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Thread: Trigonometric Identities problem from my pre-calc class that is driving me insane

  1. #1
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    Angry Trigonometric Identities problem from my pre-calc class that is driving me insane

    So the problem is a verify the equation problem where the left side has to be made equal to the right : sin-1 * (4/5) - cos-1 * (12/13) = cos-1* (56/65)

    - My confusion come from two main things
    1. I've never done a problem with negative 1 exponents and I can't find an example of one in my book
    2. I've done a hundred of these verify problems but I've never come across one that is sin - cos = cos

    Also my first time posting so I may have done something wring
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  2. #2
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    Re: Trigonometric Identities problem from my pre-calc class that is driving me insane

    The "negative one exponent" indicates the inverse function. sin^{-1}(x) and cos^{-1}(x) are also often called "arcsin(x)" and "arcos(x)". If cos(x)= y then cos^{-1}(y)= x. I find it very hard to believe that your textbook gave you a problem like this without ever having defined them!
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  3. #3
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    Re: Trigonometric Identities problem from my pre-calc class that is driving me insane

    Your Q. asks you to "verify". Note that in mathematics, that is not the same as "prove": "verify" has the same meaning as "validate".

    On those grounds, a quick use of a calculator reveals that:
    sin-1 * (4/5) = 53.130,
    cos-1 * (12/13) = 22.620
    Difference = 30.510
    And that cos-1* (56/65) = 30.510,
    Thus completing the verification - as required.

    Aside: Sorry about the "sin-1" typing. I did a cut-and-paste from the OP's Q., but that operation, strangely enough, fails here.

    Al.
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    Re: Trigonometric Identities problem from my pre-calc class that is driving me insane

    an inverse trig function value is an angle ...

    $\sin{x} = y \implies x = \sin^{-1}{y}$


    let $\theta = \sin^{-1}\left(\dfrac{4}{5}\right)$

    let $\phi = \cos^{-1}\left(\dfrac{12}{13}\right)$

    note both angles have terminal sides in quadrant I $\implies \cos{\theta} = \dfrac{3}{5}$ and $\sin{\phi} = \dfrac{5}{13}$

    $\cos(\theta - \phi) = \cos{\theta}\cos{\phi} + \sin{\theta}\sin{\phi}$

    $\cos(\theta - \phi) = \dfrac{3}{5} \cdot \dfrac{12}{13} + \dfrac{4}{5} \cdot \dfrac{5}{13}$

    $\cos(\theta - \phi) = \dfrac{36}{65} + \dfrac{20}{65}$

    $\cos(\theta - \phi) = \dfrac{56}{65}$

    $\cos^{-1}[\cos(\theta - \phi)] = \cos^{-1}\left(\dfrac{56}{65}\right)$

    $\theta - \phi = \cos^{-1}\left(\dfrac{56}{65}\right)$
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