Further Trigonometry

Further trigonometry builds on the Advanced focus areas Trigonometry and Measure of Angles and Trigonometric Identities and Equations. It has been worth about 3–10 marks a paper, and the identities and the auxiliary angle method are tools you will use again in calculus (integrating ${\sin }^{2}x$), inverse trig and vectors, so learn them well.

• 3D trigonometry: interpret a three-dimensional situation from a diagram or from words and solve it with right-angled trigonometry and the sine and cosine rules.
• Identities: derive and use the sum and difference expansions of $\sin$, $\cos$ and $\tan$, and the double angle formulas (including three forms of $\cos 2A$).
• Equations: solve trig equations over a restricted domain using factorisation and identities; write $a\cos x+b\sin x$ as $R\cos (x\pm \alpha )$ or $R\sin (x\pm \alpha )$ and use it to graph, find maximum and minimum values and solve equations; interpret solutions graphically.

The 2024 syllabus does not include the $t$-formulas, products to sums, or general solutions of trig equations. All equations are solved over a stated domain.

Most 3D questions combine two or three right-angled triangles in different planes that share an edge. A good method:

1. Draw a clear diagram with the ground as a horizontal parallelogram and vertical lines drawn vertically.
2. Mark every right angle. A vertical pole is perpendicular to every line on the ground through its foot.
3. Identify each triangle you need and redraw it flat, on its own.
4. Write each unknown length in terms of one variable (often the height $h$), then join the triangles with the sine or cosine rule or Pythagoras.

Quick example. A rectangular box has a base 6 cm by 8 cm and height 5 cm. The base diagonal is $\sqrt{6^2+8^2}=10$ cm. The space diagonal rises 5 cm over that 10 cm, so the angle it makes with the base satisfies $\tan \theta =\frac{5}{10}$, giving $\theta \approx 26.6^{\circ }$. The key step is recognising that the space diagonal, the base diagonal and a vertical edge form a right-angled triangle.

Sum and difference formulas

Note the sign swap in the cosine formula: $\cos (A+B)$ has a minus. The syllabus asks you to derive these. A common derivation finds $\cos (A-B)$ by applying the cosine rule and the distance formula to two points on the unit circle, then gets the others by substitution ($B\to -B$, and $\sin \theta =\cos (\frac{\pi }{2}-\theta )$).

Double angle formulas

Put $B=A$ in the sum formulas:

Choose the form of $\cos 2A$ that matches the rest of the problem: if the equation also has $\sin A$, use $1-2{\sin }^{2}A$. Rearranged, these give ${\sin }^{2}A=\frac{1}{2}(1-\cos 2A)$ and ${\cos }^{2}A=\frac{1}{2}(1+\cos 2A)$, which you will need for integration in Year 12.

Worked example 1: exact value

Find the exact value of $\tan 105^{\circ }$.

Write $105^{\circ }=60^{\circ }+45^{\circ }$:

Rationalise by multiplying top and bottom by $1+\sqrt{3}$:

Check: $105^{\circ }$ is in the second quadrant, where tangent is negative.

Proving identities

Start with the more complicated side and work towards the other side. Do not treat the identity as an equation (cross-multiplying, or doing the same thing to both sides), because that assumes what you are proving. Simplifying the LHS and the RHS separately until they reach the same expression is fine. For example, to prove ${\cos }^{4}A-{\sin }^{4}A=\cos 2A$:

Using factorisation and identities

When an equation mixes angles (such as $2x$ and $x$), use a double angle formula to get everything in terms of one angle and one function, then factorise. Never divide both sides by a trig expression that could be zero; you would lose solutions.

Worked example 2: equation with a double angle

Solve $\cos 2x=3\sin x-1$ for $0\le x\le 2\pi$.

Replace $\cos 2x$ with $1-2{\sin }^{2}x$ so everything is in $\sin x$:

$\sin x=-2$ has no solutions, since $-1\le \sin x\le 1$. So $\sin x=\frac{1}{2}$, which in the first and second quadrants gives

The auxiliary angle method

Any expression $a\cos x+b\sin x$ is a single wave. Expanding $R\cos (x-\alpha )=R\cos x\cos \alpha +R\sin x\sin \alpha$ and matching coefficients gives

with $R>0$ and $\alpha$ in the quadrant given by the signs of $a$ and $b$. The same matching works for $R\sin (x+\alpha )$ and the other forms; always expand the form the question asks for and match coefficients, rather than relying on a memorised version.

Because $\cos$ ranges from $-1$ to $1$, the maximum value of $a\cos x+b\sin x$ is $R$ and the minimum is $-R$.

Worked example 3: solving with the auxiliary angle

Solve $8\cos x+15\sin x=8.5$ for $0\le x\le 2\pi$, giving answers correct to 3 decimal places.

Rewrite. Let $8\cos x+15\sin x=R\cos (x-\alpha )$:

($\alpha$ is acute because $R\cos \alpha =8$ and $R\sin \alpha =15$ are both positive.)

Solve. $17\cos (x-\alpha )=8.5$, so $\cos (x-\alpha )=\frac{1}{2}$. Adjust the domain: $x-\alpha$ lies between $-1.081$ and $5.202$. In that range,

(The next value, $\frac{5\pi }{3}\approx 5.236$, is just outside.) Therefore

Both check: substituting into $8\cos x+15\sin x$ gives $8.5$. The maximum value of the left side is 17, so an equation like $8\cos x+15\sin x=20$ would have no solutions.

Interpreting solutions graphically

Solutions of $f(x)=g(x)$ are the $x$-coordinates of intersections of the two graphs. Sketching both (for example a sine curve and a line) tells you how many solutions to expect and roughly where, even when you cannot solve exactly. This is also the quickest way to answer "how many solutions" multiple choice questions.

• Writing $\sin (A+B)=\sin A+\sin B$. Expansion formulas are not distributive.
• Getting the sign of $\cos (A\pm B)$ the wrong way round.
• Dividing both sides of an equation by $\sin x$ or $\cos x$ and losing the solutions where it is zero; factorise instead.
• Forgetting to adjust the domain when solving for $2x$ or $x-\alpha$, so solutions are missed or extra ones are included.
• Putting $\alpha$ in the wrong quadrant in the auxiliary angle method; check the signs of $R\cos \alpha$ and $R\sin \alpha$.
• Calculator in degrees when the question is in radians (or the reverse).
• Proving an identity by manipulating it as an equation (cross-multiplying or starting from the result).

• Prove or show that: set out LHS $=$ … $=$ RHS, one step per line, naming the identity used when it is not obvious.
• Hence means use the previous part. If part (i) asks for $R\cos (x-\alpha )$, part (ii) almost always wants you to use it to solve an equation or find a maximum.
• Give exact answers (surds, multiples of $\pi$) unless the question asks for decimals. Keep full calculator values in intermediate steps.
• In 3D questions, a neat diagram with right angles marked often earns the first mark. State which triangle each step uses.
• For a maximum or minimum, say where it occurs as well as the value if the question asks for both.

• Advanced, Trigonometry and Measure of Angles and Trigonometric Identities and Equations: exact values, the unit circle, radians, Pythagorean identities, the sine and cosine rules and basic trig equations are assumed knowledge.
• Further work with functions (ME11-FN): graphs of $\sec x$, $\mathrm{cosec}x$ and $\cot x$.
• Inverse trigonometric functions (ME12-IT): sum and difference formulas are used to evaluate expressions such as $\sin ({\tan }^{-1}a+{\tan }^{-1}b)$.
• Further calculus skills (ME12-CS): double angle formulas turn ${\sin }^{2}x$ and ${\cos }^{2}x$ into forms you can integrate.
• Introduction to vectors (ME12-VE): 3D thinking and the cosine rule connect to angles between vectors.