We think you are located in South Africa. Is this correct?

End Of Chapter Exercises

Do you need more Practice?

Siyavula Practice gives you access to unlimited questions with answers that help you learn. Practise anywhere, anytime, and on any device!

Sign up to practise now

End of chapter exercises

Exercise 6.14

Write the following as a single trigonometric ratio: \(\dfrac{\cos (\text{90}\text{°} - A) \sin \text{20}\text{°}}{\sin (\text{180}\text{°} - A) \cos \text{70}\text{°}} + \cos (\text{180}\text{°} + A) \sin (\text{90}\text{°} + A)\)

\begin{align*} & \frac{\sin A \cdot \sin \text{20}\text{°}}{\sin A \cdot \sin( \text{90}\text{°} - \text{70}\text{°})} + (-\cos A)\cos (A) \\ &= \frac{\sin A \cdot \sin \text{20}\text{°}}{\sin A \cdot \sin \text{20}\text{°}} + (-\cos A)\cos A\\ &= 1 - \cos^2 A\\ &= \sin^2 A \end{align*}

Determine the value of the following expression without using a calculator: \(\sin\text{240}\text{°} \cos\text{210}\text{°} - \tan^2 \text{225}\text{°} \cos\text{300}\text{°} \cos\text{180}\text{°}\)

\begin{align*} & \sin(\text{180}\text{°} + \text{60}\text{°}) \cdot \cos(\text{180}\text{°} + \text{30}\text{°}) - \tan^2(\text{180}\text{°} + \text{45}\text{°}) \cdot \cos(\text{360}\text{°} - \text{60}\text{°}) \cdot (-1)\\ &= (-\sin \text{60}\text{°}) \cdot (-\cos \text{30}\text{°}) - (\tan^2 \text{45}\text{°}) \cdot (\cos \text{60}\text{°} ) \cdot (-1) \\ &= + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} - (1)^2 \cdot (\frac{1}{2}) \cdot (-1) \\ &= \frac{3}{4} + \frac{1}{2} \\ &= \frac{3 + 2}{4} \\ &= 1 \frac{1}{4} \end{align*}

Simplify: \(\dfrac{\sin (\text{180}\text{°} + \theta) \sin (\theta + \text{360}\text{°})}{\sin (-\theta) \tan (\theta - \text{360}\text{°})}\)

\begin{align*} &\frac{\sin(\text{180}\text{°} + \theta) \cdot \sin(\theta + \text{360}\text{°})}{\sin(-\theta) \cdot \tan(\theta - \text{360}\text{°})}\\ &= \frac{(- \sin \theta) \cdot \sin \theta}{(-\sin \theta) \cdot (- \tan(\text{360}\text{°} - \theta))}\\ &= \frac{\sin \theta}{\tan \theta} = \frac{\sin \theta}{1} \cdot \frac{\cos \theta}{\sin \theta}\\ &= \cos \theta \end{align*}

Without the use of a calculator, evaluate: \(\dfrac{3\sin \text{55}\text{°} \sin^2 \text{325}\text{°}}{\cos (-\text{145}\text{°})} - 3\cos \text{395}\text{°} \sin \text{125}\text{°}\)

\begin{align*} & \frac{3 \sin \text{55}\text{°} . \sin^2 (\text{360}\text{°} - (\text{35}\text{°}))}{\cos (\text{145}\text{°}} - 3\cos((\text{360}\text{°} + (\text{35}\text{°}) . \sin(\text{90}\text{°} + \text{35}\text{°})\\ &= \frac{3 \sin \text{55}\text{°} . (-\sin \text{35}\text{°})^2}{\cos (\text{90}\text{°} + \text{55}\text{°})} + 3\cos \text{35}\text{°} . \cos \text{35}\text{°}\\ &= \frac{-3 \sin \text{55}\text{°} . \sin^2 \text{35}\text{°}}{-\sin \text{55}\text{°}} + 3\cos^2 \text{35}\text{°}\\ &= 3(\sin^2 \text{35}\text{°} + \cos^2 \text{35}\text{°})\\ &= 3 \end{align*}

Prove the following identities:

\(\dfrac{1}{(\cos x - 1)(\cos x + 1)} = \dfrac{-1}{\tan^2x \cos^2x}\)
\begin{align*} \text{ RHS }&=\frac{-1}{\frac{\sin^2x}{\cos^2x}\cdot\frac{\cos^2x}{1}} \\ &=\frac{-1}{\sin^2x} \\ \text{ LHS } &=\frac{1}{\cos^2x-1} \\ &=\frac{1}{-(1-\cos^2x)} \\ &=-\frac{1}{\sin^2x} \\ &=\text{ RHS } \end{align*}
\((1 - \tan \alpha) \cos \alpha = \sin(90 + \alpha) + \cos(90 + \alpha)\)
\begin{align*} \text{LHS} &= (1 - \frac{\sin \alpha}{\cos \alpha}) . \cos \alpha \\ &= (\frac{\cos \alpha - \sin \alpha}{\cos \alpha} ) \cos \alpha \\ &= \cos \alpha - \sin \alpha \\ \text{RHS} &= \cos (-\alpha) + \sin (-\alpha) \\ &= \cos \alpha - \sin \alpha \\ &= \text{LHS} \end{align*}

Prove: \(\tan y + \dfrac{1}{\tan y} = \dfrac{1}{\cos^2y \tan y}\)

\begin{align*} \text{LHS} &= \frac{\sin y}{\cos y} + \frac{1}{\frac{1}{\sin y}{\cos y}}\\ &= \frac{\sin y}{\cos y} + \frac{\cos y}{\sin y}\\ &= \frac{\sin^2 y + \cos^2 y}{\cos y . \sin y}\\ &= \frac{1}{\cos y . \sin y}\\ \text{RHS} &= \frac{1}{\cos^2 y \frac{\sin y}{\cos y}}\\ &= \frac{\cos y}{\cos^2 y . \sin y}\\ &= \frac{1}{\cos y \sin y}\\ &= \text{LHS} \end{align*}

For which values of \(y \in [\text{0}\text{°};\text{360}\text{°}]\) is the identity above undefined?

\begin{align*} \tan y &= \text{0}\text{°}\\ y &= \text{0}\text{°}; \text{180}\text{°}; \text{360}\text{°}\\ \cos y &= \text{0}\text{°}\\ y &= \text{90}\text{°}; \text{270}\text{°} \end{align*}

Simplify: \(\dfrac{\sin(\text{180}\text{°} + \theta) \tan(\text{360}\text{°} - \theta)}{\sin(-\theta) \tan(\text{180}\text{°} + \theta)}\)

\begin{align*} & \frac{(-\sin \theta) . (- \tan \theta)}{(-\sin \theta) . \tan \theta} \\ & = -1 \end{align*}

Hence, solve the equation \(\dfrac{\sin(\text{180}\text{°} + \theta) \tan(\text{360}\text{°} - \theta)}{\sin(-\theta) \tan(\text{180}\text{°} + \theta)} = \tan \theta\) for \(\theta \in [\text{0}\text{°};\text{360}\text{°}]\).

\begin{align*} \tan \theta &= -1 \\ \therefore \theta &= \text{180}\text{°} - \text{45}\text{°} \text{ or } \theta = \text{360}\text{°} - \text{45}\text{°} \\ &= \text{135}\text{°} \text{ or } \text{315}\text{°} \end{align*}

Given \(12 \tan \theta = 5\) and \(\theta > \text{90}\text{°}\).

Draw a sketch.

c81c7f13e9e11dce1a2b9b5c69bcf165.png

Determine without using a calculator \(\sin \theta\) and \(\cos (\text{180}\text{°} + \theta)\).

\begin{align*} OP &= \sqrt{(-12)^2 + (-5)^2} \\ &= \sqrt{144 + 25} \\ &= \sqrt{169}\\ &= 13\\ \sin \theta &= - \frac{5}{13}\\ \cos (\text{180}\text{°} + \theta) &= - \cos \theta \\ &= - (-\frac{12}{13}) \\ &= \frac{12}{13} \end{align*}

Use a calculator to find \(\theta\) (correct to two decimal places).

\begin{align*} \tan \theta &= \frac{5}{12} \\ \theta &= \text{180}\text{°} + \text{22,62}\text{°}\\ &= \text{202,62}\text{°} \end{align*}
1b43c934975770805f31ae59f835b837.png

In the figure, \(P\) is a point on the Cartesian plane such that \(OP = 2\) units and \(\theta = \text{300}\text{°}\). Without the use of a calculator, determine:

the values of \(a\) and \(b\)

\begin{align*} \cos \text{300}\text{°} &= \cos \text{60}\text{°} \\ & = \frac{1}{2}\\ \therefore a &= 1 \\ b &= - \sqrt{4-1} \\ & = -\sqrt{3} \end{align*}

the value of \(\sin (\text{180}\text{°} - \theta)\)

\begin{align*} \sin (\text{180}\text{°} - \theta) &= \sin \theta \\ &= \sin \text{300}\text{°} \\ &= \sin (\text{360}\text{°} - \text{60}\text{°})\\ &= \sin (-\text{60}\text{°})\\ &= -\sin \text{60}\text{°}\\ &= - \frac{\sqrt{3}}{2} \end{align*}

Solve for \(x\) with \(x \in [-\text{180}\text{°};\text{180}\text{°}]\) (correct to one decimal place):

\(2 \sin \frac{x}{2} = \text{0,86}\)
\begin{align*} \sin \frac{1}{2}x &= \text{0,43} \\ \frac{1}{2}x &= \text{25,467}\text{°}\\ x &= \text{50,9}\text{°} \\ \text{Or } \quad \frac{1}{2}x &=\text{180}\text{°} - \text{25,467}\text{°}\\ x &= \text{309,1}\text{°} \end{align*}
\(\tan (x + \text{10}\text{°}) = \cos \text{202,6}\text{°}\)
\begin{align*} \tan (x + \text{10}\text{°}) &= -\text{0,92}\\ x + \text{10}\text{°} &= \text{180}\text{°} -\text{42,7}\text{°}\\ &= \text{137,3}\text{°}\\ x &= \text{127,3}\text{°}\\ \text{Or} \quad x + \text{10}\text{°} &= \text{360}\text{°} - \text{42,7}\text{°}\\ &= \text{317,3}\text{°}\\ \therefore x &= \text{307,3}\text{°} \end{align*}
\(\cos^2x - 4 \sin^2x = 0\)
\begin{align*} \cos x - 2\sin x &= 0\\ \cos x &= 2\sin x\\ \frac{\sin x}{\cos x} &= \frac{1}{2}\\ \tan x &= \frac{1}{2}\\ x &= \text{26,6}\text{°}\\ \text{or } x &= -\text{180}\text{°} + \text{26,6}\text{°} \\ &= \text{206,6}\text{°}\\ \text{Or } \quad \cos x + 2\sin x &= 0\\ \cos x &= -2 \sin x\\ \frac{\sin x}{\cos x} &= -\frac{1}{2}\\ \tan x &= -\frac{1}{2}\\ x &= \text{180}\text{°} - \text{26,6}\text{°}\\ &= \text{153,4}\text{°}\\ \text{or } x &= \text{360}\text{°} - \text{26,6}\text{°} \\ &= \text{333,4}\text{°} \end{align*}

Find the general solution for the following equations:

\(\frac{1}{2} \sin (x - \text{25}\text{°}) = \text{0,25}\)
\begin{align*} \sin(x - \text{25}\text{°}) &= 0.5\\ x - \text{25}\text{°} &= \text{30}\text{°} \\ \text{ or } x - \text{25}\text{°} &= \text{180}\text{°} - \text{30}\text{°}\\ x &= \text{55}\text{°} + \text{360}\text{°}n, n \in \mathbb{Z} \\ \text{ or } x &= \text{175}\text{°} + \text{360}\text{°}n, n \in \mathbb{Z} \end{align*}
\(\sin^2x + 2 \cos x = -2\)
\begin{align*} \sin^2 x + 2\cos x + 2 &= 0\\ 1 - \cos^2 x + 2\cos x + 2 &= 0\\ - \cos^2 x + 2\cos x + 3 &= 0\\ \cos^2 x - 2\cos x - 3 &= 0\\ (\cos x - 3)(\cos x + 1) &= 0\\ \cos x &= 3 \\ & \text{no solution} \\ \text{OR} \cos x &= -1\\ x &= \text{180}\text{°} + \text{360}\text{°}n, n \in \mathbb{Z} \end{align*}

Given the equation: \(\sin 2 \alpha = \text{0,84}\)

Find the general solution of the equation.
\begin{align*} \sin 2\alpha &= \text{0,84}\\ 2\alpha &= \text{57,14}\text{°} + \text{360}\text{°}n, n \in \mathbb{Z}\\ \alpha &= \text{28,6}\text{°} + \text{180}\text{°}n\\ \text{Or } 2\alpha &= \text{180}\text{°} - \text{57,14}\text{°} + \text{360}\text{°}n, n \in \mathbb{Z}\\ 2\alpha &= \text{122,86}\text{°} + \text{360}\text{°}n\\ \alpha &= \text{61,43}\text{°} + \text{180}\text{°}n \end{align*}
Illustrate how this equation could be solved graphically for \(\alpha \in [\text{0}\text{°}; \text{360}\text{°}]\).
5065612b2bde79231725f566535f680e.png
Write down the solutions for \(\sin 2 \alpha = \text{0,84}\) for \(\alpha \in [\text{0}\text{°}; \text{360}\text{°}]\).
\(\text{28,6}\text{°}; \text{61,4}\text{°}; \text{208,6}\text{°}; \text{241,4}\text{°}\)
a92d4ff13a167361af24f8da55941296.png

\(A\) is the highest point of a vertical tower \(AT\). At point \(N\) on the tower, \(n\) metres from the top of the tower, a bird has made its nest. The angle of inclination from \(G\) to point \(A\) is \(\alpha\) and the angle of inclination from \(G\) to point \(N\) is \(\beta\).

Express \(A\hat{G}N\) in terms of \(\alpha\) and \(\beta\).

\begin{align*} A\hat{G}N &= \alpha - \beta \end{align*}

Express \(\hat{A}\) in terms of \(\alpha\) and/or \(\beta\).

\begin{align*} \hat{A} &= \text{90}\text{°} - \alpha \end{align*}

Show that the height of the nest from the ground (\(H\)) can be determined by the formula \[H = \frac{n \cos \alpha \sin \beta}{\sin(\alpha - \beta)}\]

\begin{align*} \text{In } \triangle GNT, & \\ \frac{H}{GN} &= \sin \beta \\ \therefore H &= GN \sin \beta \\ \text{In } \triangle AGN, & \\ \frac{n}{\sin(\alpha - \beta)} &= \frac{GN}{\sin(\text{90}\text{°} - \alpha)}\\ \therefore GN &= \frac{n \sin(\text{90}\text{°} - \alpha)}{\sin(\alpha - \beta)} \\ & = \frac{n \cos \alpha}{\sin(\alpha - \beta)} \\ \text{Substitute for } GN, & \\ H &= \frac{n \cos \alpha \sin \beta}{\sin(\alpha-\beta)} \end{align*}

Calculate the height of the nest \(H\) if \(n = \text{10}\text{ m}\), \(\alpha = \text{68}\text{°}\) and \(\beta = \text{40}\text{°}\) (give your answer correct to the nearest metre).

\begin{align*} H &= \frac{n \cos \alpha \sin \beta}{\sin(\alpha-\beta)}\\ &= \frac{10 \cos \text{68}\text{°} \sin \text{40}\text{°}}{\sin \text{28}\text{°}}\\ &= \text{5,1}\text{ m} \end{align*}
730f1d48277a2226f28f673b68d9e691.png

Mr. Collins wants to pave his trapezium-shaped backyard, \(ABCD\). \(AB \parallel DC\) and \(\hat{B} = \text{90}\text{°}\). \(DC = \text{11}\text{ m}\), \(AB = \text{8}\text{ m}\) and \(BC = \text{5}\text{ m}\).

Calculate the length of the diagonal \(AC\).
\begin{align*} AC^2 &= 8^2 + 5^2\\ &= 64 + 25 = 89\\ AC &= \text{9,43}\text{ m} \end{align*}
Calculate the length of the side \(AD\).
\begin{align*} \text{In } \triangle ABC, &\\ \tan B\hat{A}C &= \frac{5}{8}\\ \therefore B\hat{A}C &= \text{32}\text{°}\\ A\hat{C}D = B\hat{A}C &= \text{32}\text{°} (AB \parallel DC)\\ \text{In } \triangle ADC, &\\ AD^2 &= AC^2 + DC^2 - 2.AC.AD.\cos(A\hat{C}D)\\ &= (\text{9,4})^2 + (\text{11})^2 - 2 (\text{9,4})(\text{11}) \cos(\text{32}\text{°})\\ &= 89 + 121 - 176\\ &= 34\\ \therefore AD &= \text{6,2}\text{ m} \end{align*}
Calculate the area of the patio using geometry.
\begin{align*} \text{Area } &= \frac{1}{2} \times (8+11) \times 5 \\ &= \text{47,5}\text{ m$^{2}$} \end{align*}
Calculate the area of the patio using trigonometry.
\begin{align*} \text{Area } ABCD &= \text{area } \triangle ABC + \text{area } \triangle ADC \\ &= \frac{1}{2} (8) (5) + \frac{1}{2} (9.43) (11) \sin \text{32}\text{°}\\ &= 20 + \text{27,5}\\ &= \text{47,5}\text{ m$^{2}$} \end{align*}
cf0fdbc9722f10874ea8e37c0569d6f7.png

In \(\triangle ABC\), \(AC = 2A\), \(AF = BF\), \(A\hat{F}B = \alpha\) and \(FC = 2AF\). Prove that \(\cos \alpha = \frac{1}{4}\).

\begin{align*} \text{In } \triangle ABF & \\ t^2 &= n^2 + n^2 - 2n^2 \cos \alpha \\ t^2 &= 2n^2(1 - \cos \alpha) \\ 1 - \cos \alpha &= \frac{t^2}{2n^2}\\ \cos \alpha &= 1 - \frac{t^2}{2n^2} \\ \text{In } \triangle AFC & \\ (2t)^2 &= n^2 + (2n)^2 - 4n^2 \cos (A\hat{F}C)\\ 4t^2 &= n^2 + 4n^2 - 4n^2 \cos (\text{180}\text{°} - \alpha) \\ &= 5n^2 + 4n^2 \cos \alpha\\ t^2 &= \frac{5}{4} n^2 + n^2 \cos \alpha \\ \text{Substitute for } t^2 & \\ \cos \alpha &= 1 - \frac{\frac{5}{4}n^2 + n^2 \cos \alpha}{2n^2}\\ &= \frac{2n^2 - \frac{5}{4}n^2 - n^2 \cos \alpha}{2 n^2}\\ &= \frac{\frac{3}{4}n^2 - n^2 \cos \alpha}{2n^2}\\ &= \frac{3}{8} - \frac{\cos \alpha}{2}\\ \frac{3}{2} \cos \alpha &= \frac{3}{8}\\ \therefore \cos \alpha &= \frac{1}{4} \end{align*}