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Math Help - Is it really equal? Why?

  1. #1
    Member courteous's Avatar
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    Question Is it really equal? Why?

    \displaystyle{ \cos\left\{ tan^{-1}\left(\frac{1}{R}\sqrt{\frac{L}{C}}\right) \right\}\stackrel{?}{=}\frac{1}{\sqrt{1+\frac{L}{R  ^2C}}} }

    Even though I think it is ir-relevant: R, L, C are resistance, inductance and capacitance, respectively.
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  2. #2
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    Quote Originally Posted by courteous View Post
    \displaystyle{ \cos\left\{ tan^{-1}\left(\frac{1}{R}\sqrt{\frac{L}{C}}\right) \right\}\stackrel{?}{=}\frac{1}{\sqrt{1+\frac{L}{R  ^2C}}} }

    Even though I think it is ir-relevant: R, L, C are resistance, inductance and capacitance, respectively.
    consider a right triangle ...

    \tan{y} = \frac{opp}{adj}

    \tan^{-1}\left(\frac{opp}{adj}\right) = y

    opp = \frac{1}{R} \sqrt{\frac{L}{C}}<br />

    adj = 1

    hypotenuse = \sqrt{1^2 + \left(\frac{1}{R} \sqrt{\frac{L}{C}}\right)^2}

    \cos{y} = \frac{adj}{hyp} = \frac{1}{\sqrt{1 + \frac{L}{R^2 C}}}
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