The t-formulaeEdexcel A-Level Further Maths: Revision notes

The t-formulae express $\sin\theta$, $\cos\theta$ and $\tan\theta$ as rational functions of one variable, $t=\tan\frac{\theta}{2}$. The reciprocal functions are defined by $\sec\theta=\frac{1}{\cos\theta},\quad\operatorname{cosec}\theta=\frac{1}{\sin\theta},\quad\cot\theta=\frac{1}{\tan\theta}=\frac{\cos\theta}{\sin\theta}.$ They are not defined where the function in the denominator is zero. Once $\sin\theta$, $\cos\theta$ and $\tan\theta$ are written in terms of $t$, the three reciprocals follow by taking reciprocals.

Start from the double-angle formulae with $\frac{\theta}{2}$ as the angle, and use $\sec^2\frac{\theta}{2}=1+t^2$, so $\cos^2\frac{\theta}{2}=\frac{1}{1+t^2}$. $\sin\theta=2\sin\tfrac{\theta}{2}\cos\tfrac{\theta}{2}=2\tan\tfrac{\theta}{2}\cos^2\tfrac{\theta}{2}=\frac{2t}{1+t^2}.$ $\cos\theta=\cos^2\tfrac{\theta}{2}-\sin^2\tfrac{\theta}{2}=\cos^2\tfrac{\theta}{2}\left(1-\tan^2\tfrac{\theta}{2}\right)=\frac{1-t^2}{1+t^2}.$ For the tangent, use $\tan2A=\frac{2\tan A}{1-\tan^2A}$ with $A=\frac{\theta}{2}$, or divide the sine by the cosine: $\tan\theta=\frac{2t}{1-t^2}.$ These derivations can be asked for directly.

To prove an identity, replace every trigonometric function by its expression in $t$, combine into a single fraction and simplify. Example: show $\frac{1+\operatorname{cosec}\theta}{\cot\theta}\equiv\frac{1+\tan\frac{\theta}{2}}{1-\tan\frac{\theta}{2}}$. With $\operatorname{cosec}\theta=\frac{1+t^2}{2t}$ and $\cot\theta=\frac{1-t^2}{2t}$, $\frac{1+\frac{1+t^2}{2t}}{\frac{1-t^2}{2t}}=\frac{2t+1+t^2}{1-t^2}=\frac{(1+t)^2}{(1-t)(1+t)}=\frac{1+t}{1-t}.$ Another: $\sec\theta+\tan\theta=\frac{1+t^2+2t}{1-t^2}=\frac{1+t}{1-t}$ as well. Look for a perfect square $(1+t)^2$ or a difference of squares $1-t^2=(1-t)(1+t)$; they are the usual route to the final form.

Substitute $\cos x=\frac{1-t^2}{1+t^2}$ and $\sin x=\frac{2t}{1+t^2}$, multiply through by $1+t^2$, and solve the quadratic in $t$. Example: $3\cos x+4\sin x=2$ for $0\le x<2\pi$. Then $3(1-t^2)+8t=2(1+t^2)$, giving $5t^2-8t-1=0$, so $t=\frac{8\pm\sqrt{84}}{10}=1.7165\ldots$ or $-0.1165\ldots$. Then $x=2\arctan t$: $x=2.09$. For $t=-0.1165\ldots$, $2\arctan t=-0.232$, which is outside the range, so add $2\pi$: $x=6.05$. The answers can be checked in the original equation.

$t=\tan\frac{x}{2}$ is undefined when $x=\pi$ (and every odd multiple of $\pi$), so a solution at $x=\pi$ would be lost by the substitution. Always test $x=\pi$ in the original equation. In the example, $3\cos\pi+4\sin\pi=-3\neq2$, so no solution is missed. If the range is $0\le x<2\pi$, then $0\le\frac{x}{2}<\pi$, and each value of $t$ gives exactly one $x$ in the range.