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C4 - Help with this trig equation? watch

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    This is an equation of the form Asinθ + Bcosθ, I have managed to solve it for one root but I'm not sure how to get the other.

    The question is:
    Solve 8cosθ + 15sinθ = 8.5, for 0 < θ < 360.
    I managed to get it in the form Rcos(θ + a) where R = 17 and a = 61.9 degrees, but I put the two expressions onto my graphical calculator and they produce different graphs! I'm not sure why this is, is it because I chose the wrong form?

    I put my equation equal to 8.5 and tried solving for θ but it was negative (I got -1.9 degrees) so I added 360 to that to get 358.1 degrees. I'm now not sure why I can't find the other root.

    Could someone help me? It's been happening in a few questions so I'd be very grateful.
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    (Original post by OL1V3R)
    ...
    Check your expansion of Cos(a+b), in particular the signs!

    You want to be using.....

    Spoiler:
    Show

    Either Cos(a-b)
    or Sin(a+b)


    Edit: Also, there will be more that just one solution. Go for a general solution and get it into the form "theta = ...." before you apply any restrictions due to the required interval.
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    (Original post by ghostwalker)
    Check your expansion of Cos(a+b), in particular the signs!
    Oh wait, yep you're right, when equating coefficients of sinθ to get a, I accidentally left out the minus sign! That probably explains it!

    Silly me! I do tend to make quite a few silly mistakes like that!

    In other questions where you have to find two roots of an equation like this, what geometric reasoning would you use to find the other one?
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    So you had a negative sign in your expansion and there was a positive sign in the original question.
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    (Original post by stevencarrwork)
    So you had a negative sign in your expansion and there was a positive sign in the original question.
    Where I lost a sign was here:
    8cosθ + 15sinθ = 8.5
    8cosθ + 15sinθ = Rcos(θ+a)
    R = sqrt(8² + 15²) = 17
    therefore 8cosθ + 15sinθ = 17cos(θ+a)
    therefore 8cosθ + 15sinθ = 17cosθcosa - 17sinθsina

    equating coefficients of sinθ giveS:
    15 = -17sina
    It was here that I forgot the minus sign, but I've included it here.
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    It should be cos(A-B)

    With cis you reverse the sign.
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    (Original post by OL1V3R)
    Where I lost a sign was here:
    8cosθ + 15sinθ = 8.5
    8cosθ + 15sinθ = Rcos(θ+a)
    R = sqrt(8² + 15²) = 17
    therefore 8cosθ + 15sinθ = 17cos(θ+a)
    therefore 8cosθ + 15sinθ = 17cosθcosa - 17sinθsina

    equating coefficients of sinθ giveS:
    15 = -17sina
    It was here that I forgot the minus sign, but I've included it here.
    So when you solve for a, you will get a negative angle.

    And all will then be right.
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    (Original post by OL1V3R)
    In other questions where you have to find two roots of an equation like this, what geometric reasoning would you use to find the other one?
    I wouldn't, I'd use the algebra, eg.

    If you had sin(x+40) = 1/2

    then x+40 = 30 or 150, + n360 (where n is an integer)

    so x = -10 or 110, + n360

    Now if I want to restrict the interval to 0 to 360, i get the 2 solutions 110 and 350. (the first arising with n = 0 and the second with n=1; this will vary from question to question)

    PS: Just to confirm what stevencarrwork said, your use of the negative angle like that is correct.
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    Ah okay, I see what I did there. Cheers! Rep will go to all of you!

    So I just need to remember to keep checking the sign and also always add 360 to my negative angle. Got it!
 
 
 
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