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C3 trigonometry

trig.jpg
I'm not sure where to start with this one. I think the identity I need to use is: R(cos(x-alpha) = Rcosxcos(alpha) + Rsinxsin(alpha)
Reply 1
You are correct, but from the way you are talking it sounds as if you haven't seen this done before. It's fairly straightforward, but has lots of steps.

How many lots of cos x do you have? Rcos(alpha) on the RHS, 2 on the LHS, so Rcos(alpha) = 2.

Do the same thing with sin x. You'll now have two equations.

Square both equations and add them together. This should eliminate the trig functions and let you find R.

Now go back to the two equations, and divide the sin one by the cos one. This should eliminate R and let you find alpha.
Original post by Pangol
You are correct, but from the way you are talking it sounds as if you haven't seen this done before. It's fairly straightforward, but has lots of steps.

How many lots of cos x do you have? Rcos(alpha) on the RHS, 2 on the LHS, so Rcos(alpha) = 2.

Do the same thing with sin x. You'll now have two equations.

Square both equations and add them together. This should eliminate the trig functions and let you find R.

Now go back to the two equations, and divide the sin one by the cos one. This should eliminate R and let you find alpha.


Hmm, I think it's because this identity is fairly new to me, not completely used to it yet.
So far I've got R = sqrt(29), that seem okay to you?
Reply 3
Original post by pineneedles
Hmm, I think it's because this identity is fairly new to me, not completely used to it yet.
So far I've got R = sqrt(29), that seem okay to you?


Yep, that's it. You might have spotted a shortcut while doing that which means you can always get R pretty quickly.

Now - follow the other step, and you should get tan(alpha) = ...
Original post by Pangol
Yep, that's it. You might have spotted a shortcut while doing that which means you can always get R pretty quickly.

Now - follow the other step, and you should get tan(alpha) = ...


I got tan(alpha) = 2/5
Thank you for your help :h:

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