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Math Help - Trig

  1. #1
    Member srirahulan's Avatar
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    Arrow Trig

    5cos^2{\theta}+18cos{\theta}sin{\theta}+29sin^2{\t  heta} Simplify this into a+bcos(2{\theta}+{\alpha}) a,b are constant and \alpha is completely different from \theta
    Last edited by srirahulan; October 22nd 2013 at 08:46 AM.
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  2. #2
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    Re: Trig

    Show what work you have so far.
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  3. #3
    Member srirahulan's Avatar
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    Re: Trig

    I write that like this 2cos^2{\theta}-1+1+3cos^2{\theta}+9sin2{\theta}+29sin^2{\theta} But then i could not find out the \alpha It is very difficult to arrange them

    A correction--- that 29sin^2{\theta}
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  4. #4
    Member srirahulan's Avatar
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    Re: Trig

    How can i find out the alpha and simplify like this...... please help??
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  5. #5
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    Re: Trig

    Try using the sum of angles formula for \cos and see what you get: a+b\cos(2\theta+\alpha) = a+b[\cos(2\theta)\cos(\alpha)-\sin(2\theta)\sin(\alpha)]... Keep trying manipulations until you get something close to what you started with.
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  6. #6
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    Re: Trig

    Trig-25-oct-13-3.png
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  7. #7
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    Re: Trig

    Hello, srirahulan!

    5\cos^2\theta+18\cos\theta\sin\theta+29\sin^2\thet  a

    \text{Simplify to }\,a+b\cos(2\theta+\alpha),\:\text{ where }a,b\text{ are constants and }\alpha \ne \theta

    5\cos^2\theta + 18\cos\theta\sin\theta + 29\sin^2\theta

    . . =\;5\left(\frac{1+\cos2\theta}{2}\right) + 9\sin2\theta + 29\left(\frac{1-\cos2\theta}{2}\right)

    . . =\;\tfrac{1}{2}\big(5 + 5\cos2\theta + 18\sin2\theta + 29 - 29\cos2\theta\big)

    . . =\;\tfrac{1}{2}\big[34 - 24\cos2\theta + 18\sin2\theta\big]

    . . =\;\tfrac{1}{2}\bigg[34 - 6\big(4\cos2\theta - 3\sin2\theta\big)\bigg]

    . . =\;\tfrac{1}{2}\bigg[34 - 30\left(\tfrac{4}{5}\cos2\theta - \tfrac{3}{5}\sin2\theta\right)\bigg]


    \text{Let }\cos\alpha = \tfrac{4}{5},\;\sin\alpja = \tfrac{3}{5}

    \text{We have: }\:\tfrac{1}{2}\bigg[34 - 30\left(\cos\alpha\cos2\theta - \sin\alpha\sin2\theta\right)\bigg]

    . . . . . =\;\tfrac{1}{2}\bigg[34 - 30\cps(2\theta + \alpha)\bigg]

    . . . . . =\;17 - 15\cos(2\theta + \alpha)
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