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Math Help - please help trig identities

  1. #1
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    please help trig identities

    Please can you show me how to solve
    sin2x-1=cos2x
    i have tried various methods but am getting nowhere
    this is my first post so thanks in advance for the help
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chipper View Post
    Please can you show me how to solve
    sin2x-1=cos2x
    i have tried various methods but am getting nowhere
    this is my first post so thanks in advance for the help
    just use the double angle formulas for sine and cosine:

    \sin 2x = 2 \sin x \cos x and \cos 2x = \cos^2 x - \sin^2 x
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  3. #3
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    i have tried these but think i am missing something, i can get to the point
    2sinxcosx-1=2cos^2x-1=0
    2sinxcosx-1-2cos^2x+1=0
    2sinxcosx-2cos^2x=0
    but then am abit lost how to solve, can't find a way to factorise it, or have i done something wrong
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chipper View Post
    i have tried these but think i am missing something, i can get to the point
    2sinxcosx-1=2cos^2x-1=0
    2sinxcosx-1-2cos^2x+1=0
    2sinxcosx-2cos^2x=0
    but then am abit lost how to solve, can't find a way to factorise it, or have i done something wrong
    factor out the cosine

    you get: \cos x ( \sin x - \cos x ) = 0

    \Rightarrow \cos x = 0 or \sin x - \cos x = 0

    can you continue?
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  5. #5
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    if i continue i get
    cosx=0
    x=arccos 0
    x=pi/2 (1.57)
    but am unsure how to do the next bit
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chipper View Post
    if i continue i get
    cosx=0
    x=arccos 0
    x=pi/2 (1.57)
    but am unsure how to do the next bit
    \cos x = 0 \implies x = \frac {\pi}2 + n \pi for n an integer

    \sin x - \cos x = 0

    \Rightarrow \sin x = \cos x

    \Rightarrow x = \frac {\pi}4 + k \pi for k an integer

    if you want 0 \le x \le 2 \pi, then your solutions are: x = \frac {\pi}2, \frac {3 \pi}2, \frac {\pi}4, \mbox { and } \frac {5 \pi}4

    be sure to check all these solutions in the original equation to make sure none are erroneous
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  7. #7
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    thanks so much for your help
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