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Math Help - Another identity, brain still not working

  1. #1
    Member sinewave85's Avatar
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    Another identity, brain still not working

    Ok, I know it seems like I am spunging, but there are thirty of these on this assignement, only stumped on two -- so far. The identity is:

    (sin(x) + sin(3x))/(cos(x) + cos(3x) = tan(2x)

    The only identities I have covered so far in my book are double-angle, half-angle, sum and difference, and the basics (reciprocal, quotentient, pythagorean).

    I have tried subbing the tan(2x) for 2tan(x)/(1 - tan^2(x)), but I can't figure out where to go from there. I also tried to work on the other side -- sin(2x + x), use the sum identity, etc. -- but got really confused. I am doing trig on a distance learning course, and I don't really have access to a teacher or fellow students to talk it over with, so I really appreciate the help! I have been working on these way too long.
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  2. #2
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    Quote Originally Posted by sinewave85 View Post
    Ok, I know it seems like I am spunging, but there are thirty of these on this assignement, only stumped on two -- so far. The identity is:

    (sin(x) + sin(3x))/(cos(x) + cos(3x) = tan(2x)

    The only identities I have covered so far in my book are double-angle, half-angle, sum and difference, and the basics (reciprocal, quotentient, pythagorean).

    I have tried subbing the tan(2x) for 2tan(x)/(1 - tan^2(x)), but I can't figure out where to go from there. I also tried to work on the other side -- sin(2x + x), use the sum identity, etc. -- but got really confused. I am doing trig on a distance learning course, and I don't really have access to a teacher or fellow students to talk it over with, so I really appreciate the help! I have been working on these way too long.
    Use,
    \sin x+\sin y=2\sin \left( \frac{x+y}{2}\right) \cos \left( \frac{x-y}{2} \right)
    Und,
    \cos x+\cos y=2\cos \left( \frac{x+y}{2} \right) \cos \left( \frac{x-y}{2} \right)
    Thus,
    \sin x+\sin 3x=2\sin 2x \cos x
    \cos x+\cos 3x=2\cos 2x \cos x
    Thus,
    \frac{\sin x+\sin 3x}{\cos x+\cos 3x}=\frac{2\sin 3x\cos x}{2\cos 2x\cos x}=\frac{\sin 2x}{\cos 2x}=\tan 2x
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  3. #3
    Member sinewave85's Avatar
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    I am sorry, but I am not really familiar with that identity.
    I have learned sin(x + y), cos(x + y), but not sin(x) + sin(y)
    or cos(x) + cos(y). Your solution is very efficient, but I don't
    think I can use it. I have reworked the rs to:

    (-4sin^3(a) + 4sin(a))/(4cos^(3)a - 2cos(a))

    I there anywhere to go with that?
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  4. #4
    Member sinewave85's Avatar
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    Quote Originally Posted by sinewave85 View Post
    I have reworked the rs to:

    (-4sin^3(a) + 4sin(a))/(4cos^(3)a - 2cos(a))
    Correction: I have reworked the left side.
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  5. #5
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    Hello, sinewave85!

    You know enough to derive the necessary formulas . . .
    but it takes more work than we should be expected to do!


    \frac{\sin(x)  + \sin(3x)}{\cos(x)  + \cos(3x)} \:= \:\tan(2x)

    We need the Triple-angle Identities.
    We can keep them on a handy list (I do) . . . or derive them.


    \sin(3x) \;= \;\sin(2x + x) \;=\;\sin(x)\cos(2x) + \sin(2x)\cos(x)

    . . . . . . = \;\sin(x)\left[1 - 2\sin^2(x)\right] + \left[2\sin(x)\cos(x)\right]\cos(x)

    . . . . . . = \;\sin(x) - 2\sin^3(x) + 2\sin(x)\cos^2(x)

    . . . . . . = \;\sin(x) - 2\sin^3(x) + 2\sin(x)\left[1 - \sin^2(x)\right]

    . . . . . . = \;\sin(x) - 2\sin^3(x) + 2\sin(x) - 2\sin^3(x)

    Therefore: . \boxed{\sin(3x) \;=\;3\sin(x) - 4\sin^3(x)}


    \cos(3x) \;= \;\cos(x + 2x) \;= \;\cos(x)\cos(2x) - \sin(x)\sin(2x)

    . . . . . . = \;\cos(x)\left[2\cos^2(x) - 1\right] -\sin(x)\left[2\sin(x)\cos(x)\right]

    . . . . . . = \;2\cos^3(x) - \cos(x) - 2\sin^2(x)\cos(x)

    . . . . . . = \;2\cos^3(x) - \cos(x) - 2\left[1 - \cos^2(x)\right]\cos(x)

    . . . . . . = \;2\cos^3(x) - \cos(x) - 2\cos(x) + 2\cos^3(x)

    Therefore: . \boxed{\cos(3x)\;=\;4\cos^3(x) - 3\cos(x)}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Back to the problem . . .

    We have: . \frac{\sin x + \sin(3x)}{\cos(x) + \cos(3x)} \;= \;\frac{\sin(x) + \left[3\sin(x) - 4\sin^3(x)\right]}{\cos(x) + \left[4\cos^3(x) - 3\cos(x)\right]} \;=\;\frac{4\sin(x) - 4\sin^3(x)}{4\cos^3(x) - 2\cos(x)}

    . . . . . . = \;\frac{4\sin(x)\left[1 - \sin^2(x)\right]}{2\cos(x)\left[2\cos^2(x) - 1\right]} \;=\;\frac{4\sin(x)\cos^2(x)}{2\cos(x)\left[2\cos^2(x) - 1\right]} \;=\;\frac{2\sin(x)\cos(x)}{2\cos^2(x) - 1}

    . . . . . . = \;\frac{2\sin(x)\cos(x)}{2\left[1 - \sin^2(x)\right]-1} \;=\;\frac{2\sin(x)\cos(x)}{1 - 2\sin^2(x)}


    Divide top and bottom by \cos^2(x)

    . . \frac{\frac{2\sin(x)\cos(x)}{\cos^2(x)}}{\frac{1}{  \cos^2(x)} - \frac{2\sin^2(x)}{\cos^2{x}}} \;= \;\frac{2\frac{\sin(x)}{\cos(x)}}{\sec^2(x) - 2\left(\frac{\sin(x)}{\cos(x)}\right)^2} \;= \;\frac{2\tan(x)}{\left[\tan^2(x) + 1\right] - 2\tan^2(x)}

    . . = \;\frac{2\tan(x)}{1 - \tan^2(x)} \;= \;\tan(2x) . . . There!


    I need a nap . . .
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  6. #6
    Member sinewave85's Avatar
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    Thanks for all of the help! I do appreciate all of the work you did, and I hope you got your nap.
    Last edited by sinewave85; October 30th 2006 at 10:38 AM.
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