Trigonometry
1. Radians, arcs and sectors
Radians are the natural angle unit for calculus and sector geometry.
Angle conversion
\[180^\circ=\pi\text{ rad}\]Arc length
\[s=r\theta\]Sector area
\[A=\frac12r^2\theta\]Sector problem
A sector has radius 8 and angle \(1.2\) rad. Then \[s=8(1.2)=9.6,\qquad A=\frac12(8^2)(1.2)=38.4.\]Common mistake. The formulas \(s=r heta\) and \(A= frac12r^2 heta\) require \( heta\) in radians.
2. Sine rule, cosine rule and triangle area
Choosing the correct rule is part of the assessment.
Sine rule
\[\frac a{\sin A}=\frac b{\sin B}=\frac c{\sin C}\]Cosine rule
\[a^2=b^2+c^2-2bc\cos A\]Area
\[\text{Area}=\frac12ab\sin C\]- Label each side opposite its corresponding angle.
- Use cosine rule for SSS or SAS; use sine rule when an opposite side–angle pair is available.
- For the ambiguous SSA sine-rule case, consider whether a second angle \(180^\circ-B\) is possible.
- Check angle sum and physical geometry before accepting a solution.
Exam tip. Store full calculator precision until the final answer. Triangle chains amplify early rounding errors.
3. Three-dimensional trigonometry
3D trigonometry is usually two linked 2D problems.
- Draw or isolate the relevant right triangle in one face/section.
- Find any base diagonal first using Pythagoras or cosine rule.
- Then use a second triangle for the space diagonal or required angle.
- Clearly distinguish an angle with a line from an angle with a plane.
Cuboid pattern
For cuboid side lengths \(a,b,c\), the base diagonal is \(\sqrt{a^2+b^2}\), and the space diagonal is \[d=\sqrt{a^2+b^2+c^2}.\] An angle between the space diagonal and base plane can then be found from a right triangle with vertical side \(c\) and base diagonal \(\sqrt{a^2+b^2}\).Exam tip. Redraw the 2D triangle you are actually solving. Working directly on a crowded 3D diagram causes wrong-angle errors.
4. Identities and formulae
Memorise a compact identity set, then derive variants rather than memorising dozens of separate formulas.
Pythagorean
\[\sin^2x+\cos^2x=1\]Tangent
\[\tan x=\frac{\sin x}{\cos x}\]Addition
\[\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B\]Cosine addition
\[\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B\]Double angle
\[\sin2x=2\sin x\cos x,\quad \cos2x=\cos^2x-\sin^2x=1-2\sin^2x=2\cos^2x-1\]Identity proof strategy
To prove an identity, transform one side only. Replace \(\tan x\) by \(\sin x/\cos x\), factor, use \(\sin^2x+\cos^2x=1\), and stop when the other side appears.Common mistake. Do not “prove” an identity by assuming both sides are equal and manipulating both simultaneously. Keep a clear one-side-to-the-other chain.
5. Trigonometric equations
Equation solving combines identity algebra with periodic graph knowledge.
- Reduce to one trig function where possible.
- Solve the basic equation for a reference angle/value.
- Generate all solutions in the stated interval using quadrant signs or graph periodicity.
- If the equation contains \(2x\) or another multiple angle, solve for that angle over the correspondingly enlarged interval, then divide back.
- Check endpoints and units (degrees/radians).
Double-angle interval
Solve \(\sin2x=\frac12\) for \(0^\circ\le x\le180^\circ\). Then \(0^\circ\le2x\le360^\circ\), so \[2x=30^\circ,150^\circ\Rightarrow x=15^\circ,75^\circ.\]Exam tip. Write the transformed interval next to the transformed angle. This single habit prevents many missing/extra solutions.
6. Trigonometric graphs and transformations
Graph questions assess period, amplitude, phase/shift, range and asymptotes.
Sine
\[y=\sin x:\ \text{amplitude }1,\ \text{period }2\pi\]Cosine
\[y=\cos x:\ \text{amplitude }1,\ \text{period }2\pi\]Tangent
\[y=\tan x:\ \text{period }\pi,\ \text{vertical asymptotes at }x=\frac\pi2+k\pi\]General sine/cos
\[y=A\sin(Bx+C)+D:\ \text{amplitude }|A|,\ \text{period }\frac{2\pi}{|B|}\]Transformation
For \(y=3\cos(2x)-1\), amplitude \(=3\), midline \(y=-1\), and period \(=\pi\). Mark quarter-period key points before drawing a smooth curve.Exam tip. On a graph sketch, label the scale, midline, maxima/minima or asymptotes, and enough key x-values to prove the correct period.