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Math Help - Verifying an Identity

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    Verifying an Identity

    tan (x/2 + Π/4) = sec x + tan x

    Here is my attempt:

    Substituting 1 for tan (Π/4) and applying the tangent of a sum identity gives

    [tan (x/2) +1] / [1-tan(x/2)] = 1/cos x + sin x/cos x

    Using the half angle identity for tangent, tan(x/2) = sin x /(1 + cos x) and simplifying the complex fraction gives

    (sin x + 1 + cos x) / (1 + cos x - sin x) = (1 + sin x) / cos x

    I do not know where to go from here. Please help!
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    Re: Verifying an Identity

    u have got the result (1+sin x)/cos x=1/cos x+sin x/cos x=sec x+tan x
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    Re: Verifying an Identity

    (1 + sin x )/cos x was just me changing the right side, sec x + tan x, in terms of cosine and sine. I'm trying to figure out how to get the left side of the equation (sin x + 1 + cos x) / (1 + cos x - sin x) to equal (1 + sin x )/cos x. I cannot find that relationship.
    Last edited by cdbowman42; May 29th 2014 at 12:43 PM.
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    Re: Verifying an Identity

    Hello, cdbowman42!

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

    Never ever play with both sides of an identity.
    Choose one side and show that it is equal to the other side.

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

    Multiply by \tfrac{\cos\frac{x}{2}}{\cos\frac{x}{2}}\!;\;\; \frac{\sin\frac{x}{2} + \cos\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}}

    Multlply by \tfrac{\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 \frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}}

    The numerator is: . \left(\cos\tfrac{x}{2} + \sin\tfrac{x}{2}\right)^2 \;=\; \cos^2\!\tfrac{x}{2} + 2\sin\tfrac{x}{2}\cos\tfrac{x}{2} + \sin^2\!\tfrac{x}{2} \;=\;1+\sin x

    The denominator is: . \cos^2\!\tfrac{x}{2} - \sin^2\!\tfrac{x}{2} \;=\;\cos x

    We have: . \frac{1 + \sin x}{\cos x} \;=\;\frac{1}{\cos x} + \frac{\sin x}{\cos x} \;=\;\sec x + \tan x \;=\;RHS
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  5. #5
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    Re: Verifying an Identity

    Quote Originally Posted by Soroban View Post
    Hello, cdbowman42!


    Never ever play with both sides of an identity.
    Choose one side and show that it is equal to the other side.

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


    Multiply by \tfrac{\cos\frac{x}{2}}{\cos\frac{x}{2}}\!;\;\; \frac{\sin\frac{x}{2} + \cos\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}}

    Multlply by \tfrac{\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 \frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}}

    The numerator is: . \left(\cos\tfrac{x}{2} + \sin\tfrac{x}{2}\right)^2 \;=\; \cos^2\!\tfrac{x}{2} + 2\sin\tfrac{x}{2}\cos\tfrac{x}{2} + \sin^2\!\tfrac{x}{2} \;=\;1+\sin x

    The denominator is: . \cos^2\!\tfrac{x}{2} - \sin^2\!\tfrac{x}{2} \;=\;\cos x

    We have: . \frac{1 + \sin x}{\cos x} \;=\;\frac{1}{\cos x} + \frac{\sin x}{\cos x} \;=\;\sec x + \tan x \;=\;RHS
    Sorry Soroban, I completely disagree. While it is more elegant to start with one side and go through logical steps to get to the other, it is still perfectly valid to prove the truth of an identity by working with both sides of an equation and showing that they are equivalent to the same thing. The transitive property states that if two things are equal to the same thing, then they are equal to each other.
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