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Math Help - Trigonometric equations?

  1. #1
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    Trigonometric equations?

    I have two problems that I am not entirely sure how to

    The first one is

    (1/cosx-sinx)= cosx+sinx

    Now I am not entirely sure how to this one but would I turn the right side into a fraction by placing one under it then cancelling it out?

    My second question is

    sinxcosx= 1/2

    Would I do
    sinxcosx-1/2=0 then what next as I am not sure how to proceed
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  2. #2
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    Quote Originally Posted by homeylova223 View Post

    My second question is

    sinxcosx= 1/2

    Would I do
    sinxcosx-1/2=0 then what next as I am not sure how to proceed
    Multiply both sides by 2 giving

    \displaystyle \sin x\cos x= \frac{1}{2}

    \displaystyle 2\sin x\cos x= 1

    \displaystyle \sin 2x= 1

    You have it from here?
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  3. #3
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    Quote Originally Posted by homeylova223 View Post
    I have two problems that I am not entirely sure how to

    The first one is

    (1/cosx-sinx)= cosx+sinx

    Now I am not entirely sure how to this one but would I turn the right side into a fraction by placing one under it then cancelling it out?

    My second question is

    sinxcosx= 1/2

    Would I do
    sinxcosx-1/2=0 then what next as I am not sure how to proceed
    \dfrac{1-\cos(x)\sin(x)}{\cos(x)} = \cos(x)+ \sin(x)

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

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

    Which is a nasty little equation to solve.



    Did you happen to mean \dfrac{1}{\cos(x)-\sin(x)} = \cos(x) + \sin(x) which is much easier to solve - multiply through and use the difference of two squares to get a double angel identity
    Last edited by e^(i*pi); March 9th 2011 at 04:51 PM. Reason: white space+italics
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  4. #4
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    Yes I meant the one that is easier to solve. The one you wrote on the bottom part.

    The thing is I am not sure how to multiply through. Is it like cross multiplication? Like 1 x 1 and then cosx-sinx times cosx+sinx
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  5. #5
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    Yes, it's the same as cross multiplication but in this instance it's easier to solve by multiplying both sides by \cos(x) - \sin(x)


    \dfrac{1}{\cos(x)-\sin(x)} \times (\cos(x) - \sin(x))  = (\cos(x)+ \sin(x)) \times (\cos(x)-\sin(x))

    \implies 1 = (\cos(x)+\sin(x))(\cos(x)-\sin(x))


    The RHS is of the form (a-b)(a+b) so it's time for the difference of two squares to expand
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    I have one follow up question

    I managed to reduce it to
    1= cosx^2-sinx^2 Would I do this?
    1-cosx^2=-sinx^2
    sinx^2=-sinx^2
    sinx^2+sinx^2= 0 factor sinx^2
    sinx^2=0
    sinx=0
    x= pi
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  7. #7
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by homeylova223 View Post
    I have one follow up question

    I managed to reduce it to
    1= cosx^2-sinx^2 Would I do this?
    1-cosx^2=-sinx^2
    sinx^2=-sinx^2
    sinx^2+sinx^2= 0 factor sinx^2 -e^(i*pi): You're not factoring so much as dividing by 2
    sinx^2=0
    sinx=0
    x= pi
    Yes, I was thinking of \cos(2x) = \cos^2(x)-\sin^2(x) giving \cos(2x) = 1 but your method is fine although while x = \pi is a solution but it's not the only one. Check \sin(0)
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