State the t-formula for \cos\theta.

\cos\theta=\frac{1-t^2}{1+t^2}.

Define \sec\theta, \operatorname{cosec}\theta and \cot\theta.

\sec\theta=\frac{1}{\cos\theta}, \operatorname{cosec}\theta=\frac{1}{\sin\theta}, \cot\theta=\frac{1}{\tan\theta}.

Which identity gives \cos^2\frac{\theta}{2}=\frac{1}{1+t^2}?

\sec^2A=1+\tan^2A, with A=\frac{\theta}{2}.

How is \sin\theta=\frac{2t}{1+t^2} derived?

\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}=2\tan\frac{\theta}{2}\cos^2\frac{\theta}{2}=\frac{2t}{1+t^2}.

Write 1+\cos\theta and 1-\cos\theta in terms of t.

1+\cos\theta=\frac{2}{1+t^2} and 1-\cos\theta=\frac{2t^2}{1+t^2}.

What do you get from a\cos x+b\sin x=c after substituting the t-formulae?

A quadratic in t: (c+a)t^2-2bt+(c-a)=0.

Why must x=\pi be checked separately?

\tan\frac{\pi}{2} is undefined, so the substitution cannot produce a solution at x=\pi.

If \tan\frac{\theta}{2}=\frac12, what is \cos\theta?

\frac{1-\frac14}{1+\frac14}=\frac35.

The t-formulaeEdexcel A-Level Further Maths: Flashcards

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In the t-formulae, what is $t$?

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In the t-formulae, what is $t$ ?: $t=\tan\frac{\theta}{2}$ .
State the t-formula for $\sin\theta$ .: $\sin\theta=\frac{2t}{1+t^2}$ .
State the t-formula for $\cos\theta$ .: $\cos\theta=\frac{1-t^2}{1+t^2}$ .
State the t-formula for $\tan\theta$ .: $\tan\theta=\frac{2t}{1-t^2}$ .
Define $\sec\theta$ , $\operatorname{cosec}\theta$ and $\cot\theta$ .: $\sec\theta=\frac{1}{\cos\theta}$ , $\operatorname{cosec}\theta=\frac{1}{\sin\theta}$ , $\cot\theta=\frac{1}{\tan\theta}$ .
Which identity gives $\cos^2\frac{\theta}{2}=\frac{1}{1+t^2}$ ?: $\sec^2A=1+\tan^2A$ , with $A=\frac{\theta}{2}$ .
How is $\sin\theta=\frac{2t}{1+t^2}$ derived?: $\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}=2\tan\frac{\theta}{2}\cos^2\frac{\theta}{2}=\frac{2t}{1+t^2}$ .
Write $\operatorname{cosec}\theta$ in terms of $t$ .: $\frac{1+t^2}{2t}$ .
Write $\cot\theta$ in terms of $t$ .: $\frac{1-t^2}{2t}$ .
Write $1+\cos\theta$ and $1-\cos\theta$ in terms of $t$ .: $1+\cos\theta=\frac{2}{1+t^2}$ and $1-\cos\theta=\frac{2t^2}{1+t^2}$ .
What do you get from $a\cos x+b\sin x=c$ after substituting the t-formulae?: A quadratic in $t$ : $(c+a)t^2-2bt+(c-a)=0$ .
Why must $x=\pi$ be checked separately?: $\tan\frac{\pi}{2}$ is undefined, so the substitution cannot produce a solution at $x=\pi$ .
If $\tan\frac{\theta}{2}=\frac12$ , what is $\cos\theta$ ?: $\frac{1-\frac14}{1+\frac14}=\frac35$ .
For $0\le x<2\pi$ , how do you convert a negative $t$ into $x$ ?: $x=2\arctan t+2\pi$ .

Exam questions on The t-formulae

Given that $\theta$ is acute and $\tan\frac{\theta}{2}=\frac12$ .
Find the exact value of $\operatorname{cosec}\theta+\cot\theta$ .2 marks
Let $t=\tan\frac{x}{2}$ , where $0<x<\pi$ and $x\neq\frac{\pi}{2}$ .
Show that $\sec x+\tan x=\dfrac{1+t}{1-t}$ .2 marks
Consider the equation $3\cos x+4\sin x=2$ for $0\le x<2\pi$ , and let $t=\tan\frac{x}{2}$ .
Show that the equation can be written as $5t^2-8t-1=0$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).