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    Could anybody solve this for me?? Im doin OCR pure 3, Trig and missed a lesson on it. Any help would be greatful!!

    Find the value of a between 0 and 90 for which

    3sinx + 2cosx = 13^0.5 sin (x + a )

    Thanks
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    Sin[x+h] = Sin[x]Cos[h] + Cos[x]Sin[h]

    So Cos[a] = 3/root(13), and Sin[a] = 2/root(13)

    ArcSin[2/root(13)] = 33.69... Degrees.
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    Key: 1r2 = 1/2

    3sinx + 2cosx = (13^1r2)(sin(x+a))

    sin(x+a) = sinxcosa + cosxsina From standard formula
    (13^1r2)(sin(x+a)) = (13^1r2)(sinxcosa) + (13^1r2)(cosxsina) Multiply both sides by (13^1r2)

    Therefore,

    3sinx + 2cosx = (13^1r2)(sinxcosa) + (13^1r2)(cosxsina)

    Equate co-efficients of sinx:

    3 = (13^1r2)cosa

    Therefore, a = inverse cos (3/(13^1r2))

    = 0.588002603 radians (33.69 degrees)
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    Why did you write 1r2 instead of 1/2? And he wanted it in degrees, and I already answered the question.
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    I wrote 1r2 to make it clearer. On a computer, it's not always easy to tell what someone means (for example, in enchan's inital post, he could have meant 13^(0.5 sin (x + a ))

    You hadn't answered the question when I clicked on the 'Post Reply' button. We were figuring it out at the same time, you just happened to finish 3 minutes before I did.

    And you're right, sorry, my answer should have been in degrees.
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    Fair enough .
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    Alternative method of solution.

    3sinx + 2cosx = 13^0.5 sin (x + a ) is an eqn, therfore valid for all x (within its domain)

    so let x = 0,

    then,

    2 = √(13).sin(a)
    a = arcsin(2/√(13))
    a = 33.69°
    ========
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    Is everyone here forgetting the simple P2 results, namely:

    ================================ ==========================

    asinx + bcosx = Rsin(x + a)

    where:

    R = √(a² + b²)
    a = arctan(b/a)

    ================================ ==========================

    using this for the stated equation, where a=3,b=2, we get:

    3sinx + 2cosx = √(2² + 3²)sin(x + a)
    = √(13)sin(x+a)

    a = arctan(2/3)
    a = 33.69°

    3sinx + 2cosx = √(13)sin(x + 33.69°)
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    I never bother to remember specific results like that. I find it is better to memorise the really essential identities and then apply them.
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    (Original post by AntiMagicMan)
    I never bother to remember specific results like that. I find it is better to memorise the really essential identities and then apply them.
    ur a 2nd year undergrad...alevel students will find it easier to remember these results...although i think its better to do it ur way as it promotes understanding of the issue involved rather than just pluggin values in the required results...
 
 
 
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