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End Of Chapter Exercises

End of chapter exercises

Exercise 4.7

Determine the following without using a calculator:

\(\cos \text{15} °\)

\begin{align*} \cos \text{15} ° &= \cos ( \text{60} ° - \text{45} °)\\ &= \cos \text{60} °\cos \text{45} ° + \sin \text{60} ° \sin \text{45} °\\ &= \left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{2}}\right) + \left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{\sqrt{2}}\right)\\ &=\frac{1}{2\sqrt{2}} + \frac{\sqrt{3}}{2\sqrt{2}} \\ &= \frac{1+\sqrt{3}}{2\sqrt{2}} \end{align*}

\(\cos \text{75} °\)

\begin{align*} \cos \text{75} °&= \cos ( \text{45} ° + \text{30} °)\\ &= \cos \text{45} ° \cos \text{30} ° - \sin \text{45} ° \sin \text{30} ° \\ &= \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right)\\ &= \frac{\sqrt{3}-1}{2\sqrt{2}} \end{align*}

\(\tan \text{75} °\)

\begin{align*} \tan \text{75} ° &= \frac{\sin \text{75} °}{\cos \text{75} °} \\ & \\ \sin \text{75} ° & = \sin ( \text{45} ° + \text{30} ° ) \\ &= \sin \text{45} ° \cos \text{30} ° + \cos \text{45} ° \sin \text{30} ° \\ &= \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \\ &= \frac{\sqrt{3} + 1}{2\sqrt{2}} \\ \text{And from part b) } \enspace \cos \text{75} ° &= \frac{\sqrt{3} - 1}{2\sqrt{2}} \\ \therefore \frac{\sin \text{75} ° }{\cos \text{75} ° } &= \frac{\frac{\sqrt{3} + 1}{2\sqrt{2}}}{\frac{\sqrt{3} - 1}{2\sqrt{2}}} \\ &= \frac{\sqrt{3} + 1}{2\sqrt{2}} \times \frac{2\sqrt{2}}{\sqrt{3} - 1} \\ &= \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \\ &= \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} \\ &= \frac{3 + 2\sqrt{3} + 1}{3 - 1} \\ &= \frac{4 + 2\sqrt{3} }{2} \\ &= 2 + \sqrt{3} \end{align*}

\(\cos \text{3} ° \cos \text{42} ° -\sin \text{3} ° \sin \text{42} °\)

\begin{align*} \cos \text{3} ° \cos \text{42} ° -\sin \text{3} ° \sin \text{42} ° &= \cos( \text{3} ° + \text{42} ° )\\ &= \cos \text{45} ° \\ &= \frac{1}{\sqrt{2}} \end{align*}

\(1-2{\sin}^{2}\left( \text{22,5} ° \right)\)

\begin{align*} 1-2\sin^{2}( \text{22,5} ° ) &=\cos 2( \text{22,5} ° ) \\ &= \cos \text{45} ° \\ & = \frac{1}{\sqrt{2}} \end{align*}

Given \(\cos \theta =\text{0,7}\). Using a diagram, find \(\cos 2 \theta\) and \(\cos 4\theta\).

19a3ca09bef3646fb7f3d893df326b15.png

Given \(7 \sin \alpha = 3\) for \(\alpha > \text{90} °\).

Determine the following (leave answers in surd form):

\(\cos 2 \alpha\)

Draw a sketch.

097714254d88772c18ddd1cf210631a0.png
\(\tan 2 \alpha\)
\begin{align*} \tan 2 \alpha &= \frac{\sin 2 \alpha}{\cos 2 \alpha} \\ \sin 2 \alpha &= 2 \sin \alpha \cos \alpha \\ &= 2 \left( \frac{3}{7} \right)\left( - \frac{\sqrt{40}}{7} \right) \\ &= - \frac{6\sqrt{40}}{49} \\ & \\ \tan 2 \alpha &= \frac{\sin 2 \alpha}{\cos 2 \alpha} \\ &= \frac{- \frac{6\sqrt{40}}{49} }{\frac{31}{49}} \\ &= - \frac{6\sqrt{40}}{49} \times \frac{49}{31} \\ &= - \frac{6\sqrt{40}}{31} \end{align*}

If \(4 \tan A + 3 = 0\) for \(A < \text{270} °\), determine, without the use of a calculator:

\[\left( \sin \frac{A}{2} - \cos \frac{A}{2} \right) \left( \sin \frac{A}{2} + \cos \frac{A}{2} \right)\]

Draw a sketch.

dc3e3f83196688706cf6e6a047a9ed5a.png

We are given that \(A < \text{270} °\), therefore \(A\) must lie in the second quadrant for the tangent function to be negative.

\begin{align*} r^{2} &= 3^{2} + (-4)^{2} = 25 \\ \therefore r & = 5 \\ & \\ & \left( \sin \frac{A}{2} - \cos \frac{A}{2} \right) \left( \sin \frac{A}{2} + \cos \frac{A}{2} \right) \\ &= \sin^{2} \frac{A}{2} - \cos^{2} \frac{A}{2} \\ &= - \left( \cos^{2} \frac{A}{2} - \sin^{2} \frac{A}{2} \right) \\ &= - \cos 2 \left( \frac{A}{2} \right) \\ &= - \cos A \\ &= - \left( - \frac{4}{5} \right)\\ &= \frac{4}{5} \end{align*}
Simplify: \(\cos \text{67} ° \cos \text{7} ° + \cos \text{23} ° \cos \text{83} °\)
\begin{align*} & \cos \text{67} ° \cos \text{7} ° + \cos \text{23} ° \cos \text{83} ° \\ &= \cos ( \text{90} ° - \text{23} °) \cos ( \text{90} ° - \text{83} °) + \cos \text{23} ° \cos \text{83} ° \\ &= \sin \text{23} ° \sin \text{83} ° + \cos \text{23} ° \cos \text{83} ° \\ &= \cos( \text{83} ° - \text{23} °) \\ &= \cos \text{60} ° \\ &= \frac{1}{2} \end{align*}

Solve the equation:

\(\cos3 \theta \cos \theta -\sin3 \theta \sin\theta =-\frac{1}{2}\) for \(\theta \in [- \text{90} °; \text{90} °]\).

\begin{align*} \cos3 \theta \cos \theta -\sin3 \theta \sin \theta &= -\frac{1}{2} \\ \cos( 3\theta + \theta ) &= -\frac{1}{2} \\ \cos 4 \theta &= -\frac{1}{2} \\ \text{ref } \angle &= \text{60} ° \\ \text{Second quadrant: } \enspace 4 \theta &= \text{180} ° - \text{60} ° + k \cdot \text{360} ° \\ &= \text{120} ° + k \cdot \text{360} ° \\ \therefore \theta &= \text{30} ° + k \cdot \text{90} ° \\ \therefore \theta &= - \text{60} ° \text{ or } \text{30} ° \\ \text{Third quadrant: } \enspace 4 \theta &= \text{180} ° + \text{60} ° + k \cdot \text{360} ° \\ &= \text{240} ° + k \cdot \text{360} ° \\ \therefore \theta &= \text{60} ° + k \cdot \text{90} ° \\ \therefore \theta &= - \text{30} ° \text{ or } \text{60} ° \\ \text{Final answer: } \enspace \theta &\in \{ - \text{60} °; - \text{30} °; \text{30} °; \text{60} ° \} \end{align*}

Find the general solution, without a calculator, for the following equations:

\(3\sin\theta =2\cos^{2}\theta\)

\begin{align*} 3\sin\theta &= 2\cos^{2}\theta\\ 3\sin\theta &= 2(1 - \sin^{2}\theta)\\ 3\sin\theta &= 2 - 2\sin^{2}\theta\\ 2\sin^{2}\theta + 3\sin\theta - 2 &= 0\\ \text{let }k &= \sin\theta\\ 2k^{2} + 3k - 2 &= 0\\ (k + 2)(2k - 1) &= 0 \\ \text{So }k = -2 &\text{ or }k = \frac{1}{2}\\ \text{if }k = -2 \text{, }&\sin\theta = -2 \text{ which has no solution.}\\ \text{if }k = \frac{1}{2} &\text{, }\sin\theta = \frac{1}{2}\\ \text{ref } \angle &= \text{30} ° \\ \text{First quadrant: } \enspace \theta &= \text{30} ° + k \cdot \text{360} ° \\ \text{Second quadrant: } \enspace \theta &= ( \text{180} ° - \text{30} °) + k \cdot \text{360} ° \\ &= \text{150} ° + k \cdot \text{360} °, k \in \mathbb{Z} \\ \text{Final answer: } \theta = \text{30} ° + k \cdot \text{360} ° &\text{ or } \theta = \text{150} ° + k \cdot \text{360} °, k \in \mathbb{Z} \end{align*}

\(2 \sin 2x - 2 \cos x = \sqrt{2} - 2 \sqrt{2} \sin x\)

\begin{align*} 2 \sin 2x - 2 \cos x &= \sqrt{2} - 2\sqrt{2} \sin x \\ 0&= 2 (2\sin x \cos x) - 2 \cos x + 2\sqrt{2} \sin x - \sqrt{2} \\ 0&= 4\sin x \cos x - 2 \cos x + 2\sqrt{2} \sin x - \sqrt{2} \\ 0&= 2\cos x (2\sin x - 1) + \sqrt{2} ( 2\sin x - 1) \\ 0&= (2\sin x - 1)(2\cos x + \sqrt{2}) \\ \text{If } 2\sin x - 1 &= 0 \\ \therefore \sin x &= \frac{1}{2} \\ \text{ref } \angle &= \text{30} ° \\ \text{First quadrant: } \enspace x &= \text{30} ° + k \cdot \text{360} ° \\ \text{Second quadrant: } \enspace x &= \text{150} ° + k \cdot \text{360} °, k \in \mathbb{Z} \\ & \\ \text{If } 2\cos x + \sqrt{2} &= 0 \\ \therefore \cos x &= - \frac{\sqrt{2}}{2} \\ &= - \frac{\sqrt{2}}{2} \times \frac{\sqrt{2}}{\sqrt{2}} \\ &= - \frac{1}{\sqrt{2}} \\ \text{ref } \angle &= \text{45} ° \\ \text{Second quadrant: } \enspace x &= ( \text{180} ° - \text{45} °) + k \cdot \text{360} ° \\ &= \text{135} ° + k \cdot \text{360} ° \\ \text{Third quadrant: } \enspace x &= ( \text{180} ° + \text{45} °) + k \cdot \text{360} ° \\ &= \text{225} ° + k \cdot \text{360} °, k \in \mathbb{Z} \end{align*}

\(\cos x \cos \text{10} ° + \sin x \cos \text{100} ° = 1 - 2 \sin^{2} x\)
\begin{align*} \cos x \cos \text{10} ° + \sin x \cos \text{100} ° &= 1 - 2 \sin^{2} x \\ \cos x \cos \text{10} ° + \sin x \cos ( \text{90} ° + \text{10} °) &= \cos 2x \\ \cos x \cos \text{10} ° - \sin x \sin \text{10} ° &= \cos 2x \\ \cos (x + \text{10} °) &= \cos 2x \\ \text{First quadrant: } \enspace x + \text{10} ° &= 2x + k \cdot \text{360} ° \\ x &= \text{10} ° + k \cdot \text{360} ° \\ & \\ \text{Fourth quadrant: } \enspace x + \text{10} ° &= ( \text{360} ° - 2x) + k \cdot \text{360} ° \\ 3x &= \text{350} ° + k \cdot \text{360} ° \\ \therefore x &= \text{116,7} ° + k \cdot \text{120} ° \\ & \\ \text{Final answer: } \enspace x &= \text{10} ° + k \cdot \text{360} ° \\ x &= \text{116,7} ° + k \cdot \text{120} °, k \in \mathbb{Z} \end{align*}
\(6 \sin^{2} \alpha + 2 \sin 2 \alpha - 1 = 0\)
\begin{align*} 6 \sin^{2} \alpha + 2 \sin 2 \alpha - 1 &= 0 \\ 6 \sin^{2} \alpha + 2 (2 \sin \alpha \cos \alpha) - 1 &= 0 \\ 6 \sin^{2} \alpha + 4\sin \alpha \cos \alpha - (\sin^{2} \alpha + \cos^{2} \alpha) &= 0 \\ 6 \sin^{2} \alpha + 4\sin \alpha \cos \alpha - \sin^{2} \alpha - \cos^{2} \alpha &= 0 \\ 5 \sin^{2} \alpha + 4\sin \alpha \cos \alpha - \cos^{2} \alpha &= 0 \\ (5 \sin \alpha - \cos \alpha)(\sin \alpha + \cos \alpha) &= 0 \\ \text{If } 5 \sin \alpha - \cos \alpha &= 0 \\ 5 \sin \alpha &= \cos \alpha \\ \therefore \tan \alpha &= \frac{1}{5} \\ \therefore \alpha &= \text{11,3} ° + k \cdot \text{180} ° \\ & \\ \text{If } \sin \alpha + \cos \alpha &= 0 \\ \sin \alpha &= - \cos \alpha \\ \therefore \tan \alpha &= - 1 \\ \text{ref } \angle &= \text{45} ° \\ \text{Second quadrant: } \alpha &= ( \text{180} ° - \text{45} °) + k \cdot \text{180} ° \\ &= \text{135} ° + k \cdot \text{180} °, k \in \mathbb{Z} \\ & \\ \text{Final answer: } \enspace \alpha &= \text{11,3} ° + k \cdot \text{180} ° \\ \alpha &= \text{135} ° + k \cdot \text{180} °, k \in \mathbb{Z} \end{align*}
Prove: \({\sin}^{3}\theta =\frac{3\sin\theta -\sin3\theta }{4}\)

\begin{align*} \text{RHS } &= \frac{1}{4} \left( 3\sin\theta -\sin[2\theta + \theta] \right) \\ &= \frac{1}{4}(3 \sin\theta - (\sin2\theta \cos\theta + \cos2\theta \sin\theta))\\ &= \frac{1}{4}(3 \sin\theta - \sin2\theta \cos\theta - \cos2\theta \sin\theta)\\ &= \frac{1}{4}(3 \sin\theta - 2 \sin\theta \cos\theta \cos\theta - \sin\theta (1 - 2 \sin^{2}\theta))\\ &= \frac{1}{4}(3 \sin\theta - 2 \sin\theta \cos^{2}\theta - \sin\theta + 2\sin^{3}\theta)\\ &= \frac{1}{4}(2 \sin\theta - 2 \sin\theta \cos^{2}\theta + 2\sin^{3}\theta)\\ &= \frac{1}{4}(2 \sin\theta - 2 \sin\theta (1 - \sin^{2}\theta) + 2\sin^{3}\theta)\\ &= \frac{1}{4}(2 \sin\theta - 2 \sin\theta + 2\sin^{3}\theta + 2\sin^{3}\theta)\\ &= \frac{1}{4}(4\sin^{3}\theta)\\ &= \sin^{3}\theta\\ &= \text{LHS} \end{align*}

Hence, solve the equation \(3 \sin \theta - \sin 3 \theta = 2\) for \(\theta \in [ \text{0} °; \text{360} °]\).
\begin{align*} 3 \sin \theta - \sin 3 \theta &= 2 \\ \frac{ 3 \sin \theta - \sin 3 \theta}{4} &= \frac{2}{4} \\ \frac{ 3 \sin \theta - \sin 3 \theta}{4} &= \frac{1}{2} \\ \therefore \sin^{3} \theta &= \frac{1}{2} \\ \therefore \sin \theta &= \text{0,793} \ldots \\ \text{ref } \angle &= \text{52,53} ° \\ \text{First quadrant: } \theta &= \text{52,53} ° + k \cdot \text{360} °, k \in \mathbb{Z} \\ & \\ \text{Second quadrant: } \theta &= ( \text{180} ° - \text{52,53} °) + k \cdot \text{180} ° \\ &= \text{127,47} ° + k \cdot \text{360} °, k \in \mathbb{Z} \\ & \\ \text{Final answer: } \theta &= \text{52,53} ° \\ \theta &= \text{127,47} ° \end{align*}

Prove the following identities:

\({\cos}^{2}\alpha \left(1-{\tan}^{2}\alpha \right)=\cos2\alpha\)

\begin{align*} \text{LHS }&= \cos^{2}\alpha(1 - \tan^{2}\alpha)\\ &= \cos^{2}\alpha - \cos^{2}\alpha \left( \frac{\sin^{2}\alpha}{\cos^{2}\alpha} \right) \\ &= \cos^{2}\alpha - \sin^{2}\alpha\\ &= \cos2\alpha\\ &= \text{ RHS} \end{align*}

Restrictions: \(\alpha \ne \text{90} ° + k \cdot \text{180} °, \quad k \in \mathbb{Z}\).

\(4\sin\theta \cos\theta \cos2\theta =\sin4\theta\)

\begin{align*} \text{RHS }&= \sin4\theta\\ &= \sin2(2\theta)\\ &= 2\sin2\theta\cos2\theta\\ &= 2(2\sin\theta\cos\theta)\cos2\theta\\ &= 4\sin\theta\cos\theta\cos2\theta\\ &= \text{ LHS} \end{align*}

\(4\cos^{3}x-3 \cos x=\cos{3x}\)

\begin{align*} \text{RHS }&=\cos 3x \\ &= \cos(2x+x)\\ & =\cos 2x \cos x - \sin 2x \sin x\\ & =\cos2x\cos x - 2 \sin^{2}x \cos x\\ & =\cos x( \cos2x - 2 \sin^{2}x) \\ & =\cos x( 2\cos^{2}x - 1 - 2 [1 - \cos^{2}x]) \\ & =\cos x( 2\cos^{2}x - 1 - 2 + 2\cos^{2}x) \\ & =\cos x( 4\cos^{2}x - 3) \\ & = 4\cos^{3}x - 3 \cos x \\ & =\text{ LHS} \end{align*}

\(\cos 2 A + 2 \sin 2A + 2 = (3 \cos A + \sin A)(\cos A + \sin A)\)

\begin{align*} \text{LHS } &= \cos 2 A + 2 \sin 2A + 2 \\ &= (\cos^{2} A - \sin^{2} A) + (4 \sin A \cos A) + 2(\cos^{2} A + \sin^{2} A) \\ &= \cos^{2} A - \sin^{2} A + 4 \sin A \cos A + 2\cos^{2} A + 2\sin^{2} A) \\ &= 3\cos^{2} A + 4 \sin A \cos A + \sin^{2} A \\ &= (3 \cos A + \sin A)(\cos A + \sin A) \\ & =\text{ RHS} \end{align*}

\(\dfrac{\cos 2x}{(\cos x + \sin x)^{3}} = \dfrac{\cos x - \sin x}{1 + \sin 2x}\)

\begin{align*} \text{LHS }&= \frac{\cos 2x}{(\cos x + \sin x)^{3}} \\ &= \frac{\cos^{2}x - \sin^{2}x}{(\cos x + \sin x)^{3}} \\ &= \frac{(\cos x - \sin x)(\cos x + \sin x)}{(\cos x + \sin x)^{2} (\cos x + \sin x)} \\ &= \frac{\cos x - \sin x }{(\cos x + \sin x)^{2}} \\ &= \frac{\cos x - \sin x }{\cos^{2} x + 2 \cos x \sin x+ \sin^{2} x} \\ &= \frac{\cos x - \sin x }{1 + 2 \cos x \sin x } \\ &= \frac{\cos x - \sin x }{1 + \sin 2x } \\ & =\text{ RHS} \end{align*}

Prove: \(\tan{y}=\dfrac{\sin2y}{\cos2y+1}\)

\begin{align*} \text{RHS } &= \frac{\sin2y}{\cos2y+1}\\ &= \frac{\sin 2y}{(2\cos^{2}y - 1) + 1}\\ &= \frac{2\sin y\cos y}{2 \cos^{2}y}\\ &= \frac{\sin y}{\cos y}\\ &= \tan y \\ &= \text{ LHS} \end{align*}

For which values of \(y\) is the identity undefined?

The identity is undefined for the values of \(y\) such that \(\cos2y+1 = 0\) since division by zero is not permitted.

\begin{align*} \text{If } \cos2y+1 &= 0 \\ \cos2y &= - 1 \\ \therefore 2y &= \text{180} ° + k \cdot \text{360} ° \\ \therefore y &= \text{90} ° + k \cdot \text{180} °, k \in \mathbb{Z} \end{align*}

Restricted values are: \(y = \text{90} ° + k \cdot \text{180} °, k \in \mathbb{Z}\)

Given: \(1 + \tan^{2} 3 \theta - 3 \tan 3 \theta = 5\)

Find the general solution.
\begin{align*} 1 + \tan^{2} 3 \theta - 3 \tan 3 \theta &= 5 \\ \tan^{2} 3 \theta - 3 \tan 3 \theta - 4 &= 0 \\ (\tan 3 \theta - 4)(\tan 3 \theta + 1) &= 0 \\ \text{If } \tan 3 \theta & = -1 \\ \text{ref } \angle &= \text{45} ° \\ \therefore 3 \theta &= ( \text{180} ° - \text{45} °)+ k \cdot \text{180} ° \\ 3 \theta &= \text{135} ° + k \cdot \text{180} ° \\ \therefore \theta &= \text{45} ° + k \cdot \text{60} °, k \in \mathbb{Z} \\ & \\ \text{If } \tan 3 \theta & = 4 \\ \text{ref } \angle &= \text{75,96} ° \\ \therefore 3 \theta &= \text{75,96} ° + k \cdot \text{180} ° \\ \therefore \theta &= \text{25,32} ° + k \cdot \text{60} °, k \in \mathbb{Z} \end{align*}
Find the solution for \(\theta \in [ \text{0} °; \text{90} °]\).
\begin{align*} k=0: \quad \theta \theta &= \text{45} ° \\ &= \text{25,32} ° \\ k=1: \quad \theta &= \text{85,32} ° \end{align*}
Draw a graph of \(y = \tan 3 \theta\) for \(\theta \in [ \text{0} °; \text{90} °]\) and indicate the solutions to the equation on the graph.
f83881f38b179a61f5ec42e11e059942.png
Use the graph to determine where \(\tan 3 \theta < -1\).
b583062b094e4e434f44177f34d4afea.png

\(\text{30} ° < \theta < \text{45} °\)

Show that: \[\sin (A + B) - \sin (A - B) = 2 \cos A \sin B\]
\begin{align*} \text{LHS } &= \sin (A + B) - \sin (A - B) \\ &= \sin A \cos B + \cos A \sin B - [ \sin A \cos B - \cos A \sin B ] \\ &= \sin A \cos B + \cos A \sin B - \sin A \cos B + \cos A \sin B \\ &= 2 \cos A \sin B \\ &= \text{RHS} \end{align*}
Use this result to solve \(\sin 3x - \sin x = 0\) for \(x \in [- \text{180} °; \text{360} ° ]\).
\begin{align*} \sin 3x - \sin x &= 0 \\ \sin (2x + x) - \sin (2x - x) &= 0 \\ \text{So } A = 2x & \text{ and } B = x \\ \therefore \text{we can write } 2 \cos 2x \sin x &= 0 \\ & \\ \text{If } \cos 2x &= 0 \\ 2x &= \text{90} ° + k \cdot \text{360} ° \\ \therefore x &= \text{45} ° + k \cdot \text{180} °, k \in \mathbb{Z} \\ & \\ 2x &= \text{270} ° + k \cdot \text{360} ° \\ \therefore x &= \text{135} ° + k \cdot \text{180} °, k \in \mathbb{Z} \\ & \\ \text{If } \sin x &= 0 \\ x &= \text{0} ° + k \cdot \text{360} °, k \in \mathbb{Z} \\ \text{ or } x &= \text{180} ° + k \cdot \text{180} °, k \in \mathbb{Z} \\ & \\ \text{Final answer: } k=-2: \quad x &= - \text{180} ° \\ k=-1: \quad x &= - \text{135} °; - \text{45} ° \\ k=0: \quad x &= \text{45} °; \text{135} °; \text{0} °; \text{180} ° \\ k=1: \quad x &= \text{225} °; \text{315} °; \text{360} ° \\ \text{Final answer: } x &= - \text{180} °; - \text{135} °; - \text{45} °; \\ & \text{0} °; \text{45} °; \text{135} °; \text{180} °; \\ & \text{225} °; \text{315} °; \text{360} ° \end{align*}
On the same system of axes, draw two graphs to solve graphically: \(\sin 3x - \sin x = 0\) for \(x \in [ \text{0} °; \text{360} ° ]\). Indicate the solutions on the graph using the letters \(A, B, \ldots\) etc.
56504850ea17922503a4239c6ec2abc9.png

Given: \(\cos 2 x = \sin x\) for \(x \in [ \text{0} °; \text{360} °]\)

Solve for \(x\) algebraically.
\begin{align*} \cos 2x &= \sin x \\ \cos 2x &= \cos \left( \text{90} ° - x \right) \\ 2x &= \text{90} ° - x + k \cdot \text{360} ° \\ 3x &= \text{90} ° + k \cdot \text{360} ° \\ x &= \text{30} ° + k \cdot \text{120} ° \\ & \\ 2x &= [ \text{360} ° - ( \text{90} ° - x )] + k \cdot \text{360} ° \\ &= \text{270} ° + x + k \cdot \text{360} ° \\ x &= \text{270} ° + k \cdot \text{360} ° \\ & \\ k=0: \quad x &= \text{30} ° \\ x &= \text{270} ° \\ k=1: \quad x &= \text{150} ° \\ x &= \text{270} ° \end{align*}
Verify the solution graphically by sketching two graphs on the same system of axes.
33455391af102a3878aaed7c23d4e448.png

The following graphs are given below:

\begin{align*} f: &= a \sin x \\ g: &= \cos bx \qquad \left( x \in [ \text{0} °; \text{360} °] \right) \end{align*}0ed8485408a55bc0edf23db78512b285.png
Explain why \(a = 2\) and \(b = \frac{1}{2}\).

From the graph we can see that the amplitude of the sine graph is \(\text{2}\), therefore \(a = 2\).

For the cosine graph, the period is \(\text{720} °\), therefore \(b = \frac{ \text{360} °}{ \text{720} °} = \frac{1}{2}\).

For how many \(x\)-values in \([ \text{0} °; \text{360} °]\) will \(f(x) - g(x) = 0\)?

For:

\begin{align*} f(x) - g(x) &= 0 \\ f(x) &= g(x) \end{align*}

For the interval \([ \text{0} °; \text{360} °]\), the diagram shows that the two graphs intersect at three places.

Use the graph to solve \(f(x) - g(x) = 1\).

From the graph we have the following solution:

\begin{align*} f( \text{360} °) - g( \text{360} °) &= 0 - (-1) = 1 \\ \therefore x &= \text{360} ° \end{align*}
Solve \(a \sin x = \cos bx\) for \(x \in [ \text{0} °; \text{360} °]\) using trigonometric identities.
\begin{align*} 2 \sin x &= \cos \frac{1}{2}x \\ 2 \sin 2 \left( \frac{x}{2} \right) &= \cos \frac{1}{2}x \\ 2 \left( 2 \sin \frac{x}{2} \cos \frac{x}{2} \right) &= \cos \frac{1}{2}x \\ 4 \sin \frac{x}{2} \cos \frac{x}{2} - \cos \frac{1}{2}x &= 0 \\ \cos \frac{x}{2} \left( 4 \sin \frac{x}{2} - 1 \right) &= 0 \end{align*} \begin{align*} \text{If } \cos \frac{x}{2} &= 0 \\ \frac{x}{2} &= \text{90} ° \\ \therefore x &= \text{180} ° \\ & \\ \text{If } 4 \sin \frac{x}{2} - 1 &= 0 \\ 4 \sin \frac{x}{2} &= 1 \\ \sin \frac{x}{2} &= \frac{1}{4} \\ \text{ref } \angle &= \text{14,47} ° \\ \therefore \frac{x}{2} &= \text{14,47} ° \\ x &= \text{29} ° \\ \text{ or } \frac{x}{2} &= \text{180} ° - \text{14,47} ° \\ &= \text{165,53} ° \\ \therefore x &= \text{331} ° \\ & \\ \text{Final answer: } x &= \text{180} °, \text{29} °, \text{331} ° \end{align*}
For which values of \(x\) will \(\frac{1}{2} \cos \left( \frac{x}{2} \right) \leq \sin x\) for \(x \in [ \text{0} °; \text{360} °]\)?
\begin{align*} \frac{1}{2} \cos \left( \frac{x}{2} \right) & \leq \sin x \\ \cos \left( \frac{x}{2} \right) & \leq 2\sin x \\ \text{where } g(x) & \leq f(x) \\ \text{Answer: } \text{29} ° \leq x \leq \text{180} ° &\text{ or } \text{331} ° \leq x \leq \text{360} ° \end{align*}

In \(\triangle ABC\), \(AB = c, BC = a, CA = b \text{ and } \hat{C} = \text{90} °\)

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Prove that \(\sin 2A = \frac{2ab}{c^{2}}\).
\begin{align*} \text{LHS} &= \sin 2A \\ &= 2 \sin A \cos A \\ &= 2 \left( \frac{a}{c} \right) \left( \frac{b}{c} \right) \\ &= \frac{2ab}{c^{2}} \\ &= \text{RHS} \end{align*}
Show that \(\cos 2A = \frac{b^{2} - a^{2}}{c^{2}}\).
\begin{align*} \text{LHS} &= \cos 2A \\ &= \cos^{2} A - \sin^{2} A \\ &= \left( \frac{b}{c} \right)^{2} - \left( \frac{a}{c} \right)^{2} \\ &= \frac{b^{2} - a^{2}}{c^{2}} \\ &= \text{RHS} \end{align*}

Given the graphs of \(f(\theta) = p \sin k\theta\) and \(g(\theta) = q \tan \theta\), determine the values of \(p, k\) and \(q\).

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\(f(\theta) = \frac{3}{2} \sin 2\theta\) and \(g(\theta) = -\frac{3}{2} \tan \theta\)
\(\triangle RST\) is an acute angled triangle with \(RS = ST = t\). Show that area \(\triangle RST = t^{2} \sin \hat{T} \cos \hat{T}\).
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\(RSTU\) is a cyclic quadrilateral with \(RU = \text{6}\text{ cm}, UT = \text{7,5}\text{ cm}, RT = \text{11}\text{ cm} \text{ and } RS = \text{9,5}\text{ cm}\).

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Calculate \(\hat{U}\).
\begin{align*} \text{In } \triangle RUT \quad RT^{2} &= RU^{2} + UT^{2} - 2 RU \cdot UT \cos \hat{U} \\ \therefore \cos \hat{U} &= \frac{RU^{2} + UT^{2} - RT^{2}}{2 RU \cdot UT } \\ &= \frac{6^{2} + (\text{7,5})^{2} - 11^{2}}{2(6)(\text{7,5}) } \\ &= - \text{0,3194} \ldots \\ \text{ref } \angle &= \text{71,4} ° \\ \hat{U} &= \text{180} ° - \text{71,4} ° \\ &= \text{108,6} ° \end{align*}
Determine \(\hat{S}\).
\begin{align*} \hat{S} &= \text{180} ° - \text{108,6} ° \quad (\text{opp. } \angle \text{s of cyclic. quad. suppl.}) \\ &= \text{71,4} ° \end{align*}
Find \(R\hat{T}S\).
\begin{align*} \frac{\sin R\hat{T}S}{RS} &= \frac{\sin R\hat{S}T}{RT} \\ \frac{\sin R\hat{T}S}{\text{9,5}} &= \frac{\sin \text{71,4}}{\text{11}} \\ \sin R\hat{T}S &= \frac{\text{9,5} \sin \text{71,4}}{\text{11}} \\ &= \text{0,8185} \ldots \\ \therefore R\hat{T}S &= \text{54,9} ° \end{align*}

\(BCDE\) is a cyclic quadrilateral that lies in a horizontal plane. \(AB\) is a vertical pole with base \(B\). The angle of elevation from \(E\) to \(A\) is \(x°\) and \(C\hat{D}E = y°\). \(\triangle BEC\) is an isosceles triangle with \(BE = BC\).

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Show that \(B\hat{C}E = \frac{1}{2}y\).

In cyclic quadrilateral \(BCDE\):

\begin{align*} \hat{B} &= \text{180} ° - y \\ BE &= BC \qquad \text{(given)} \\ B\hat{C}E &= B\hat{E}C \\ &= \frac{ \text{180} ° - ( \text{180} ° - y)}{2} \\ &= \frac{y}{2} \end{align*}
Show that \(CE = 2BE \cos \left( \frac{y}{2} \right)\)
\begin{align*} \text{In } \triangle BCE: \quad \frac{CE}{\sin C\hat{B}E} &= \frac{BE}{\sin B\hat{C}E} \\ \therefore CE &= \frac{BE \sin C\hat{B}E }{\sin B\hat{C}E} \\ &= \frac{BE \sin ( \text{180} ° - y) }{\sin \left( \frac{y}{2} \right)} \\ &= \frac{BE \sin y}{\sin \left( \frac{y}{2} \right)} \\ &= \frac{BE (2 \sin \left( \frac{y}{2} \right) \cos \left( \frac{y}{2} \right) )}{\sin \left( \frac{y}{2} \right) } \\ &= 2BE \cos \left( \frac{y}{2} \right) \end{align*}
If \(AB = \text{2,6}\text{ m}, x = \text{37} ° \text{ and } y = \text{109} °\), calculate the length of \(CE\).
\begin{align*} \text{In } \triangle ABE: \quad \frac{AB}{BE} &= \tan x \qquad (A\hat{B}E = \text{90} °) \\ \therefore BE &= \frac{AB}{\tan x} \\ &= \frac{\text{2,6} }{\tan \text{37} °} \\ &= \text{3,45} \\ \therefore CE &= 2BE \cos \left( \frac{y}{2} \right) \\ &= 2 (\text{3,45} ) \cos \left( \frac{ \text{109} °}{2} \right) \\ &= \text{4}\text{ m} \end{align*}

The first diagram shows a rectangular box with \(SR = \text{8}\text{ cm}, PS = \text{6}\text{ cm}\) and \(PA = \text{4}\text{ cm}\). The lid of the box, \(ABCD\), is opened \(\text{30} °\) to the position \(XYCD\), as shown in the second diagram.

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Write down the dimensions (length, breadth and diagonal) of the lid \(XYCD\).
\begin{align*} \text{length } &= XY = DC = \text{8}\text{ cm} \\ \text{breadth } &= XD = YC = \text{6}\text{ cm} \\ \text{diagonal } &= AC = XC \\ &= \sqrt{8^{2} + 6^{2}} \qquad (\text{Pythagoras})\\ &= \text{10}\text{ cm} \end{align*}
Calculate \(XZ\), the perpendicular height of \(X\) above the base of the box.
\begin{align*} TZ &= AP = \text{4}\text{ cm} \\ \text{In } \triangle XTD: \quad \frac{XT}{XD} &= \sin \text{30} ° \\ \frac{XT}{6} &= \frac{1}{2} \\ \therefore XT &= \text{3}\text{ cm} \\ \therefore XZ &= 4 + 3 = \text{7}\text{ cm} \end{align*}
Calculate the ratio \(\frac{\sin X\hat{Z}C}{\sin X\hat{C}Z}\).
\begin{align*} \text{In } \triangle XTC: \quad \frac{\sin X\hat{Z}C}{XC} &= \frac{\sin X\hat{C}Z}{XZ} \\ \therefore \frac{\sin X\hat{Z}C}{\sin X\hat{C}Z} &= \frac{XC}{XZ} \\ &= \frac{10}{7} \end{align*}

\(AB\) is a vertical pole on a horizontal plane \(BCD\). \(DC\) is \(a\) metres and the angle of elevation from \(D\) to \(A\) is \(\theta\). \(A\hat{C}D = \alpha\) and \(A\hat{D}C = \beta\).

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Name the two right angles in the diagram.
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Show that \(AB = \frac{a \sin \alpha \sin \theta}{\sin (\alpha + \beta)}\).
\begin{align*} \text{In } \triangle ABC: \quad \frac{AB}{AD} &= \sin \theta \\ \therefore AB &= AD \sin \theta \\ \text{In } \triangle ADC: \quad \frac{AD}{\sin \alpha} &= \frac{DC}{\sin D\hat{A}C} \\ AD &= \frac{a \sin \alpha}{\sin [ \text{180} ° - (\alpha + \beta)]} \\ &= \frac{a \sin \alpha}{\sin (\alpha + \beta)} \\ & \\ \therefore AB &= \frac{a \sin \alpha \sin \theta }{\sin (\alpha + \beta)} \end{align*}
If it is given that \(AD = AC\), show that the height of the pole is given by \(AB = \frac{a \sin \theta}{2 \cos \alpha}\).
\begin{align*} AD &= AC \qquad (\text{given}) \\ \therefore \beta &= \alpha \qquad (\text{isos. } \triangle ADC) \\ \therefore AB &= \frac{a \sin \alpha \sin \theta }{\sin (\alpha + \alpha)} \\ &= \frac{a \sin \alpha \sin \theta }{\sin 2 \alpha} \\ &= \frac{a \sin \alpha \sin \theta }{2 \sin \alpha \cos \alpha} \\ &= \frac{a \sin \theta }{2 \cos \alpha} \end{align*}
Calculate the height of the pole if \(a = \text{13}\text{ m}, \theta = \text{33} °, \alpha = \beta = \text{65} °\).
\begin{align*} AB &= \frac{a \sin \theta }{2 \cos \alpha} \\ &= \frac{13 \times \sin \text{33} ° }{2 \cos \text{65} °} \\ &= \text{8,4}\text{ m} \end{align*}

\(AB\) is a flagpole on top of a government building \(BC\). \(AB = f\) units and \(D\) is a point on the ground in the same horizontal plane as the base of the building, \(C\). The angle of elevation from \(D\) to \(A\) and \(B\) is \(\alpha\) and \(\beta\), respectively.

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Show that \(f = \frac{BC \sin (\alpha - \beta)}{\sin \beta \cos \alpha}\)
\begin{align*} \text{In } \triangle ABD: \quad \frac{f}{\sin (\alpha - \beta)} &= \frac{DB}{\sin ( \text{90} ° - \alpha)} \\ \therefore f &= \frac{DB \sin (\alpha - \beta)}{\cos \alpha} \\ \text{In } \triangle BDC: \quad \frac{BC}{DB} &= \sin \beta \\ \therefore DB &= \frac{BC}{\sin \beta} \\ & \\ \therefore f &= \frac{BC \sin (\alpha - \beta)}{\sin \beta \cos \alpha} \end{align*}
Calculate the height of the flagpole (to the nearest metre) if the building is \(\text{7}\text{ m}\), \(\alpha = \text{63} °\) and \(\beta = \text{57} °\).
\begin{align*} f &= \frac{BC \sin (\alpha - \beta)}{\sin \beta \cos \alpha} \\ &= \frac{7 \times \sin ( \text{63} ° - \text{57} °)}{\sin \text{57} ° \cos \text{63} °} \\ &= \text{1,92} \\ \therefore \text{height} &\approx \text{2}\text{ m} \qquad (\text{nearest metre}) \end{align*}