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Math Help - Trig identity

  1. #1
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    Trig identity

    We got this question the other day and I'm not exactly sure how to do it.

    Given A+B+C= pi

    Prove:

    cos(2A)+cos(2B)+cos(2C)+1 = -4cos(A)cos(B)cos(C)

    The hint was to use product formulas.

    Can anybody help me with how I would go about doing this?
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  2. #2
    Behold, the power of SARDINES!
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    Here is the idea

    -4\cos(A)\cos(B)\cos(C)=-2\cos(A)[2\cos(B)\cos(C)]

    using your hint the product identity is

    \cos(x+y)+\cos(x-y)=2\cos(x)\cos(y)

    -2\cos(A)[2\cos(B)\cos(C)]= -2\cos(A)[\cos(B+C)+\cos(B-C)]=

    -2\cos(A)\cos(B+C)-2\cos(A)\cos(B-C)

    using the product identiy again (twice) we end up with

    -\cos(A+B+C)-\cos(-A-B+C)-\cos(A-B+C)-\cos(-A+B+C)

    now we use the fact that A+B+C=\pi and

    notice that -A-B=C-\pi
    A+C=\pi -B
    and finally B+C=\pi -A

    sub these into the above and use the sum identity for cosine and you should be done.

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  3. #3
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    Quote Originally Posted by Shapeshift View Post
    We got this question the other day and I'm not exactly sure how to do it.

    Given A+B+C= pi

    Prove:

    cos(2A)+cos(2B)+cos(2C)+1 = -4cos(A)cos(B)cos(C)

    The hint was to use product formulas.

    Can anybody help me with how I would go about doing this?
    Given, A+B+C=\pi

    \Rightarrow A+B=\pi-C

    Now,

    LS = (\cos 2A + \cos 2B) + (\cos 2C) + 1

    = 2 \cos (A+B) \cos (A-B) + (2 \cos^2 C - 1) + 1

    [since,\;\; \cos A + \cos B = 2\cos \left( \frac{A+B}{2}\right) \cos \left( \frac{A-B}{2}\right)\;\;\;and \;\;\; \cos 2\theta=2 \cos^2 \theta - 1 \;\;]

    =  2 \cos (A+B) \cos (A-B) + 2 \cos^2 C

    =  2 \cos (\pi-C) \cos (A-B) + 2 \cos^2 C

    =  -2 \cos C \cos (A-B) + 2 \cos^2 C

    =-2 \cos C \left[ \cos (A-B) - \cos C \right]

    =-2 \cos C \left[ \cos (A-B) - \cos (\pi-(A+B)) \right]

    =-2 \cos C \left[ \cos (A-B) + \cos (A+B) \right]

    =-2 \cos C (2 \cos A \cos B)

    =-4 \cos A \cos B \cos C

    Did you get it now ???
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