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Math Help - tricky trig equations

  1. #1
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    tricky trig equations

    Express 3cosx-4sinx in the form Rcos(x+a) for some R>0 and 0<a<90 giving the value of a correct to the nearest minute

    Hence solve the equation 3cosx-4sinx=5 for in the domain of [0,360]
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  2. #2
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    Hi requal

    I'm not sure what you mean by "nearest minute". Maybe it means nearest degree?

    Let :
    p cos x - q sin x = R cos (x+a)

    R=\sqrt{p^2+q^2}

    Then, expand RHS and compare it with LHS to find cos(a) and sin(a). Hence, find a
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  3. #3
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    Quote Originally Posted by songoku View Post
    Hi requal

    I'm not sure what you mean by "nearest minute". Maybe it means nearest degree?

    [snip]
    60 minutes = 1 degree: 60 \, ' = 1^0.

    60 seconds = 1 minute: 60 \, '' = 1 \, '
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  4. #4
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    Hi mr fantastic

    Oh I see. Find (a) in degree first then convert it to minute.

    Thanks
    Last edited by songoku; September 14th 2009 at 03:02 AM.
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  5. #5
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    sorry, to bump but I'm still confused. Why does R equal the square root of p^2+q^2
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  6. #6
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    Hi requal

    p cos x - q sin x = R cos (x+a)
    p cos x - q sin x = R cos(x) cos(a) - R sin(x) sin(a)

    Comparing coefficient :
    p cos x = R cos(x) cos(a)
    p = R cos(a) .........................(1)

    q sin x = R sin(x) sin(a)
    q = R sin(a)...........................(2)

    Square (1) and (2) and add them :
    p^2 = R^2 [cos(a)]^2
    q^2 = R^2 [sin(a)]^2
    --------------------------- +
    p^2 + q^2 = R^2
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