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Math Help - trigonometric identity

  1. #1
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    trigonometric identity

    Been 20 years since I've identified trig identities. We are stuck on a particular type and would appreciate some explanation. 1/sin(-x) * (1-cos^2x) that's cos(squared x) and 1+1/cos(-x) / -sinx-tanx
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by singleton2787 View Post
    Been 20 years since I've identified trig identities. We are stuck on a particular type and would appreciate some explanation. 1/sin(-x) * (1-cos^2x) that's cos(squared x) and 1+1/cos(-x) / -sinx-tanx
    well the right side is indescernible bu

    \frac{1}{\sin(-x)}=\frac{-1}{\sin(x)}=\csc(x)

    and 1-\cos^2(x)=\sin^2(x)

    So \sin^2(x)\cdot\csc(x)=\sin(x)

    rewrite the other one in a more hospitable form
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  3. #3
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    problem re-written

    Problem 1 (simplify) 1/sin(-x) * (1-cos^2x)
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by singleton2787 View Post
    Problem 1 (simplify) 1/sin(-x) * (1-cos^2x)
    I am very sorry I was unclear...the above work was the solution to that problem...I meant the other one...once again sorry
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  5. #5
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    i dont understand..

    how does 1-\cos^2(x)=\sin^2(x)?
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by singleton2787 View Post
    how does 1-\cos^2(x)=\sin^2(x)?
    Because \cos^2(x)+\sin^2(x)=1

    This can be derived multiple ways...the most intuitive being

    On the unit circle which has a radius of one any of the points on the circle have coordinates

    (\cos(x),\sin(x))


    So the distance from the origin to any point on the circle which I stated earlier is 1

    So setting up the distance equation we have

    \sqrt{(0-\cos(x))^2+(0-\sin(x))^2}=1^2

    Simplifying we get

    \sqrt{\cos^2(x)+\sin^2(x)}=1^2\Rightarrow{\cos^2(x  )+\sin^2(x)=1}

    Usually in trig classes this indentity is just taken to be true
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  7. #7
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    second problem

    thanks for your help can you explain this one? \frac{1+sec(-x)}{sin(-x)+tan(-x)}
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    Quote Originally Posted by singleton2787 View Post
    thanks for your help can you explain this one? \frac{1+sec(-x)}{sin(-x)+tan(-x)}
    First, convert everything to sines and cosines using the identities

    \sec x = \frac1{\cos x}

    \tan x = \frac{\sin x}{\cos x}

    Then, combine the fractions in the numerator and denominator, reduce, factor and cancel, and you should be left with -\csc x if you do it right. When working this one, it may be helpful to observe that, since sine is an odd function and cosine is even,

    \sin\left(-x\right) = -\sin x

    \cos\left(-x\right) = \cos x

    If you have difficulty with the simplification, let us know.
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