{(\cos^2\tfrac{x}{2} + \sin^2\tfrac{x}{2})}^{\text{This is 1}} \;+\; \overbrace{2\sin\tfrac{x}{2}\cos\tfrac{x}{2}}^{\te  xt{This is }\sin x}}  {\underbrace{\cos^2\tfrac{x}{2} - \sin^2\tfrac{x}{2}}_{\text{This is }\cos x}}

    . . . . . . =\;\;\frac{1 + \sin x}{\cos x} \;\;=\;\;\frac{1}{\cos x} + \frac{\sin x}{\cos x} \;\;=\;\; \sec x + \tan x

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