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Math Help - tan and cos

  1. #1
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    tan and cos

    if (a+b)tan(theta-phi) = (a-b)tan(theta + phi) and a cos2 phi + b cos2 theta =c then prove that a square - b square + c square = 2 ac cos2 phi.

    my soln.
    a+b/a-b = tan(t + p)/tan(t-p)
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: tan and cos

    Rewrite:
    a^2-b^2+c^2=2ac\cos(2\phi) as
    \left[(a-b)(a+b)\right]+c^2=2ac\cos(2\phi)

    Maybe you can do something with this.
    ...
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  3. #3
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    Re: tan and cos

    Quote Originally Posted by Siron View Post
    Rewrite:
    a^2-b^2+c^2=2ac\cos(2\phi) as
    \left[(a-b)(a+b)\right]+c^2=2ac\cos(2\phi)

    Maybe you can do something with this.
    ...
    not being able to proceed.
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  4. #4
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    Opalg's Avatar
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    Re: tan and cos

    Quote Originally Posted by saha.subham View Post
    if (a+b)\tan(\theta-\phi) = (a-b)\tan(\theta + \phi) and a \cos(2 \phi) + b \cos(2 \theta) =c then prove that a^2-b^2+c^2 = 2ac\cos(2\phi).
    In the first equation, write the tans as sin/cos and then use the trig formulas for sums and products, to get

    (a+b)\sin(\theta-\phi)\cos(\theta + \phi) = (a-b)\sin(\theta+\phi)\cos(\theta - \phi),

    (a+b)\bigl(\sin(2\theta)-\sin(2\phi)\bigr) = (a-b)\bigl(\sin(2\theta) + \sin(2\phi)\bigr),

    b\sin(2\theta) = a\sin(2\phi).\qquad(1)

    You also have the equation a \cos(2 \phi) + b \cos(2 \theta) =c, from which

    b \cos(2 \theta) = c - a \cos(2 \phi).\qquad(2)

    Square the equations (1) and (2) and add:

    b^2 = a^2\sin^2(2\phi) + (c - a \cos(2 \phi))^2.

    That simplifies to give the result that you want.

    Notice that this is really a problem about a triangle, with sides a, b, c and angles 2\theta (opposite the side a) and 2\phi (opposite the side b). The equation (1) is then the sine rule, and the equation a^2-b^2+c^2 = 2ac\cos(2\phi) is the cosine rule.
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