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Math Help - Trig identities! Help please.

  1. #1
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    Trig identities! Help please.

    I need to reduce trig identities to the simplest form. i was wondering if you could take me further.



    (secx)(cscx)-tanx
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    cotx(1/secx)+sinx


    i changed the top to (1/cosx)(1/sinx)-(sinx/cosx)


    Thats as far as i can do it
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  2. #2
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    Keep going ... Convert everything into terms of sinx and cosx and combine the fractions by finding the common denominator. For example in the numerator:

    \sec x \csc x - \tan x

     = \frac{1}{\cos x \sin x} - \frac{\sin x}{\cos x}

     = \frac{1}{\cos x \sin x} - \frac{\sin^{2} x}{\cos x \sin x}


     = \frac{\overbrace{1 - \sin^{2} x}^{\mbox{Look familiar?}}}{\cos x \sin x}
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  3. #3
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    so i get :


    (cos^2x)/(cosxsinx)


    for the numerator?
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  4. #4
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    Hello, Airjunkie!

    Reduce to the simplest form: . \frac{\sec x\csc x -\tan x}{\cot x\!\cdot\!\frac{1}{\sec x}+\sin x}

    i changed the top to: . \frac{1}{\cos x}\!\cdot\!\frac{1}{\sin x} -\frac{\sin x}{\cos x} . . . . Good!
    Get a common denominator and combine the fractions.

    . . \frac{1}{\cos x\sin x} - \frac{\sin x}{\cos x}\cdot\frac{\sin x}{\sin x} \;\;=\;\;\frac{1}{\cos x\sin x} - \frac{\sin^2\!x}{\cos x\sin x}

    . . = \;\;\frac{1-\sin^2\!x}{\cos x\sin x} \;\;=\;\;\frac{\cos^2\!x}{\cos x\sin x} \;\;=\;\;\frac{\cos x}{\sin x}


    The denominator is: . \frac{\cos x}{\sin x}\!\cdot\!\frac{\cos x}{1} + \sin x \;\;=\;\;\frac{\cos^2\!x}{\sin x} + \frac{\sin^2\!x}{\sin x}
    . . =\;\;\frac{\overbrace{\cos^2\!x + \sin^2\!x}^{\text{This is 1}}}{\sin x} \;\;=\;\;\frac{1}{\sin x}


    The problem becomes: . \frac{\dfrac{\cos x}{\sin x}}{\dfrac{1}{\sin x}} \;\;=\;\;\boxed{\cos x}

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  5. #5
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    Wow, i see what i did wrong. I understand it better. But not I have to solve a much more complicated one, and im quite confused still.




    i didn't fall for the 1/1 at least :P
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  6. #6
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    Let's concentrate piece by piece.

    1. \frac{\cos \left(\frac{\pi}{2} - \theta\right)}{sin^{2}\left(\frac{\pi}{2} - \theta\right) + cos^{2}\left(\frac{\pi}{2} - \theta\right)}

    First off, you should immediately recognize the identity found in the denominator. As for the numerator, if you know your sin
    and cos graphs like the back of your hand:
    \cos \left(\frac{\pi}{2} -  \theta \right) = \overbrace{\cos \left(\theta - \frac{\pi}{2}\right)}^{ \cos (- x ) = \cos x }

    which is just y = cos x shifted to the right by pi/2. What would that be?

    2. \frac{\frac{1}{\frac{1}{\sin\theta}}}{\sec \theta} = \frac{\sin \theta}{\frac{1}{\cos \theta}} = \mbox{etc.}

    3. - \frac{1}{\cot (\pi + \theta)} = - \tan(\theta + \pi)
    Ignoring the negative sign for now, if you shift the graph y = tanx to the left by pi, what do you get?

    4.
    \sin^{2}(- \theta) + \cos^{2}(\theta) = (\sin \left(-\theta\right))^{2} + \cos^{2}(\theta) = \underbrace{(-\sin (\theta))^{2}}_{\sin(-\theta) = -\sin (\theta)} + \cos^{2} (\theta) = \sin^{2}\theta + \cos^{2} \theta
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  7. #7
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    Im not trying to be mean but unless i see steps it cant help me.

    By the way this isnt really for a grade, but she said these will be on the tests. So id really like to study and memorize the rules with them. I need to see the whole thing. Ill try to look over it again
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