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Math Help - Trigonometric Identities

  1. #1
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    Trigonometric Identities

    (sin2θ-cos2θ)/(1-tan2θ)=-cos2θ

    So, the above is the equation that I have to verify.
    I think the first step to this equation is

    -1 (sin2θ-cos2θ)/(1-tan2θ)=-cos2θ
    -1 (-(1))/(-1(sec2θ)=-cos2θ <-- So... I hope you can kinda get my train of thought here, but I feel like it's completely wrong ()

    1/(-1(sec2θ) = -cos2θ
    Then by simplifying the above you get -cos2θ ~ however... the blackboard answer I get to select from... doesn't even have an answer close to what I did above...

    Please help
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  2. #2
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    Re: Trigonometric Identities

    Hello, Ashz!

    \text{Verify: }\:\frac{\sin^2\!\theta - \cos^2\!\theta}{1 - \tan^2\!\theta} \:=\:-\cos^2\!\theta

    We have: . \frac{\sin^2\!\theta - \cos^2\!\theta}{1 - \frac{\sin^2\!\theta}{\cos^2\!\theta}}


    \text{Multiply by }\tfrac {\cos^2\theta}{\cos^2\theta}: \;\;\frac{\cos^2\!\theta\;\!(\sin^2\!\theta - \cos^2\!\theta)}{\cos^2\!\theta\left(1 - \frac{\sin^2\!\theta}{\cos^2\!\theta}\right)} \;=\;\frac{\cos^2\!\theta\;\!(\sin^2\!\theta - \cos^2\!\theta)}{\cos^2\!\theta - \sin^2\!\theta}

    . . . . . . . . . . . =\;\frac{-\cos^2\!\theta\;\!(\cos^2\!\theta - \sin^2\!\theta)}{\cos^2\!\theta - \sin^2\!\theta} \;=\; -\cos^2\!\theta

    Thanks from Sudharaka and Ashz
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  3. #3
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    Re: Trigonometric Identities

    Quote Originally Posted by Ashz View Post
    (sin2θ-cos2θ)/(1-tan2θ)=-cos2θ

    So, the above is the equation that I have to verify.
    I think the first step to this equation is

    -1 (sin2θ-cos2θ)/(1-tan2θ)=-cos2θ

    This is incorrect. You have only multiplied the left hand side by -1.


    -1 (-(1))/(-1(sec2θ)=-cos2θ <-- So... I hope you can kinda get my train of thought here, but I feel like it's completely wrong ()

    You have been given (sin^{2}\theta-cos^{2}\theta)/(1-tan^{2}\theta)=-cos^{2}\theta If you want to multiply by -1 you have to multiply both the left had side and the right hand side.

    1/(-1(sec2θ) = -cos2θ
    Then by simplifying the above you get -cos2θ ~ however... the blackboard answer I get to select from... doesn't even have an answer close to what I did above...

    Please help
    Hi Ashz,

    Consider the left hand side of this equation. Substitute, \tan^{2}\theta=\frac{\sin^{2}\theta}{\cos^{2} \theta} and simplify.
    Thanks from Ashz
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  4. #4
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    Re: Trigonometric Identities

    Quote Originally Posted by Ashz View Post
    (sin2θ-cos2θ)/(1-tan2θ)=-cos2θ

    So, the above is the equation that I have to verify.
    I think the first step to this equation is

    -1 (sin2θ-cos2θ)/(1-tan2θ)=-cos2θ
    -1 (-(1))/(-1(sec2θ)=-cos2θ <-- So... I hope you can kinda get my train of thought here, but I feel like it's completely wrong ()

    1/(-1(sec2θ) = -cos2θ
    Then by simplifying the above you get -cos2θ ~ however... the blackboard answer I get to select from... doesn't even have an answer close to what I did above...

    Please help
    \displaystyle \begin{align*} \frac{\sin^2{x} - \cos^2{x}}{1 - \tan^2{x}} &\equiv \frac{\sin^2{x} - \cos^2{x}}{1 - \frac{\sin^2{x}}{\cos^2{x}}} \\ &\equiv \frac{\sin^2{x} - \cos^2{x}}{\frac{\cos^2{x} - \sin^2{x}}{\cos^2{x}}} \\ &\equiv \left(\sin^2{x} - \cos^2{x}\right) \left( \frac{\cos^2{x}}{\cos^2{x} - \sin^2{x}} \right) \\ &\equiv -\left(\cos^2{x} - \sin^2{x}\right)\left(\frac{\cos^2{x}}{\cos^2{x} - \sin^2{x}}\right) \\ &\equiv -\cos^2{x} \end{align*}

    Q.E.D.
    Thanks from Sudharaka and Ashz
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  5. #5
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    Re: Trigonometric Identities

    ahhh ~ that makes so much sense now!!! Thank you!!!!
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