Using t = tan(θ)/(2), write cosθ in terms of t. [1 mark]

Give the general solution of sinθ = (1)/(2). [2 marks]

Write sinθ + sin 2θ + sin 3θ as the imaginary part of a geometric series. [2 marks]

Pearson Edexcel Edexcel 2019-style practice: Use the substitution t = tan(θ)/(2) to find int (1)/(1 + sinθ)\,dθ.

Substitute the standard t-formulae, simplify the rational integrand, then integrate. With t = tan(θ)/(2): sinθ = (2t)/(1 + t^2) and dθ = (2)/(1 + t^2)\,dt (M1). 1 + sinθ = 1 + (2t)/(1 + t^2) = (1 + t^2 + 2t)/(1 + t^2) = ((1 + t)^2)/(1 + t^2) (A1). So the integral becomes int (1 + t^2)/((1 + t)^2)·(2)/(1 + t^2)\,dt = int (2)/((1 + t)^2)\,dt (M1 A1). = -(2)/(1 + t) + c = -(2)/(1 + tan(θ)/(2)) + c (A1 A1).

Pearson Edexcel Edexcel 2022-style practice: Show that _k=0^n-1cos(kθ) can be written using a geometric series of complex exponentials, and hence find a closed form for the sum when θ ≠ 2mπ.

Treat the cosine sum as the real part of a geometric series in e^iθ. _k=0^n-1cos(kθ) = Re_k=0^n-1 e^ikθ (M1). The sum is geometric with first term 1 and ratio e^iθ (A1). _k=0^n-1 e^ikθ = e^inθ - 1e^iθ - 1 (M1). Factoring e^i nθ/2 from the top and e^iθ/2 from the bottom gives e^inθ/2e^iθ/2·e^inθ/2 - e^-inθ/2e^iθ/2 - e^-iθ/2 = e^i(n-1)θ/2·(sin(nθ/2))/(sin(θ/2)) (M1 A1). Taking the real part: _k=0^n-1cos(kθ) = (sin(nθ/2))/(sin(θ/2))cos((n-1)θ)/(2) (A1 A1).

EnglandFurther MathsSyllabus dot point

How do you sum trigonometric series and work with the t-substitution and inverse trig functions?

The t-substitution for trigonometric integrals and equations, summing series of sines and cosines, the general solution of trigonometric equations, and inverse trigonometric functions.

A focused answer to the Edexcel A-Level Further Mathematics Further Pure further trigonometry content, covering the t-substitution for trigonometric integrals and equations, summing finite series of sines and cosines, finding general solutions of trigonometric equations, and working with the inverse trigonometric functions.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The t-substitution
Summing trigonometric series
General solutions and inverse functions
Examples in context
Try this

What this dot point is asking

Edexcel Further Pure wants you to use the $t = \tan\frac{\theta}{2}$ substitution to turn trigonometric integrals and equations into rational ones, sum finite series of sines and cosines (typically via complex exponentials), find general solutions of trigonometric equations, and handle the inverse trigonometric functions with their restricted ranges. These techniques bring together calculus, complex numbers and trigonometric identities.

The t-substitution

Setting $t = \tan\frac{\theta}{2}$ converts every trigonometric function of $\theta$ into a rational function of $t$ , which can then be integrated by partial fractions or recognised standard forms. The substitution is the standard route for integrals like $\int \frac{1}{a + b\cos\theta}\,d\theta$ that resist other methods.

Integrate with the t-substitution

Find $\int \frac{1}{1 + \cos\theta}\,d\theta$ .

Step 1: Replace every trig expression using the standard formulae

Set $t = \tan\frac{\theta}{2}$ . The substitution gives $\cos\theta = \frac{1 - t^2}{1 + t^2}$ and $d\theta = \frac{2}{1 + t^2}\,dt$ . Both facts come directly from the double-angle formulae applied to the half-angle. Writing everything in terms of $t$ turns the trigonometric integral into a rational one.

Step 2: Simplify the denominator

Combine the constant $1$ and the expression for $\cos\theta$ over a common denominator:

1 + \cos\theta = 1 + \frac{1 - t^2}{1 + t^2} = \frac{(1 + t^2) + (1 - t^2)}{1 + t^2} = \frac{2}{1 + t^2}.

The $t^2$ terms cancel, leaving just $\frac{2}{1 + t^2}$ .

Step 3: Substitute and integrate

The integrand becomes $\frac{1}{2/(1 + t^2)} = \frac{1 + t^2}{2}$ , and after multiplying by $d\theta = \frac{2}{1 + t^2}\,dt$ the $(1 + t^2)$ factors cancel completely:

\int \frac{1}{1 + \cos\theta}\,d\theta = \int \frac{1 + t^2}{2} \cdot \frac{2}{1 + t^2}\,dt = \int 1\,dt = t + c.

Step 4: Convert back to $\theta$

Since $t = \tan\frac{\theta}{2}$ , the final result is $\tan\frac{\theta}{2} + c$ .

Final answer: $\displaystyle\int \frac{1}{1 + \cos\theta}\,d\theta = \tan\frac{\theta}{2} + c$ .

Summing trigonometric series

A sum of sines or cosines whose angles are in arithmetic progression is the imaginary or real part of a geometric series of complex exponentials $e^{ik\theta}$ . Sum the geometric series with the standard formula, then extract the real part (for cosines) or imaginary part (for sines), simplifying with the factor-out trick that produces $\sin$ ratios.

Sum a cosine series with complex exponentials

Express $\cos\theta + \cos 3\theta + \cos 5\theta$ as the real part of a geometric series and find its closed form.

The sum equals $\operatorname{Re}\left(e^{i\theta} + e^{3i\theta} + e^{5i\theta}\right)$ , a geometric series with first term $e^{i\theta}$ and common ratio $e^{2i\theta}$ .

Summing, $\dfrac{e^{i\theta}(e^{6i\theta} - 1)}{e^{2i\theta} - 1}$ . Factoring $e^{3i\theta}$ from the numerator bracket and $e^{i\theta}$ from the denominator bracket leads to $\dfrac{\sin 3\theta}{\sin\theta}\cdot\dfrac{1}{2}$ type expressions; taking the real part gives the closed form $\dfrac{\sin 3\theta}{2\sin\theta}\cdot 2\cos 3\theta$ which simplifies using product-to-sum identities.

General solutions and inverse functions

A trigonometric equation has infinitely many solutions, captured by a general formula built from one principal solution. The inverse trigonometric functions, by contrast, return a single value within a restricted range, so you must adjust for the quadrant you actually need.

Examples in context

Further trigonometry sits at the crossroads of the course. Summing trig series is a direct application of de Moivre's theorem and geometric series from complex numbers, the single most important technique here. The $t$ -substitution is one of several integration tools alongside the partial fractions and standard integrals of further calculus. Osborn's rule for hyperbolic identities (in hyperbolic functions) is the mirror image of the trig identities used throughout this topic. The inverse trig functions reappear as the results of the standard integrals $\int \frac{1}{a^2 + x^2}\,dx$ and $\int \frac{1}{\sqrt{a^2 - x^2}}\,dx$ .

Common traps

Forgetting the $d\theta$ factor: The $t$ -substitution changes $d\theta$ to $\frac{2}{1 + t^2}\,dt$ ; omitting this factor gives the wrong integral entirely.
Dropping the general solution: A trig equation usually needs the full family (such as $n\pi + (-1)^n\alpha$ ), not just the principal value, unless a range is specified.
Ignoring inverse function ranges: $\arccos$ returns a value in $[0, \pi]$ only and $\arcsin$ in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ ; adjust for the quadrant the problem actually requires.
Taking the wrong part of the complex sum: Cosine series are the real part and sine series the imaginary part of $\sum e^{ik\theta}$ ; confusing the two flips the answer.

Try this

Q1. Using $t = \tan\frac{\theta}{2}$ , write $\cos\theta$ in terms of $t$ . [1 mark]

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

Q2. Give the general solution of $\sin\theta = \frac{1}{2}$ . [2 marks]

Cue. $\theta = n\pi + (-1)^n\frac{\pi}{6}$ .

Q3. Write $\sin\theta + \sin 2\theta + \sin 3\theta$ as the imaginary part of a geometric series. [2 marks]

Cue. $\operatorname{Im}\left(e^{i\theta} + e^{2i\theta} + e^{3i\theta}\right)$ , first term $e^{i\theta}$ , ratio $e^{i\theta}$ .

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20196 marksUse the substitution

t = \tan\frac{\theta}{2}

to find

\int \frac{1}{1 + \sin\theta}\,d\theta

Show worked answer →

Substitute the standard $t$ -formulae, simplify the rational integrand, then integrate.

With $t = \tan\frac{\theta}{2}$ : $\sin\theta = \frac{2t}{1 + t^2}$ and $d\theta = \frac{2}{1 + t^2}\,dt$ (M1).

$1 + \sin\theta = 1 + \frac{2t}{1 + t^2} = \frac{1 + t^2 + 2t}{1 + t^2} = \frac{(1 + t)^2}{1 + t^2}$ (A1).

So the integral becomes $\int \frac{1 + t^2}{(1 + t)^2}\cdot\frac{2}{1 + t^2}\,dt = \int \frac{2}{(1 + t)^2}\,dt$ (M1 A1).

$= -\frac{2}{1 + t} + c = -\frac{2}{1 + \tan\frac{\theta}{2}} + c$ (A1 A1).

Edexcel 20227 marksShow that

\sum_{k=0}^{n-1}\cos(k\theta)

can be written using a geometric series of complex exponentials, and hence find a closed form for the sum when

\theta \neq 2m\pi

Show worked answer →

Treat the cosine sum as the real part of a geometric series in $e^{i\theta}$ .

$\sum_{k=0}^{n-1}\cos(k\theta) = \operatorname{Re}\sum_{k=0}^{n-1} e^{ik\theta}$ (M1). The sum is geometric with first term $1$ and ratio $e^{i\theta}$ (A1).

$\sum_{k=0}^{n-1} e^{ik\theta} = \frac{e^{in\theta} - 1}{e^{i\theta} - 1}$ (M1). Factoring $e^{i n\theta/2}$ from the top and $e^{i\theta/2}$ from the bottom gives $\frac{e^{in\theta/2}}{e^{i\theta/2}}\cdot\frac{e^{in\theta/2} - e^{-in\theta/2}}{e^{i\theta/2} - e^{-i\theta/2}} = e^{i(n-1)\theta/2}\cdot\frac{\sin(n\theta/2)}{\sin(\theta/2)}$ (M1 A1).

Taking the real part: $\sum_{k=0}^{n-1}\cos(k\theta) = \frac{\sin(n\theta/2)}{\sin(\theta/2)}\cos\frac{(n-1)\theta}{2}$ (A1 A1).

Related dot points

Sources & how we know this

Pearson Edexcel A-Level Further Mathematics (9FM0) specification — Pearson Edexcel (2017)