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Rcosalpha, how do I know to use double angle formula and not the factor formula? Watch

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    For example:

    Solve between 0 and 360, 8cosθ + 15sinθ = 10

    How do I know to use the double angle formula and not the factor formula?

    Also, I've seen the solutions so I know it's the double angle formula. However, none of the formulas (eg. Cos(A+-B) = CosACosB -+ sinAsinB) have just a cos and a sin in them, they have a coscos and a sinsin (just like the one I just said as an example), so how do you know which of the double angle formulas to use?

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    (Original post by vector12)
    For example:

    Solve between 0 and 360, 8cosθ + 15sinθ = 10

    How do I know to use the double angle formula and not the factor formula?

    Also, I've seen the solutions so I know it's the double angle formula. However, none of the formulas (eg. Cos(A+-B) = CosACosB -+ sinAsinB) have just a cos and a sin in them, they have a coscos and a sinsin (just like the one I just said as an example), so how do you know which of the double angle formulas to use?

    Thanks
    I would go with the latter, actually. Consider

    R\sin(\theta + \alpha) = R\sin(a)\cos\theta + R\cos(a)\sin\theta.

    You can compare the terms,

    R\sin(a)\cos\theta = 8\cos\theta \implies R\sin(a) = 8

    and

    R\cos(a)\sin\theta = 15\sin\theta \implies R\cos(a) = 15 .

    Are you familiar with determining R and \alpha here?
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    (Original post by _gcx)
    I would go with the latter, actually. Consider

    R\sin(\theta + \alpha) = R\sin(a)\cos\theta + R\cos(a)\sin\theta.

    You can compare the terms,

    R\sin(a)\cos\theta = 8\cos\theta \implies R\sin(a) = 8

    and

    R\cos(a)\sin\theta = 15\sin\theta \implies R\cos(a) = 15 .

    Are you familiar with determining R and \alpha here?
    What do you mean by you would go with the latter?

    Yes, I do know how to compare terms to determine R and alpha. Looking at the mark scheme, I don't see how your way would work since tan alpha = Rsin(a)/Rcos(a) which would be 8/15, like the mark scheme, rather than your way which would give tan alpha to be 15/8, which would mess up the value for alpha wouldn't it?
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    (Original post by vector12)
    What do you mean by you would go with the latter?

    Yes, I do know how to compare terms to determine R and alpha. Looking at the mark scheme, I don't see how your way would work since tan alpha = Rsin(a)/Rcos(a) which would be 8/15, like the mark scheme, rather than your way which would give tan alpha to be 15/8, which would mess up the value for alpha wouldn't it?
    How did you get \tan \alpha = \frac {15} 8 from my post? Are you sure you didn't misread R\cos(a) for R\sin(a) and vice versa?
 
 
 
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