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Math Help - 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)
     <br />
\cos\left(\frac{\pi}{3}\right) \sin(\theta) + \sin\left(\frac{\pi}{3}\right) \cos(\theta)<br />

    or

     <br />
\sin\left(\frac{\pi}{6}\right) \sin(\theta) + \cos\left(\frac{\pi}{6}\right) \cos(\theta)<br />

    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 \tfrac12 can be written as \cos\tfrac{\pi}{3} and that \tfrac{\sqrt3}{2} is the sine of the same angle ( \tfrac{\pi}{3}), here's the more general method for handling this type of expression:

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

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

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

    \tfrac12 = r \cos\phi (1)

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

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

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

    Square (1) and (2) and add:

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

    \Rightarrow r = 1 (taking the positive root)

    So there are the values of r and \phi.

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

    Grandad
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