Trig identities using t-formulae (Edexcel A Level Further Maths: Further Pure 1): Revision Note

Trig identities using t-formulae

What are the t-formulae?

• The three t-formulae state that if $t=\tan \frac{\theta }{2}$ then

  • $\sin \theta =\frac{2t}{1+t^2}$

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

  • $\tan \theta =\frac{2t}{1-t^2}$

• They express $\sin \theta$, $\cos \theta$ and $\tan \theta$ in terms of one variable only, $t$

• From these, you can see the reciprocals

  • $\mathrm{cosec}\theta =\frac{1+t^2}{2t}$

  • $\sec \theta =\frac{1+t^2}{1-t^2}$

  • $\cot \theta =\frac{1-t^2}{2t}$

How do I derive the t-formulae?

• To derive $\sin \theta =\frac{2t}{1+t^2}$

  • first use the double-angle formula $\sin 2A\equiv 2\sin A\cos A$

    • $\sin \theta =2\sin \frac{\theta }{2}\cos \frac{\theta }{2}$

  • then multiply by $\frac{{\sec }^2\frac{\theta }{2}}{{\sec }^2\frac{\theta }{2}}$ and use $1+{\tan }^{2}A\equiv {\sec }^{2}A$ in the denominator

    • $\sin \theta \equiv 2\sin \frac{\theta }{2}\cos \frac{\theta }{2}\times \frac{{\sec }^2\frac{\theta }{2}}{{\sec }^2\frac{\theta }{2}}$

    • $\sin \theta \equiv \frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}{\sec }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}$

  • Simplify the numerator, $2sc\times \frac{1}{c^2}=\frac{2s}{c}=2t$

    • $\sin \theta \equiv \frac{2\tan \frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}=\frac{2t}{1+t^2}$

• To derive $\cos \theta =\frac{1-t^2}{1+t^2}$

  • first use the double-angle formula $\cos 2A\equiv {\cos }^{2}A-{\sin }^{2}A$

    • $\cos \theta \equiv {\cos }^2\frac{\theta }{2}-{\sin }^2\frac{\theta }{2}$

  • then multiply by $\frac{{\sec }^2\frac{\theta }{2}}{{\sec }^2\frac{\theta }{2}}$ and use $1+{\tan }^{2}A\equiv {\sec }^{2}A$ in the denominator

    • $\cos \theta \equiv \left({\cos }^2\frac{\theta }{2}-{\sin }^2\frac{\theta }{2}\right)\times \frac{{\sec }^2\frac{\theta }{2}}{{\sec }^2\frac{\theta }{2}}$

    • $\cos \theta \equiv \frac{\left({\cos }^2\frac{\theta }{2}-{\sin }^2\frac{\theta }{2}\right){\sec }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}$

  • Simplify the numerator, $\left(c^2-s^2\right)\times \frac{1}{c^2}=1-\frac{s^2}{c^2}=1-t^2$

    • $\cos \theta \equiv \frac{1-{\tan }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}=\frac{1-t^2}{1+t^2}$

• To derive $\tan \theta =\frac{2t}{1-t^2}$

  • either use the double-angle formula $\tan 2A\equiv \frac{2\tan A}{1-{\tan }^{2}A}$

    • $\tan \theta \equiv \frac{2\tan \frac{\theta }{2}}{1-{\tan }^2\frac{\theta }{2}}=\frac{2t}{1-t^2}$

  • or substitute the results for $\sin \theta$ and $\cos \theta$above into $\tan \theta \equiv \frac{\sin \theta }{\cos \theta }$

    • $\tan \theta \equiv \frac{\sin \theta }{\cos \theta }=\frac{2t}{1+t^2}÷\frac{1-t^2}{1+t^2}=\frac{2t}{\cancel{1+t^2}}\times \frac{\cancel{1+t^2}}{1-t^2}=\frac{2t}{1-t^2}$

How do I prove trigonometric identities using the t-formulae?

• To prove trigonometric identities using t-formulae

  • let $t=\tan \frac{\theta }{2}$

  • convert the trig functions on one or both sides of the identity into algebraic fractions in $t$ using

    • $\sin \theta =\frac{2t}{1+t^2}$, $\cos \theta =\frac{1-t^2}{1+t^2}$, $\tan \theta =\frac{2t}{1-t^2}$

    • $\mathrm{cosec}\theta =\frac{1+t^2}{2t}$, $\sec \theta =\frac{1+t^2}{1-t^2}$, $\cot \theta =\frac{1-t^2}{2t}$

  • rearrange the algebraic fractions in $t$ to prove the identity

    • e.g. adding, subtracting, multiplying, dividing

  • convert back to $\theta$ for the final step

    • using $t=\tan \frac{\theta }{2}$