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Math Help - Find constants A and B

  1. #1
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    Find constants A and B

    Given that
    A(sin\theta+cos\theta)+B(cos\theta-sin\theta)=4sin\theta
    Find the constants A and B.

    Can't see what i need to do here.
    Thanks
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  2. #2
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    Hello, George321!

    Given that: . A(\sin\theta+\cos\theta)+B(\cos\theta-\sin\theta)\:=\:4\sin\theta

    Find the constants A and B

    The left side is: . A\sin\theta + A\cos\theta + B\cos\theta - B\sin\theta \;=\;(A-B)\sin\theta + (A+B)\cos\theta


    The equaton becomes: . (A-B)\sin\theta + (A+B)\cos\theta \;=\;4\sin\theta


    Equate coefficients: . \begin{Bmatrix}A - B &=& 4 \\ A+B &=& 0 \end{Bmatrix} . . . . and solve the system.

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  3. #3
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    Thanks for your help, im now stuck on the second bit though...
     <br />
\int_{0}^{\frac{1}{4}\pi}\frac{4sin\theta}{sin\the  ta+cos\theta}d\theta
    giving the answer in the form a\pi-lnb

    Can't remember how to integrate this kind of equation...
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  4. #4
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    \int_{0}^{\frac{\pi}{4}}{\dfrac{4\sin{\theta}}{\si  n{\theta}+\cos{\theta}}\;{d\theta}} = \int_{0}^{\frac{\pi}{4}}{\dfrac{2\left(\sin{\theta  }+\cos{\theta}\right)-2\left(\cos{\theta}-\sin{\theta}\right)}{\sin{\theta}+\cos{\theta}}\;{  d\theta}}  = \int_{0}^{\frac{\pi}{4}}\left(\dfrac{2\left(\sin{\  theta}+\cos{\theta}\right)}{\sin{\theta}+\cos{\the  ta}}-\dfrac{2\left(\cos{\theta}-\sin{\theta}\right)}{\sin{\theta}+\cos{\theta}}\ri  ght)\;{d\theta} = \int_{0}^{\frac{\pi}{4}}\left(2-\dfrac{2\left(\cos{\theta}-\sin{\theta}\right)}{\sin{\theta}+\cos{\theta}}\ri  ght)\;{d\theta} = \int_{0}^{\frac{\pi}{4}}\left(2-2\left\{\dfrac{\cos{\theta}-\sin{\theta}}{\sin{\theta}+\cos{\theta}}\right\}\r  ight)\;{d\theta} = \bigg[2\theta-2\log\left(\sin{\theta}+\cos{\theta}\right)\bigg]_{0}^{\frac{\pi}{4}} = \boxed{\dfrac{1}{2}\pi-\ln(2)}.
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