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Math Help - Trig - Double Angles Questions, Please Help!!

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    Trig - Double Angles Questions, Please Help!!

    Please close this thread, solved all problems...
    Last edited by Macleef; November 17th 2007 at 07:27 PM.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Macleef View Post

    1) Prove:

    sin2x / 1 - cos2x = cotx

    My Attempt:
    LS:
    = 2sinxcosx / - 1 - (1 - 2sin^2x)
    = 2sinxcosx / - 1 + 2sin^2x
    = cosx / sinx - 1
    Please use parenthesis!!

    Is there a typo? -1 - 1 = -2, not -1. And if you can't factor a 2 out of everything, then you can't cancel the 2, so the last line here is wrong. The same goes for sin(x).

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

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

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

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

    \frac{2sin(x)cos(x)}{2sin^2(x)} = \frac{cos(x)}{sin(x)}

    \frac{cos(x)}{sin(x)} = \frac{cos(x)}{sin(x)}

    You had the right idea, but you didn't have the details right.

    -Dan
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    Thanks I see where I went wrong now, can you also please help with the rest?

    Edit:

    I don't know how to use the parenthesis?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Macleef View Post
    I don't know how to use the parenthesis?
    The kid is crying so I can't.

    You don't know how to use parenthesis??? I'm talking about the difference between
    a + b/c + d = a + \frac{b}{c} + d
    vs.
    (a + b)/(c + d) = \frac{a + b}{c + d}

    You use them to group terms together.

    -Dan
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    Quote Originally Posted by topsquark View Post
    The kid is crying so I can't.

    You don't know how to use parenthesis??? I'm talking about the difference between
    a + b/c + d = a + \frac{b}{c} + d
    vs.
    (a + b)/(c + d) = \frac{a + b}{c + d}

    You use them to group terms together.

    -Dan
    What? The kid is crying so you can't????

    I can group them...but not with the latex feature because I don't know how to use it
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    Hello, Macleef!

    4) Prove: . 8\cos^4\!x \:= \:\cos4x + 4\cos2x + 3
    We'll use the identity: . \cos^2\!\theta \:=\:\frac{1 + \cos2\theta}{2} . . . twice.


    We have: . 8\left(\cos^2\!x\right)^2 \;=\;8\left(\frac{1 + \cos2x}{2}\right)^2 \;=\;\;8\left(\frac{1 + 2\cos2x + \cos^2\!2x}{4}\right)

    . . = \;\;2\left(1 + 2\cos2x + \cos^2\!2x\right) \;\;=\;\;2\left(1 + 2\cos2x + \frac{1 + \cos4x}{2}\right)

    . . = \;\;2 + 4\cos2x + 1 + \cos4x \;\;=\;\;{\color{blue}\cos4x + 4\cos2x + 3}

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    Quote Originally Posted by Soroban View Post
    Hello, Macleef!

    We'll use the identity: . \cos^2\!\theta \:=\:\frac{1 + \cos2\theta}{2} . . . twice.


    We have: . 8\left(\cos^2\!x\right)^2 \;=\;8\left(\frac{1 + \cos2x}{2}\right)^2 \;=\;\;8\left(\frac{1 + 2\cos2x + \cos^2\!2x}{4}\right)

    . . = \;\;2\left(1 + 2\cos2x + \cos^2\!2x\right) \;\;=\;\;2\left(1 + 2\cos2x + \frac{1 + \cos4x}{2}\right)

    . . = \;\;2 + 4\cos2x + 1 + \cos4x \;\;=\;\;{\color{blue}\cos4x + 4\cos2x + 3}

    wow, you lost me -_- from the beginning when you wrote, "we'll use this identity"
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Macleef View Post
    wow, you lost me -_- from the beginning when you wrote, "we'll use this identity"
    There are two ways to get at this identity. I'll use the one you will have been likely to see:
    cos(2 \theta) = 2cos^2(\theta) - 1

    2cos^2(\theta) = cos(2\theta) + 1

    cos^2(\theta) = \frac{cos(2\theta) + 1}{2}

    (The quick way to do this is using a half-angle identity.)

    -Dan
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Macleef View Post
    Please close this thread, solved all problems...
    do not delete your question even after it is answered. it may help someone else
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