Trigonometry (Cambridge International IGCSE Further Maths): Exam Questions

In this question, all angles are in radians.

Write down the amplitude of $2\mathrm{cos}\frac{x}{3}-1$.

Write down the period of $2\mathrm{cos}\frac{x}{3}-1$.

Extract:

Write down the amplitude of 2 cos(x/3) - 1.

For any function in the form a cos(bx) + c, the amplitude is given by a.

2

Write down the period of 2 cos(x/3) - 1.

For any function in the form a cos(bx) + c, the period is 2π/b in radians or 360°/b in degrees.

6π or 1080°

On the axes below, sketch the graph of $y=2\mathrm{cos}\frac{x}{3}-1$ for $- pi \text{less} - \text{than} or \text{slanted} \text{equal} to x \text{less} - \text{than} or \text{slanted} \text{equal} to 3 pi$ .

(i) Write $6xy+3y+4x+2$ in the form $(ax+b)(cy+d)$, where $a,b,c$ and $d$ are positive integers.

(ii) Hence solve the equation $6\mathrm{sin}θ\mathrm{cos}θ+3\mathrm{cos}θ+4\mathrm{sin}θ+2=0$ for $0^{\circ}<θ<360^{\circ}$.

The graph of $y=a+2\mathrm{tan}bx$, where $a$ and $b$ are constants, passes through the point $(0,-4)$ and has period $480^{\circ}$.

Find the value of $a$ and of $b$.

Extract:

The graph of y = a + 2 tan bx, where a and b are constants, passes through the point (0, -4) and has period 480°.

On the axes, sketch the graph of $y$ for values of $x$ between 0° and 480°.

Solve tan ( alpha + 45 ^ deg ) = - (1)/(sqrt( 2) for 0 ^ deg <= alpha <= 360 ^ deg.

The diagram shows the graph of $\mathrm{f} ( x ) = a \cos b x + c \text{for} 0 \text{less} \text{or} \text{equal} \text{than} x \text{less} \text{or} \text{equal} \text{than} \frac{8 \pi }{3}$radians.

Find the value of each of the constants $a,b$ and $c$.

The curve $y=a\mathrm{sin}bx+c$ has a period of 180°, an amplitude of 20 and passes through the point $(90^{\circ},-3)$. Find the value of each of the constants $a,b\mathrm{and}c$.

The function $g$ is defined, for$-135^{\circ}\leq x\leq135^{\circ},\mathrm{by}g(x)=3\mathrm{tan}\frac{x}{2}-4$. Sketch the graph of $y=g(x)$ on the axes below, stating the coordinates of the point where the graph crosses the $y$-axis.

Solve the equation 5 sin ( 4 B - pi / 8 ) + 2 = 0 for - pi / 4 <= B <= pi / 4 radians

Write down the amplitude of $1 + 4 \cos ( \frac{x}{3} )$

Write down the period of $1 + 4 \cos ( \frac{x}{3} )$

On the axes below, sketch the graph of $y = 1 + 4 \cos ( \frac{x}{3} ) \text{for} - 180^{\circ} \text{less} \text{or} \text{equal} \text{than} x^{\circ} \text{less} \text{or} \text{equal} \text{than} 180^{\circ}$.

Solve the equation $\cot ( y - \frac{\pi }{2} ) = \sqrt{3}$, where $y$ is in radians and $0 \text{less} - \text{than} or \text{slanted} \text{equal} to y \text{less} - \text{than} or \text{slanted} \text{equal} to pi$ .

Solve the equation $1 \frac{1}{2} \sec ( 2 ϕ + \frac{\pi }{4} ) = \frac{1}{ \sqrt{3} }$ for $-π<ϕ<π$, where $ϕ$ is in radians. Give your answers in terms of $π$.

In this question all lengths are in centimetres.

The diagram shows the isosceles triangle $ABC$, where $AB=AC$ and $BC=2+4\sqrt{3}$. The height, $AD$, of the triangle is $5-\sqrt{3}$.

Find tan $ABC$, giving your answer in the form$c+d\sqrt{3}$, where $c\mathrm{and}d$ are integers.

Extract:

The diagram shows the isosceles triangle ABC, where AB = AC and BC = 2 + 4√3. The height, AD, of the triangle is 5 - √3.

tan ABC = √3 - 1

Find $\mathrm{sec}^{2}ABC$, giving your answer in the form $e+f\sqrt{3}$, where $e\mathrm{and}f$ are integers.

Solve cosec ( y + pi / 3 ) = 2 for 0 <= y <= 2 pi radians, giving your answers in terms of $π$.

Solve $\mathrm{tan}3x=-1\mathrm{for}-\frac{π}{2}\leq x\leq\frac{π}{2}$ radians, giving your answers in terms of $π$.

Extract:

Solve tan 3x = -1 for -π/2 ≤ x ≤ π/2 radians, giving your answers in terms of π.

x = π/4, -π/12, -5π/12

Use your answers to part (a) to sketch the graph of $y=4\mathrm{tan}3x+4\mathrm{for}-\frac{π}{2}\leq x\leq\frac{π}{2}$ radians on the axes below. Show the coordinates of the points where the curve meets the axes.

Solve $3\mathrm{cot}^{2}x-14\mathrm{cosec}x-2=0\mathrm{for}0^{\circ}<x<360^{\circ}$.

Solve the equation

$5\mathrm{sec}^{2}A+14\mathrm{tan}A-8=0\mathrm{for}0^{\circ}\leq A\leq180^{\circ}$,

Solve $\sin ( 3 ϕ + \frac{2 \mathrm{\pi } }{3} ) = \cos ( 3 ϕ + \frac{2 \mathrm{\pi } }{3} ) \text{for} 0 \text{less} \text{or} \text{equal} \text{than} ϕ \text{less} \text{or} \text{equal} \text{than} \frac{2 \mathrm{\pi } }{3} \text{radians}$, giving your answers in terms of $π$.

In this question, all angles are measured in radians.

The graph of $y=a\mathrm{sin}bx+c$ has an amplitude of $10$, a period of $16π$ and passes through the point with coordinates $( 12 \pi , 4 )$.

Find the constants $a$, $b$ and $c$.

Sketch the graph of $y = \left| \tan ( \frac{x}{2} ) + 1 \right|$ for $-π<x<π$.

Label clearly any points of intersection with the coordinate axes and any asymptotes.

(i) Show that $\frac{\mathrm{cos}^{2}2x}{1+\mathrm{sin}2x}=1-\mathrm{sin}2x$.

(ii) Hence solve the equation for $\frac{3\mathrm{cos}^{2}2x}{1+\mathrm{sin}2x}=1$ for $0 ^{\circ} \text{less} - \text{than} or \text{slanted} \text{equal} to x \text{less} - \text{than} or \text{slanted} \text{equal} to 90 ^{\circ}$ .

(i) Show that $\frac{1}{\mathrm{sin}θ-1}-\frac{1}{\mathrm{sin}θ+1}=a\mathrm{sec}^{2}θ$, where $a$ is a constant to be found.

(ii) Hence solve $\frac{1}{\mathrm{sin}3ϕ-1}-\frac{1}{\mathrm{sin}3ϕ+1}=-8\mathrm{for}-\frac{π}{3}\leqϕ\leq\frac{π}{3}\mathrm{radians}.$

(i) Show that $\frac{1}{\mathrm{sec}θ-1}-\frac{1}{\mathrm{sec}θ+1}=2\mathrm{cot}^{2}θ$.

(ii) Hence solve $\frac{1}{\mathrm{sec}2x-1}-\frac{1}{\mathrm{sec}2x+1}=6$ for $-90^{\circ}<x<90^{\circ}$.

Show that $\frac{\mathrm{sin}^{4}y-\mathrm{cos}^{4}y}{\mathrm{cot}y}=\mathrm{tan}y-2\mathrm{cos}y\mathrm{sin}y$.

(i) Show that $\frac{1}{( 1 + \text{cosec} \\theta ) ( \sin \\theta - \sin ^{2 } \\theta )} = \sec ^{2} \\theta$.

(ii) Hence solve $( 1 + \text{cosec} \\theta ) ( \sin \\theta - \sin ^{2} \\theta ) = \frac{3}{4} \text{for} - 180^{\circ} \text{less} \text{or} \text{equal} \text{than} \\theta \text{less} \text{or} \text{equal} \text{than} 180^{\circ}$.

Show that $\frac{\mathrm{sin}x\mathrm{tan}x}{1-\mathrm{cos}x}=1+\mathrm{sec}x$.

Given that $2\mathrm{cos}x=3\mathrm{tan}x$, show that $2\mathrm{sin}^{2}x+3\mathrm{sin}x-2=0$.

Extract:

Given that 2 cos x = 3 tan x, show that 2 sin²x + 3 sin x - 2 = 0.

Substitute tan x = sin x / cos x into the equation. 2 cos x = 3(sin x / cos x)

Multiply through by cos x. 2 cos²x = 3 sin x

Substitute cos²x = 1 - sin²x into the equation. 2(1 - sin²x) = 3 sin x

Expand the brackets and rearrange. 2 - 2 sin²x = 3 sin x -2 sin²x - 3 sin x + 2 = 0

Multiply through by -1. 2 sin²x + 3 sin x - 2 = 0

Hence solve 2 cos ( 2 alpha + pi / 4 ) = 3 tan ( 2 alpha + pi / 4 ) for 0 less than alpha less than pi radians, giving your answers in terms of $π$.

Solve the equation $5\mathrm{tan}x-3\mathrm{cot}x=2\mathrm{sec}x\mathrm{for}0^{\circ}\leq x\leq360^{\circ}$.

Throughout this question, $A=\frac{\mathrm{tan}x}{1+\mathrm{sec}x}$ and $B=\frac{\mathrm{tan}x}{1-\mathrm{sec}x}$.

Show that $AB=k$ where $k$ is a constant to be found.

Show that $A+B=-2\mathrm{cot}x$.

Use your answers to part (a) and part (b) to simplify the expression $( A + 1 ) ( B + 1 )$.

Hence, or otherwise, solve

$( \frac{\tan ( x - 60 )}{1 + \sec ( x - 60 )} + 1 ) ( \frac{\tan ( x - 60 )}{1 - \sec ( x - 60 )} + 1 ) = 2 \sqrt{3}$

for $-180^{\circ}<x<180^{\circ}$.