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Thread: reduction of a cosΘ+/_ b sinΘ

  1. #1
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    reduction of a cosΘ+/_ b sinΘ

    reduct this problem..

    1/2 sinθ + /2 cos Θ

    to the form

    a. r sin (Θ+Φ)
    b. r cos(Θ+Φ)
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  2. #2
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    Quote Originally Posted by jasonlewiz View Post
    reduct this problem..

    1/2 sinθ + /2 cos Θ

    to the form

    a. r sin (Θ+Φ)
    b. r cos(Θ+Φ)
    $\displaystyle
    \cos\left(\frac{\pi}{3}\right) \sin(\theta) + \sin\left(\frac{\pi}{3}\right) \cos(\theta)
    $

    or

    $\displaystyle
    \sin\left(\frac{\pi}{6}\right) \sin(\theta) + \cos\left(\frac{\pi}{6}\right) \cos(\theta)
    $

    finish up by using the sum/difference identities for sine and cosine
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  3. #3
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    Handling expressions of the form a sin x + b cos x

    Hello jasonlewiz
    Quote Originally Posted by jasonlewiz View Post
    reduct this problem..

    1/2 sinθ + /2 cos Θ

    to the form

    a. r sin (Θ+Φ)
    b. r cos(Θ+Φ)
    skeeter's solution is all you need for this particular problem. But, in case you don't spot that $\displaystyle \tfrac12$ can be written as $\displaystyle \cos\tfrac{\pi}{3}$ and that $\displaystyle \tfrac{\sqrt3}{2}$ is the sine of the same angle ($\displaystyle \tfrac{\pi}{3}$), here's the more general method for handling this type of expression:

    $\displaystyle \tfrac12\sin\theta + \tfrac{\sqrt3}{2}\cos\theta = r \sin(\theta + \phi)$

    $\displaystyle = r\sin\theta\cos\phi + r\cos\theta\sin\phi$

    Compare coefficients of $\displaystyle \sin\theta$ and $\displaystyle \cos\theta$:

    $\displaystyle \tfrac12 = r \cos\phi$ (1)

    $\displaystyle \tfrac{\sqrt3}{2} = r\sin\phi$ (2)

    Divide (2) by (1): $\displaystyle \tan\phi = \sqrt3$

    $\displaystyle \Rightarrow \phi = \tfrac{\pi}{3}$

    Square (1) and (2) and add:

    $\displaystyle \tfrac14 + \tfrac34 = r^2(\sin^2\phi + \cos^2\phi)=r^2$

    $\displaystyle \Rightarrow r = 1$ (taking the positive root)

    So there are the values of $\displaystyle r$ and $\displaystyle \phi$.

    You can do part (b) in exactly the same way, but starting with $\displaystyle \tfrac12\sin\theta + \tfrac{\sqrt3}{2}\cos\theta = r \cos(\theta + \phi)$. (In fact, it's slightly easier to start with $\displaystyle \tfrac12\sin\theta + \tfrac{\sqrt3}{2}\cos\theta = r \cos(\theta - \phi)$ since $\displaystyle \phi$ then turns out to be a positive angle.)

    Grandad
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