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Thread: Find constants A and B

  1. #1
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    Find constants A and B

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

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

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

    Find the constants $\displaystyle A$ and $\displaystyle B$

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


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


    Equate coefficients: .$\displaystyle \begin{Bmatrix}A - B &=& 4 \\ A+B &=& 0 \end{Bmatrix}$ . . . . 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...
    $\displaystyle
    \int_{0}^{\frac{1}{4}\pi}\frac{4sin\theta}{sin\the ta+cos\theta}d\theta$
    giving the answer in the form $\displaystyle a\pi-lnb$

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