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

End Of Chapter Exercises

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*}