Trigonometric proofs and applicationsAQA A-Level Maths: Revision notes

Proofs use a small toolkit of identities, all true wherever both sides are defined:

• $\sin^2\theta+\cos^2\theta\equiv1$, $\;1+\tan^2\theta\equiv\sec^2\theta$, $\;1+\cot^2\theta\equiv\operatorname{cosec}^2\theta$
• $\tan\theta\equiv\frac{\sin\theta}{\cos\theta}$, $\;\sec\theta\equiv\frac{1}{\cos\theta}$, $\;\operatorname{cosec}\theta\equiv\frac{1}{\sin\theta}$
• Addition formulae: $\sin(A\pm B)\equiv\sin A\cos B\pm\cos A\sin B$ and $\cos(A\pm B)\equiv\cos A\cos B\mp\sin A\sin B$
• Double angle formulae: $\sin2A\equiv2\sin A\cos A$, $\;\cos2A\equiv\cos^2A-\sin^2A\equiv2\cos^2A-1\equiv1-2\sin^2A$, $\;\tan2A\equiv\frac{2\tan A}{1-\tan^2A}$ Choose the form of $\cos2A$ that matches what you want to cancel: $2\cos^2A-1$ to leave only cosines, $1-2\sin^2A$ to leave only sines.

To prove an identity, start from the more complicated side and transform it step by step until it equals the other side. Do not work on both sides at once and do not cross-multiply as if it were an equation, because that assumes the result you are proving. Useful moves: write everything in terms of $\sin$ and $\cos$; use a Pythagorean identity to convert squares; replace $\sin2A$ or $\cos2A$; factorise (difference of two squares is common); combine fractions over a common denominator. Finish with a clear concluding statement. Example: prove $\frac{\sin2\theta}{1+\cos2\theta}\equiv\tan\theta$. $\text{LHS}=\frac{2\sin\theta\cos\theta}{1+(2\cos^2\theta-1)}=\frac{2\sin\theta\cos\theta}{2\cos^2\theta}=\frac{\sin\theta}{\cos\theta}=\tan\theta=\text{RHS}.$ The identity holds except where $\cos\theta=0$, where both sides are undefined.

Example: $\sin^4x-\cos^4x\equiv(\sin^2x-\cos^2x)(\sin^2x+\cos^2x)\equiv\sin^2x-\cos^2x\equiv-\cos2x$. Example: prove $\cos\theta+\sin\theta\equiv\sqrt2\cos\left(\theta-\frac{\pi}{4}\right)$. Expand the right-hand side: $\sqrt2\left(\cos\theta\cos\frac{\pi}{4}+\sin\theta\sin\frac{\pi}{4}\right)=\sqrt2\cdot\frac{\sqrt2}{2}(\cos\theta+\sin\theta)=\cos\theta+\sin\theta.$ This shows why $\cos\theta+\sin\theta$ has greatest value $\sqrt2$ and least value $-\sqrt2$. A disproof needs only one counter-example. To show $\sin2\theta\equiv2\sin\theta$ is false, take $\theta=30^{\circ}$: $\sin60^{\circ}=\frac{\sqrt3}{2}\ne2\times\frac12=1$.

A projectile launched at speed $u$ at angle $\theta$ to the horizontal has components $u\cos\theta$ (horizontal, constant) and $u\sin\theta$ (initial vertical). Landing on level ground needs $0=(u\sin\theta)T-\frac12gT^2$, so $T=\frac{2u\sin\theta}{g}$. $\text{Range}=u\cos\theta\times T=\frac{2u^2\sin\theta\cos\theta}{g}=\frac{u^2\sin2\theta}{g}.$ The range is greatest when $\sin2\theta=1$, so $\theta=45^{\circ}$. For a range less than the maximum, $\sin2\theta=\sin(180^{\circ}-2\theta)$ gives two angles that sum to $90^{\circ}$. Example: $u=20$, range 30 m: $\sin2\theta=\frac{30\times9.8}{400}=0.735$, so $\theta=23.7^{\circ}$ or $66.3^{\circ}$.

Resolve a vector of magnitude $F$ at angle $\alpha$ to the $x$-axis into components $F\cos\alpha$ and $F\sin\alpha$. For two forces, add the components, then find the resultant magnitude with Pythagoras and its direction with $\tan\phi=\frac{\text{vertical}}{\text{horizontal}}$. Example: forces of 8 N along the $x$-axis and 7 N at $60^{\circ}$ to it. Horizontal: $8+7\cos60^{\circ}=11.5$. Vertical: $7\sin60^{\circ}=\frac{7\sqrt3}{2}$. Magnitude $=\sqrt{11.5^2+36.75}=13$ N, at $\tan^{-1}\left(\frac{3.5\sqrt3}{11.5}\right)=27.8^{\circ}$ to the 8 N force. A force that balances them has the same magnitude and the opposite direction. On a slope at angle $\alpha$, weight $mg$ has components $mg\sin\alpha$ down the slope and $mg\cos\alpha$ into it.

In context questions state what the model assumes (no air resistance, a particle, a smooth surface), keep calculator values unrounded until the end, and give answers to 3 s.f. unless told otherwise. Check that the calculator is in radians or degrees as the question requires. After solving, return to the context: a range must be positive and a launch angle must lie between $0^{\circ}$ and $90^{\circ}$, so reject solutions that do not fit.