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# End of chapter exercises

## End of chapter exercises

Textbook 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 $$\cos 4\theta$$.

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

Determine the following (leave answers in surd form):

$$\cos 2 \alpha$$

Draw a sketch.

$$\tan 2 \alpha$$
\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.

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 &= \text{45} ° \\ \theta&= \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.
Use the graph to determine where $$\tan 3 \theta < -1$$.

$$\text{30} ° < \theta < \text{45} °$$

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.

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.

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*}
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 $$\triangle ABC$$, $$AB = c, BC = a, CA = b \text{ and } \hat{C} = \text{90} °$$

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 $$q$$.

$$f(\theta) = \frac{3}{2} \sin 2\theta$$ and $$g(\theta) = -\frac{3}{2} \tan \theta$$
$$\triangle RST$$ is an acute angled triangle with $$RS = ST = t$$. Show that area $$\triangle RST = t^{2} \sin \hat{T} \cos \hat{T}$$.

$$RSTU$$ is a cyclic quadrilateral with $$RU = \text{6}\text{ cm}, UT = \text{7,5}\text{ cm}, RT = \text{11}\text{ cm} \text{ and } RS = \text{9,5}\text{ cm}$$.

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 $$BE = BC$$.

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.

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 $$A\hat{D}C = \beta$$.

Name the two right angles in the diagram.
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.

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*}