Trigonometry

Solve the equation $\sin^{-1}(3x) = -\frac{1}{3}\pi$, and give the solution exactly.

Feb/March 2016

Solve the equation $3\sin^2 2\theta + 8\cos 2\theta = 0$ for $0^\circ \leq \theta \leq 180^\circ$.

Feb/March 2019

Solve the equation $3\tan^2 x - 5\tan x - 2 = 0$ in the interval $0^\circ \leq x \leq 180^\circ$.

Feb/March 2020

Find the values of $\theta$ that satisfy $\frac{\tan\theta + 3\sin\theta + 2}{\tan\theta - 3\sin\theta + 1} = 2$ for $0^\circ \leq \theta \leq 90^\circ$.

Feb/March 2020

Solve the equation $\dfrac{\tan\theta + 2\sin\theta}{\tan\theta - 2\sin\theta} = 3$, with $0^{\circ} < \theta < 180^{\circ}$.

Feb/March 2021

Show that the expression $\dfrac{\sin \theta + 2\cos \theta}{\cos \theta - 2\sin \theta} - \dfrac{\sin \theta - 2\cos \theta}{\cos \theta + 2\sin \theta}$ is equal to $\dfrac{4}{5\cos^2 \theta - 4}$.

Feb/March 2022

First form a quadratic equation in $\cos\theta$, then solve the equation $\tan\theta\,\sin\theta = 1$ for $0^\circ < \theta < 360^\circ$.

Feb/March 2023

Show that $\frac{(\sin\theta + \cos\theta)^2 - 1}{\cos^2\theta} = 2\tan\theta$.

Feb/March 2024

Show that $3\tan^2\theta + 5\sin^2\theta$ can be written as $\dfrac{8\sin^2\theta - 5\sin^4\theta}{1 - \sin^2\theta}$.

Feb/March 2025

The acute angle $x$ radians satisfies $\tan x = k$, with $k$ as a positive constant.

May/June 2010

The function $f$ is defined by $f(x) = 2\sin^2 x - 3\cos^2 x$ for $0 \le x \le \pi$.

May/June 2010

Show that the equation $3(2\sin x - \cos x) = 2(\sin x - 3\cos x)$ may be rearranged into the form $\tan x = -\tfrac{3}{4}$.

May/June 2010

The function $f : x \mapsto a + b\cos x$ is defined for $0 \leq x \leq 2\pi$. It is given that $f(0) = 10$ and that $f(\frac{2}{3}\pi) = 1$.

May/June 2010

Demonstrate that the equation $2\sin x \tan x + 3 = 0$ can be rewritten as $2\cos^2 x - 3\cos x - 2 = 0$.

May/June 2010

Show that $2\tan^2\theta\sin^2\theta = 1$ may be rewritten in the form $2\sin^4\theta + \sin^2\theta - 1 = 0$.

May/June 2011

Prove the identity $\frac{\cos\theta}{\tan\theta(1 - \sin\theta)} = 1 + \frac{1}{\sin\theta}$.

May/June 2011

Prove that the identity $\left( \frac{1}{\sin \theta} - \frac{1}{\tan \theta} \right)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}$ holds.

May/June 2011

Solve the equation $\sin 2x = 2\cos 2x$, for $0^\circ \leq x \leq 180^\circ$.

May/June 2012

Prove that $\tan x + \frac{1}{\tan x} \equiv \frac{1}{\sin x \cos x}$.

May/June 2012

Prove the identity $\tan^2\theta - \sin^2\theta = \tan^2\theta\sin^2\theta$.

May/June 2012

Solve $\sin 2x + 3\cos 2x = 0$ within the interval $0^\circ \leq x \leq 360^\circ$.

May/June 2012

Show that $\dfrac{\sin\theta}{\sin\theta + \cos\theta} + \dfrac{\cos\theta}{\sin\theta - \cos\theta} = \dfrac{1}{\sin^2\theta - \cos^2\theta}$.

May/June 2013

We are told that $a = \sin \theta - 3 \cos \theta$ and $b = 3 \sin \theta + \cos \theta$, where $0^\circ \leq \theta \leq 360^\circ$.

May/June 2013

Rewrite $2\cos^2\theta = \tan^2\theta$ as a quadratic in $\cos^2\theta$.

May/June 2013

Sketch the curves $y = \sin 2x$ and $y = \cos x - 1$ together on the same diagram for $0 \le x \le 2\pi$.

May/June 2013

The diagram displays a section of the graph of $y = a + b \sin x$.

May/June 2014

Prove the identity $\frac{\sin \theta}{1 - \cos \theta} - \frac{1}{\sin \theta} = \frac{1}{\tan \theta}$.

May/June 2014

The reflex angle $\theta$ satisfies $\cos\theta = k$, where $0 \le k \le 1$.

May/June 2014

Prove that $\frac{1}{\cos\theta} - \frac{\cos\theta}{1 + \sin\theta} \equiv \tan\theta$.

May/June 2014

Prove that $\frac{\tan x + 1}{\sin x \tan x + \cos x} \equiv \sin x + \cos x$.

May/June 2014

The position vectors of points $A$, $B$ and $C$ with respect to an origin $O$ are $\overrightarrow{OA} = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}$, $\overrightarrow{OB} = \begin{pmatrix} 6 \\ -1 \\ 7 \end{pmatrix}$ and $\overrightarrow{OC} = \begin{pmatrix} 2 \\ 4 \\ 7 \end{pmatrix}$.

May/June 2014

If $\theta$ is an obtuse angle in radians and $\sin\theta = k$, determine, in terms of $k$, an expression for

May/June 2015

Prove the identity $\frac{\sin \theta - \cos \theta}{\sin \theta + \cos \theta} = \frac{\tan \theta - 1}{\tan \theta + 1}$.

May/June 2015

A city-centre tourist attraction is a large vertical wheel that passengers can travel on. Its motion is modelled by the formula $h = 60(1 - \cos kt)$, where $h$ m is the passenger’s height above the ground, $k$ is a constant, $t$ is the number of minutes since the passenger began the ride at ground level, and $kt$ is measured in radians.

May/June 2015

Write $3\sin\theta = \cos\theta$ in the form $\tan\theta = k$ and find the solutions when $0^{\circ} < \theta < 180^{\circ}$.

May/June 2015

Solve the equation $3\sin^2\theta = 4\cos\theta - 1$ for values of $\theta$ in the interval $0^\circ \leq \theta \leq 360^\circ$.

May/June 2016

From the diagram, triangle $ABC$ is right-angled at $C$, and $M$ is the midpoint of $BC$. You are told that angle $ABC = \frac{\pi}{3}$ radians and angle $BAM = \theta$ radians. If the lengths of $BM$ and $MC$ are each called $x$,

May/June 2016

Prove that the identity $\frac{1 + \cos\theta}{1 - \cos\theta} - \frac{1 - \cos\theta}{1 + \cos\theta} = \frac{4}{\sin\theta \tan\theta}$ holds.

May/June 2016

Show that $3 \sin x \tan x - \cos x + 1 = 0$ may be transformed into a quadratic equation in $\cos x$ and hence solve $3 \sin x \tan x - \cos x + 1 = 0$ for $0 \leq x \leq \pi$.

May/June 2016

Prove that $\frac{1 + \cos \theta}{\sin \theta} + \frac{\sin \theta}{1 + \cos \theta} = \frac{2}{\sin \theta}$.

May/June 2017

A curve is defined by the equation $y = 2\cos x$.

May/June 2017

Prove the identity $(\frac{1}{\cos \theta} - \tan \theta)^2 \equiv \frac{1 - \sin \theta}{1 + \sin \theta}$.

May/June 2017

Show that the equation $\dfrac{2\sin\theta + \cos\theta}{\sin\theta + \cos\theta} = 2\tan\theta$ can be rearranged into the form $\cos^2\theta = 2\sin^2\theta$.

May/June 2017

Show that $(\sin \theta + \cos \theta)(1 - \sin \theta \cos \theta) \equiv \sin^3 \theta + \cos^3 \theta$.

May/June 2018

Solve the equation $2\cos x + 3\sin x = 0$, for $0^\circ \leq x \leq 360^\circ$.

May/June 2018

Express $\dfrac{\tan^2 \theta - 1}{\tan^2 \theta + 1}$ as $a\sin^2 \theta + b$, with $a$ and $b$ as the constants to determine.

May/June 2018

Prove that $\left(\frac{1}{\cos x} - \tan x\right)^2 = \frac{1 - \sin x}{1 + \sin x}$.

May/June 2019

The angle $x$ satisfies $\sin x = a + b$ and $\cos x = a - b$, with $a$ and $b$ being constants.

May/June 2019

In the diagram, a semicircle has diameter $AB$, centre $O$ and radius $r$. Point $C$ is on the circumference, and $AOC = \theta$ radians. The perimeter of sector $BOC$ is twice the perimeter of sector $AOC$.

May/June 2019

The curve is given by $y = 3\cos 2x$, and the line is given by $2y + \frac{3x}{\pi} = 5$.

May/June 2019

Prove that $\displaystyle \frac{1 + \sin\theta}{\cos\theta} + \frac{\cos\theta}{1 + \sin\theta} = \frac{2}{\cos\theta}$.

May/June 2020

Express the equation $3 \cos \theta = 8 \tan \theta$ in quadratic form with respect to $\sin \theta$.

May/June 2020

Show that $\frac{\tan\theta}{1 + \cos\theta} + \frac{\tan\theta}{1 - \cos\theta}$ is equal to $\frac{2}{\sin\theta \, \cos\theta}$.

May/June 2020

The diagram illustrates a section of the graph of $y = a\tan(x - b) + c$.

May/June 2021

Prove the identity by using trig identities to show that $\frac{1 - 2\sin^2\theta}{1 - \sin^2\theta} = 1 - \tan^2\theta$.

May/June 2021

Prove that $\dfrac{1 + \sin x}{1 - \sin x} - \dfrac{1 - \sin x}{1 + \sin x} = \dfrac{4 \tan x}{\cos x}$.

May/June 2021

Show that the equation $\dfrac{\tan x + \sin x}{\tan x - \sin x} = k$, where $k$ is a constant, can be rewritten in the form $\dfrac{1 + \cos x}{1 - \cos x} = k$.

May/June 2021

Prove the identity $$\frac{\sin^3 \theta}{\sin \theta - 1} - \frac{\sin^2 \theta}{1 + \sin \theta} = -\tan^2 \theta (1 + \sin^2 \theta).$$

May/June 2022

The graph of $y = \sin x$ is mapped onto the graph $y = 4\sin\left(\tfrac{1}{2}x - 30^\circ\right)$.

May/June 2022

The diagram displays a section of the curve whose equation is $y = p \sin(q\theta) + r$, with $p$, $q$ and $r$ as constants.

May/June 2022

Solve $6\sqrt{y} + \frac{2}{\sqrt{y}} - 7 = 0$.

May/June 2022

Solve the equation $4\sin\theta + \tan\theta = 0$ for $0^\circ < \theta < 180^\circ$.

May/June 2023

The curve is given by $y = 2 + 3\sin \frac{1}{2}x$ for $0 \leq x \leq 4\pi$.

May/June 2023

By expanding $(\cos\theta + \sin\theta)^2$ first, find the three solutions of $(\cos\theta + \sin\theta)^2 = 1$ when $0 \leq \theta \leq \pi$.

May/June 2023

Demonstrate that the equation $3\tan^2 x - 3\sin^2 x - 4 = 0$ can be rewritten in the form $a\cos^4 x + b\cos^2 x + c = 0$, with $a$, $b$ and $c$ as the constants to determine.

May/June 2023

Prove the identity $\dfrac{\sin^2 x - \cos x - 1}{1 + \cos x} = -\cos x$.

May/June 2024

Show that $\frac{7\tan\theta}{\cos\theta} + 12 = 0$ may be rewritten as $12\sin^2\theta - 7\sin\theta - 12 = 0$.

May/June 2024

The diagram depicts the curve $y = k\cos(x - \frac{1}{6}\pi)$, where $k$ is a positive constant and $x$ is measured in radians. The curve meets the $x$-axis at point $A$, and $B$ is a minimum point.

May/June 2024

Show that $\cos\theta(7\tan\theta - 5\cos\theta) = 1$ can be transformed into $a\sin^2\theta + b\sin\theta + c = 0$, with integer values for $a$, $b$ and $c$ to be determined.

May/June 2024

Solve $6\sin\theta = 1 + \frac{2}{\sin\theta}$ over the range $-180^{\circ} < \theta < 180^{\circ}$.

May/June 2025

The curve is given by $y = 4\cos 2x + 3$ for $0 \leq x \leq 2\pi$.

May/June 2025

Prove that $$\frac{\tan \theta + 7}{\tan^2 \theta - 3} = \frac{\sin \theta \cos \theta + 7 \cos^2 \theta}{1 - 4 \cos^2 \theta}$$ is an identity.

May/June 2025

Solve the equation $4 \sin \theta \tan \theta = 1 + 5 \cos \theta$ for $-180^\circ < \theta < 180^\circ$.

May/June 2025

Prove that $\frac{\sin x \tan x}{1 - \cos x} = 1 + \frac{1}{\cos x}$.

Oct/Nov 2010

Prove that $\tan^2 x - \sin^2 x = \tan^2 x\sin^2 x$ holds.

Oct/Nov 2010

Solve for $x$ in the equation $15\sin^2 x = 13 + \cos x$ over $0^{\circ} \leq x \leq 180^{\circ}$.

Oct/Nov 2010

Sketch the graph of $y = 2\sin x$ for $0 \leq x \leq 2\pi$.

Oct/Nov 2010

On one set of axes, sketch the graphs of $y = \cos 2\theta$ and $y = \frac{1}{2}$ for $0 \leq \theta \leq 2\pi$.

Oct/Nov 2011

On the same set of axes, sketch the graphs of $y = \sin x$ and $y = \cos 2x$ for $0^{\circ} \leq x \leq 180^{\circ}$.

Oct/Nov 2011

The diagram depicts circle $C_1$ touching circle $C_2$ at point $X$. Circle $C_1$ is centred at $A$ and has radius $6\text{ cm}$, whereas circle $C_2$ is centred at $B$ and has radius $10\text{ cm}$. Points $D$ and $E$ are on $C_1$ and $C_2$ respectively, and $DE$ is parallel to $AB$. Angle $DAX = \frac{\pi}{3}$ radians and angle $EBX = \theta$ radians.

Oct/Nov 2011

Starting from $3\sin^2 x - 8\cos x - 7 = 0$, show that, for real values of $x$, $\cos x = \frac{2}{3}$.

Oct/Nov 2011

Solve $2\cos^2\theta = 3\sin\theta$ for $0^\circ \leq \theta \leq 360^\circ$.

Oct/Nov 2012

Show that the equation $2\cos x = 3\tan x$ can be rearranged into a quadratic equation in $\sin x$.

Oct/Nov 2012

Solve $7\cos x + 5 = 2\sin^2 x$, for $0^\circ \leq x \leq 360^\circ$.

Oct/Nov 2012

Solve for $x$ in the equation $4\sin^2 x + 8\cos x - 7 = 0$ where $0^\circ \leq x \leq 360^\circ$.

Oct/Nov 2013

Given that $\cos x = p$, where $x$ is an acute angle in degrees, find the expression in terms of $p$.

Oct/Nov 2013

For $0 \leq x \leq 2\pi$, the function $f$ is defined by $f : x \mapsto 3\cos x - 2$.

Oct/Nov 2013

Find the possible values of $x$ satisfying $\sin^{-1}(x^2 - 1) = \frac{1}{3}\pi$, and give each answer correct to $3$ decimal places.

Oct/Nov 2013

Determine the value of $x$ for which $\sin^{-1}(x - 1) = \tan^{-1}(3)$.

Oct/Nov 2014

Find the values of $\theta$ that satisfy $\frac{13\sin^2\theta}{2 + \cos\theta} + \cos\theta = 2$ for $0^\circ \leq \theta \leq 180^\circ$.

Oct/Nov 2014

Demonstrate that $1 + \sin x \tan x = 5\cos x$ can be rearranged into $6\cos^2 x - \cos x - 1 = 0$.

Oct/Nov 2014

Show that $\sin^4 \theta - \cos^4 \theta \equiv 2\sin^2 \theta - 1$ by simplifying the left-hand side.

Oct/Nov 2014

Solve the equation $\sin^{-1}(4x^4 + x^2) = \frac{1}{6}\pi$.

Oct/Nov 2015

Show that the equation $\frac{4\cos\theta}{\tan\theta} + 15 = 0$ may be rewritten as $4\sin^2\theta - 15\sin\theta - 4 = 0$.

Oct/Nov 2015

Prove the identity $\left(\frac{1}{\sin x} - \frac{1}{\tan x}\right)^2 = \frac{1 - \cos x}{1 + \cos x}$.

Oct/Nov 2015

Show that the equation $\frac{1}{\cos \theta} + 3 \sin \theta \tan \theta + 4 = 0$ can be rewritten as $3 \cos^2 \theta - 4 \cos \theta - 4 = 0$, and hence solve the equation $\frac{1}{\cos \theta} + 3 \sin \theta \tan \theta + 4 = 0$ for $0^\circ \leq \theta \leq 360^\circ$.

Oct/Nov 2015

Show that, after expansion, $\cos^4 x = 1 - 2\sin^2 x + \sin^4 x$.

Oct/Nov 2016

The function $f$ is defined as $f : x \mapsto 5 - 2\sin 2x$ on $0 \leq x \leq \pi$.

Oct/Nov 2016

Rewrite the equation $\sin 2x + 3\cos 2x = 3(\sin 2x - \cos 2x)$ so that it takes the form $\tan 2x = k$, where $k$ is a constant.

Oct/Nov 2016

By presenting every required step,

Oct/Nov 2016