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Math Help - trigonometric identities

  1. #1
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    trigonometric identities

    I think I have this problem down but then again I'm not sure. If you could check my work, please:

    cot^2x = 1-2cscx+csc^2x
    csc^2x-1 = 1-2cscx+csc^2x (from the pythag. identity)
    0 = 2-2cscx + csc^2x-csc^2x
    0 = 2-2cscx
    0 = -1+cscx (divided all by -2)
    cscx = 1

    Also, I am so stuck on the following two. I have to use the double (or half, but I'm sure it's double) angle formulas but I don't know how to properly incorporate them:

    1. 2cos^2x-1 = cosx

    and

    2. cos2x + 6sin^2x = 2

    Help would be very much appreciated. Thank you!
    Last edited by augustfai; April 30th 2008 at 01:47 PM.
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  2. #2
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    Quote Originally Posted by augustfai View Post
    Also, I am so stuck on the following two. I have to use the double angle formulas but I don't know how to properly incorporate them:

    1. 2cos^2x-1 = cosx

    and

    2. cos2x + 6sin^2x = 2

    Help would be very much appreciated. Thank you!
    For the first one this may help. let u=\cos(x)

    \underbrace{2\cos^2(x)}_{2u^2}-1=\underbrace{\cos(x)}_{u}

    2u^2-1=u \iff 2u^2-u-1=0 \iff (2u+1)(u-1)

    so u=-\frac{1}{2} and u=1

    but u=cos(x) so we get

    \cos(x)=-\frac{1}{2} and \cos(x)=1

    All that is left is to solve these.

    for the 2nd one use this identity
    using the identity \cos(2x)=1-2\sin^{2}(x)

    and try the sub u=\sin(x)

    Good luck.
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  3. #3
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    Quote Originally Posted by augustfai View Post
    I think I have this problem down but then again I'm not sure. If you could check my work, please:

    cot^2x = 1-2cscx+csc^2x
    csc^2x-1 = 1-2cscx+csc^2x (from the pythag. identity)
    0 = 2-2cscx + csc^2x-csc^2x
    0 = 2-2cscx
    0 = -1+cscx (divided all by -2)
    cscx = 1

    Also, I am so stuck on the following two. I have to use the double angle formulas but I don't know how to properly incorporate them:

    1. 2cos^2x-1 = cosx

    and

    2. cos2x + 6sin^2x = 2

    Help would be very much appreciated. Thank you!
    Your work on the problem above is correct.

    For 1. 2 \cos{^2}{x} - 1 = \cos{(2x)}. I would show this using \cos{(2x)} = \cos{(x+x)} = \cos{x}\cdot{\cos{x}} - \sin{x}\cdot{\sin{x}} = \cos{^2}{x} - \sin{^2}{x}.
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  4. #4
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    Quote Originally Posted by icemanfan View Post
    Your work on the problem above is correct.

    For 1. 2 \cos{^2}{x} - 1 = \cos{(2x)}. I would show this using \cos{(2x)} = \cos{(x+x)} = \cos{x}\cdot{\cos{x}} - \sin{x}\cdot{\sin{x}} = \cos{^2}{x} - \sin{^2}{x}.
    Thank you, but I'm a little confused on this bit: how did 2 \cos{^2}{x} - 1 = \cos{(x)} become 2 \cos{^2}{x} - 1 = \cos{(2x)}?
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  5. #5
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    If the equation in 1 is written correctly, you may disregard my first post.

    For 2) use the double-angle identity that I worked out in the first post and substitute 2 \cos{^2}{x} + 6 \sin{^2}{x} = 3. Then:

    2 \cos{^2}{x} + 2 \sin{^2}{x} + 4 \sin{^2}{x} = 3

    2 + 4 \sin{^2}{x} = 3

    4 \sin{^2}{x} = 1

    \sin{^2}{x} = \frac{1}{4} and so

    \sin{x} = \frac{1}{2} or \sin{x} = -\frac{1}{2}
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  6. #6
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    Quote Originally Posted by augustfai View Post
    Thank you, but I'm a little confused on this bit: how did 2 \cos{^2}{x} - 1 = \cos{(x)} become 2 \cos{^2}{x} - 1 = \cos{(2x)}?
    The equation that I worked out is an identity. It wasn't actually solving an equation. Since it looked so close to the identity I just assumed you had copied it incorrectly, but that was my mistake.
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  7. #7
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    Quote Originally Posted by icemanfan View Post
    The equation that I worked out is an identity. It wasn't actually solving an equation. Since it looked so close to the identity I just assumed you had copied it incorrectly, but that was my mistake.
    Ah, alright. Thank you. I actually don't know whether I'm supposed to use the double or the half-angle formulas, I assumed double since half doesn't seem to work out.

    EDIT:
    I think I sort of figured out the first one--

    2 \cos{^2}{x} - 1 = \cos{(x)}

    turns into cos2x = cos x

    What, then, from there?
    Last edited by augustfai; April 30th 2008 at 02:03 PM.
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  8. #8
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    Hello,

    Quote Originally Posted by augustfai View Post
    Ah, alright. Thank you. I actually don't know whether I'm supposed to use the double or the half-angle formulas, I assumed double since half doesn't seem to work out.

    EDIT:
    I think I sort of figured out the first one--

    2 \cos{^2}{x} - 1 = \cos{(x)}

    turns into cos2x = cos x

    What, then, from there?
    cos(a)=cos(b) <=> a=b+2k pi, OR a=-b+2k pi

    So... ?
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  9. #9
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    I figured it out! Thanks, everyone.
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