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Math Help - Trig problem

  1. #1
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    Trig problem

    Hi just started trig for university and stuck on this problem highly appreciate if someone can point me in the right direction thanks

    Find "R" and "a" such that:

    9sin% + 4cos% = Rsin(% + a)

    % = angle theta in radian.
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  2. #2
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    Re: Trig problem

    Hello, ZKG9744!

    If you "just started trig", this is a very fancy problem.


    \text{Find }R\text{ and }a\text{ such that: }\:9\sin\theta + 4\cos\theta \:=\: R\sin(\theta + a)

    We have: . 9\sin\theta + 4\cos\theta

    Multiply by \tfrac{\sqrt{97}}{\sqrt{97}}\!:\;\;\tfrac{\sqrt{97  }}{\sqrt{97}}(9\sin\theta + 4\cos\theta) \;=\;\sqrt{97}\left(\tfrac{9}{\sqrt{97}}\sin\theta + \tfrac{4}{\sqrt{97}}\cos\theta\right)

    \text{Let }\,\sin a \,=\,\tfrac{4}{\sqrt{97}},\:\cos a \,=\,\tfrac{9}{\sqrt{97}}


    We have: . \sqrt{97}(\cos a\sin\theta + \sin a\cos\theta) \;=\;\sqrt{97}\sin(\theta + a)

    \text{Therefore: }\:R \,=\,\sqrt{97},\;\; a \,=\,\sin^{\text{-}1}\!\left(\tfrac{4}{\sqrt{97}}\right)
    Thanks from ZKG9744
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  3. #3
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    Re: Trig problem

    Thanks understanding is better less fuzzy now. Sqr(97/97) still confused where that came from sorry
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  4. #4
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    Re: Trig problem

    If you think of a right angled triangle with the side opposite a having length equal to 4 and the side adjacent to a having the length 9. The hypotenuse, by pythagoras' theorem is \sqrt{4^2+9^2}=\sqrt{97}
    Then with the basic definition of sin: \sin{a}=\frac{opposite}{hypotenuse}=\frac{4}{\sqrt  {97}}. Similar thing with cos.

    You might find it easier to do the problem by first expanding sin(\theta+a)
    Last edited by Shakarri; August 27th 2013 at 05:24 PM.
    Thanks from ZKG9744
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  5. #5
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    Re: Trig problem

    @ Shakarri
    Thank you very very much
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