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Please help - c4 parametrics and double angle?

http://www.mei.org.uk/files/papers/2012_Jan_c4.pdf

its all of q8 but especially part 2, can someone give me the step by step as even mark scheme isnt making sense
Reply 1
You'll need to be a little more specific.
i get up to 3rd point of the mark scheme, then i dont get how we get to tan2 theta double angle from there, and from that i have no idea about the next part from that
Reply 3
Original post by liverpool2044
i get up to 3rd point of the mark scheme, then i dont get how we get to tan2 theta double angle from there, and from that i have no idea about the next part from that


Remember the identity that tan2θ=2tanθ1tan2θ\displaystyle \tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta}.

Now you know that your gradient of QP is 4t2t22=2tt21\frac{4t}{2t^2 - 2} = \frac{2t}{t^2 - 1}. But from part (i) you know that tanθ=1tt=1tanθ\tan \theta = \frac{1}{t} \Rightarrow t = \frac{1}{\tan \theta}.

So substitute this into your QP:

Unparseable latex formula:

\displaystyle[br]\begin{equation*} \frac{\frac{2}{\tan \theta}}{\frac{1}{\tan^2 \theta} - 1} = \cdots = \frac{2 \tan \theta}{1 - \tan^2 \theta}\end{equation*}



After some simplification and common denominators and cancelling common factors, etc...

But you know exactly what that identity is, so QP=tan2θQP = \tan 2\theta

Does this help at all?
Original post by Zacken
Remember the identity that tan2θ=2tanθ1tan2θ\displaystyle \tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta}.

Now you know that your gradient of QP is 4t2t22=2tt21\frac{4t}{2t^2 - 2} = \frac{2t}{t^2 - 1}. But from part (i) you know that tanθ=1tt=1tanθ\tan \theta = \frac{1}{t} \Rightarrow t = \frac{1}{\tan \theta}.

So substitute this into your QP:

Unparseable latex formula:

\displaystyle[br]\begin{equation*} \frac{\frac{2}{\tan \theta}}{\frac{1}{\tan^2 \theta} - 1} = \cdots = \frac{2 \tan \theta}{1 - \tan^2 \theta}\end{equation*}



After some simplification and common denominators and cancelling common factors, etc...

But you know exactly what that identity is, so QP=tan2θQP = \tan 2\theta


Does this help at all?


thank very helpful, its the few marks after too where it begins to ask about tpq can you help with that?
doesnt make sense
Original post by liverpool2044
thank very helpful, its the few marks after too where it begins to ask about tpq can you help with that?


if you extend the tangent line as far as the x axis and then consider the angles in the resulting triangle...

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