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Thread: [SOLVED] Trigonometry Help!?

  1. #1
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    [SOLVED] Trigonometry Help!?

    Prove the identity $\displaystyle \frac{1-tan^2x}{1+tan^2x} = 1-2sin^2x$
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    Moo
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    Hello,

    $\displaystyle \tan(x)=\frac{\sin(x)}{\cos(x)}$



    $\displaystyle \frac{1-tan^2x}{1+tan^2x} = \frac{1-\frac{\sin^2x}{\cos^2x}}{1+\frac{\sin^2x}{\cos^2x} }$ $\displaystyle = \frac{\frac{\cos^2x-\sin^2x}{\cos^2x}}{\frac{\cos^2x+\sin^2x}{cos^2x}}$ $\displaystyle = \frac{\cos^2x-\sin^2x}{\underbrace{\cos^2x+\sin^2x}_{1}}$ $\displaystyle = \underbrace{\cos^2x}_{1-\sin^2x}-\sin^2x = 1-2 \sin^2x$
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    Hello, looi76!

    A slightly different approach . . .


    $\displaystyle \frac{1-\tan^2\!x}{1+\tan^2\!x} = 1-2\sin^2\!x$
    We have: .$\displaystyle \frac{1-\tan^2\!x}{\sec^2\!x} \;=\;\frac{1 - \dfrac{\sin^2\!x}{\cos^2\!x}}{\dfrac{1}{\cos^2\!x} }$


    Multiply by $\displaystyle \frac{\cos^2\!x}{\cos^2\!x}\!:\quad\frac{\cos^2\!x \left(1 - \dfrac{\sin^2\!x}{\cos^2\!x}\right)} {\cos^2\!x\left(\dfrac{1}{\cos^2\!x}\right)} \;=\;\frac{\cos^2\!x - \sin^2\!x}{1}$


    Then: .$\displaystyle \cos^2\!x - \sin^2\!x \;\;=\;\;(1-\sin^2\!x) - \sin^2\!x\;\;=\;\;1-2\sin^2\!x$


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Hello, moo!

    Beautiful job with the LaTeX!

    Take a look at how I use \dfrac to make enlarged fractions.
    (I learned about it only a few months ago.)

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  4. #4
    Moo
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    Hi Soroban !

    Actually, i've learnt the use of \underbrace while reading one of your messages

    $\displaystyle \sum_{k=0}^\infty \dfrac{x^n \ \text{which \ is \ the \ nth \ power \ of \ x}}{n!}$


    $\displaystyle \sum_{k=0}^\infty \frac{x^n \ \text{which \ is \ the \ nth \ power \ of \ x}}{n!}$


    erm... can't see the difference :s


    $\displaystyle \frac{1-\tan^2\!x}{\sec^2\!x} \;=\;\frac{1 - \dfrac{\sin^2\!x}{\cos^2\!x}}{\dfrac{1}{\cos^2\!x} }$

    $\displaystyle \frac{1-\tan^2\!x}{\sec^2\!x} \;=\;\frac{1 - \frac{\sin^2\!x}{\cos^2\!x}}{\frac{1}{\cos^2\!x}}$


    Ow, i see :-)

    Thanks !
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    what is sec ?
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    $\displaystyle sec(x)$ is just another name for $\displaystyle \frac{1}{cos(x)}$
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  7. #7
    Moo
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    And to explain :

    $\displaystyle \sec^2x=\frac{1}{\cos^2x}=\dfrac{\cos^2x+\sin^2x}{ \cos^2x}=1+\dfrac{\sin^2x}{cos^2x}$ $\displaystyle =1+\tan^2x$
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