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Math Help - trigonometric equations

  1. #1
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    trigonometric equations

    I can't seem to do (a)(ii) of this question, there are probably two ways to do it but i get stuck using my way.
    Given 4cos\theta+3sin\theta=Rcos(\theta-\alpha) find the value of E and the value of \alpha where R>0 and 0< \alpha<\frac{\pi}{2}
    [I got R = 5 and [tex]\alpha=0.6435 rads which is right]

    Hence ind all values between 0 and 2 \pi satisfying
    (a)(i) 4cos \theta+3sin\theta=2
    [i also got this right its 1.8 and 5.77 rads]
    (ii) 4cos 2\theta+3sin2\theta=5cos\theta
    i expanded again and got 5cos( (2\theta-0.64350)=5cos\theta but i dont know where to go from here,,,ive come across a similar scenario before and cant remember what I did, If anybody could help that'd be great, thnx
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  2. #2
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    If two angles α and β (between 0 and 2π) have the same cosine then either α = β or α + β = 2π.
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  3. #3
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    hmm but the answer givwen is 0.21,0.64,2.31 and 4.4 rads how did u get at least two of them exactly,,,do you use sina +sinb=2sin(a+b/2)cos(a+b/2) or is that possible?'', oh i see what you mean now but that only gives me 0.6435 ans 0.231
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  4. #4
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    You know that 5\cos(2\theta - 0.64350) = 5\cos\theta. Cancel the 5s, then you see that you have two angles with the same cosine. there are then four possibilities:

    1. 2\theta - 0.6435 = \theta.

    2. 2\theta - 0.6435 + \theta = 2\pi \approx 6.2632 .

    3. 2\theta - 0.6435 + \theta = 0 .

    4. 2\theta - 0.6435 + \theta = 4\pi \approx 12.5264 .

    You could add any multiple of 2π to the right-hand side of 1. or 2., but the only ways to get solutions for θ in the interval from 0 to 2π are the four given above.
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  5. #5
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    thanks!! I get the right answer now, Is that some rule to remember i.e for the limit of 2 \pi you can only go to twice i.e 4pi or is there some kinda proof i could search in the internet thanks!?
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