The Student Room Group

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

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
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

Rsin(θ+α)=Rsin(a)cosθ+Rcos(a)sinθR\sin(\theta + \alpha) = R\sin(a)\cos\theta + R\cos(a)\sin\theta.

You can compare the terms,

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

and

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

Are you familiar with determining RR and α\alpha here?
Reply 2
Original post by _gcx
I would go with the latter, actually. Consider

Rsin(θ+α)=Rsin(a)cosθ+Rcos(a)sinθR\sin(\theta + \alpha) = R\sin(a)\cos\theta + R\cos(a)\sin\theta.

You can compare the terms,

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

and

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

Are you familiar with determining RR and α\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?
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 tanα=158\tan \alpha = \frac {15} 8 from my post? Are you sure you didn't misread Rcos(a)R\cos(a) for Rsin(a)R\sin(a) and vice versa? :smile:

Quick Reply

Latest