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Math Help - trigonometric equation?

  1. #1
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    trigonometric equation?

    hello i would like to find the smallest angle in degrees from the equation

    3sinθ+2cosθ=3

    i think that i need the identity sin^2θ=cos^2θ=1 to solve this and i was then unsure where to head to next either squaring the equation to match the identity or because there is no powers on either side working it out as sin^2θ=1-cos^2θ which i put like
    3(1-cosθ)+2cosθ=3.

    could someone please tell me if i am going right in any way and what the next steps are that i have to take....without to many calculations so i can have another go to understand thanks olie
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  2. #2
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    Re: trigonometric equation?

    Quote Originally Posted by ibiza007 View Post
    hello i would like to find the smallest angle in degrees from the equation

    3sinθ+2cosθ=3

    i think that i need the identity sin^2θ=cos^2θ=1 to solve this and i was then unsure where to head to next either squaring the equation to match the identity or because there is no powers on either side working it out as sin^2θ=1-cos^2θ which i put like
    3(1-cosθ)+2cosθ=3.

    could someone please tell me if i am going right in any way and what the next steps are that i have to take....without to many calculations so i can have another go to understand thanks olie
    first off , \sin{t} \ne 1 - \cos{t} ... so get that idea out of your head.

    note that ...

    a\sin{t} + b\sin{t} = R\sin(x + \phi)

    R = \sqrt{a^2+b^2}

    \tan{\phi} = \frac{b}{a}

    for your problem ...

    3\sin{t} + 2\sin{t} = 3 = \sqrt{13}\sin(x + \phi)

    \sin(x + \phi) = \frac{3}{\sqrt{13}}

    x + \phi = \arcsin\left(\frac{3}{\sqrt{13}}\right)

    x = \arcsin\left(\frac{3}{\sqrt{13}}\right) - \arctan\left(\frac{2}{3}\right) \approx 22.6^\circ
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  3. #3
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    Re: trigonometric equation?

    You can also try another approach.
     3\sin{\theta}+2\cos{\theta}=3

    \Rightarrow (3\sin{\theta}+2\cos{\theta})^2=9...........[Square both the sides]

    \Rightarrow 9\sin^2{\theta}+4\cos^2{\theta}+12\sin{\theta}\cos  {\theta}=9

    \Rightarrow 5\sin^2{\theta}+4(\sin^2{\theta}+\cos^2{\theta})+1  2\sin{\theta}\cos{\theta}=9

    \Rightarrow 5\sin^2{\theta}+4+12\sin{\theta}\cos{\theta}=9

    \Rightarrow 5\sin^2{\theta}+12\sin{\theta}\cos{\theta}=5

    \Rightarrow 12\sin{\theta}\cos{\theta}=5-5\sin^2{\theta}

    \Rightarrow 12\sin{\theta}\cos{\theta}=5(1-\sin^2{\theta})

    \Rightarrow 12\sin{\theta}\cos{\theta}=5(\cos^2{\theta})

    \Rightarrow 12\sin{\theta}=5(\cos{\theta})

    \Rightarrow \frac{\sin{\theta}}{\cos{\theta}}=\frac{5}{12}

    \Rightarrow \tan{\theta}=\frac{5}{12}

    \Rightarrow \theta=\tan^{-1}(\frac{5}{12})

    \Rightarrow \theta \approx 22.62^{\circ}
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  4. #4
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    Re: trigonometric equation?

    thankyou for your help. you use t in your formula what does it represent? also i have read up on this method but i keep getting stuck im trying to get two angles within the range 0-360 degrees i looked at a formula close to this which is to write the formula as Rcos(x-a)(or Rsin(x-a) i get the almost there when
    x-a = arcsin(3/square root 13) then the results don't match yours, i also don't understand why you subtracted arctan(2/3) at the end.
    in the example i have read they get an angle closer to 0 and an angle closer to 360 please can you help me understand a bit further please i thought i was nearly there a few times but don't quite make it each time.
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  5. #5
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    Re: trigonometric equation?

    Quote Originally Posted by ibiza007 View Post
    thankyou for your help. you use t in your formula what does it represent? also i have read up on this method but i keep getting stuck im trying to get two angles within the range 0-360 degrees i looked at a formula close to this which is to write the formula as Rcos(x-a)(or Rsin(x-a) i get the almost there when
    x-a = arcsin(3/square root 13) then the results don't match yours, i also don't understand why you subtracted arctan(2/3) at the end.
    in the example i have read they get an angle closer to 0 and an angle closer to 360 please can you help me understand a bit further please i thought i was nearly there a few times but don't quite make it each time.
    t is just short for \theta ... cause I'm too lazy to type it out all the time.

    go back and look at my solution again ... if \tan{\phi} = \frac{b}{a} = \frac{2}{3} , then \phi = \arctan\left(\frac{2}{3}\right)
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