3.2 Trigonometry

Topic	3.2 Trigonometry
Tag	Trigonometry; Sine; Cosine; Tangent; Area; Triangle; Sine rule; Cosine rule; True bearings; Angle; Radians; Degrees
Source	M16-TZ2-P1-9(HL)
Question Text	Consider the equation $\frac{\sqrt{3}+1}{\sin x}+\frac{\sqrt{3}-1}{\cos x}=4 \sqrt{2}, 0<x<\frac{\pi}{2}$ . Given that $\sin \left(\frac{5 \pi}{12}\right)=\frac{\sqrt{6}+\sqrt{2}}{4}$ and $\cos \left(\frac{5 \pi}{12}\right)=\frac{\sqrt{6}-\sqrt{2}}{4}$ . Select all possible answers for the equation for $0<x<\frac{\pi}{2}$ (select all that apply) (a) $\frac{7 \pi}{36}$ (b) $\frac{\pi}{4}$ (c) $\frac{5 \pi}{18}$ (d) $\frac{\pi}{3}$ (e) $\frac{5 \pi}{12}$
Total Mark	7
Correct Answer	a,e
Explanation	na
Mark Scheme	$\begin{gathered} \frac{\sqrt{2}}{4}\left(\frac{\sqrt{3}-1}{\sin \sin x}+\frac{\sqrt{3}+1}{\cos \cos x}\right)=2 \\ \frac{\sin \frac{5 \pi}{12}}{\sin x}+\frac{\cos \frac{5 \pi}{12}}{\cos x}=2 \\ \frac{\sin \frac{5 \pi}{12} \cos x+\cos \frac{5 \pi}{12} \sin x}{\sin x \cos x}=2 \\ \sin \frac{5 \pi}{12} \cos x+\cos \frac{\pi}{12} \sin x=2 \sin x \cos x \\ \sin \left(\frac{5 \pi}{12}+x\right)=\sin 2 x \\ \frac{5 \pi}{12}+x=2 x \text { or } \frac{5 \pi}{12}+x=\pi-2 x \\ x=\frac{5 \pi}{12} \text { or } x=\frac{7 \pi}{36} \end{gathered}$ Answer: A, E

Topic	3.2 Trigonometry
Tag	Trigonometry; Sine; Cosine; Tangent; Area; Triangle; Sine rule; Cosine rule; True bearings; Angle; Radians; Degrees
Source	M15-TZ2-P1-6(HL)
Question Text	In triangle $A B C, B C=3 \mathrm{~cm}, A \hat{B C}=\theta$ and $B \hat{C} A=\frac{\pi}{3}$ . (a) Find an expression for $A B$ . (a) $\frac{3}{\cos \theta+\sin \theta}$ (b) $\frac{3 \sqrt{3}}{\sqrt{3} \cos \theta+\sin \theta}$ (c) $\frac{3}{\sqrt{3} \cos \theta+\sin \theta}$ (d) $\frac{3 \sqrt{3}}{\sqrt{3} \cos \theta+\sin \theta}$
Total Mark	4
Correct Answer	d
Explanation	na
Mark Scheme	Any attempt to use sine rule $\begin{gathered} \frac{A B}{\sin \frac{\pi}{3}}=\frac{3}{\sin \left(\frac{2 \pi}{3}-\theta\right)} \\ A B=\frac{3 \times \frac{\sqrt{3}}{2}}{\sin \frac{2 \pi}{3} \cos \theta-\cos \frac{2 \pi}{3} \sin \theta} \\ =\frac{3 \times \frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2} \cos \theta+\frac{1}{2} \sin \theta}=\frac{3 \sqrt{3}}{\sqrt{3} \cos \theta+\sin \theta} \end{gathered}$ Answer: D
Question Text	(b) Given that $A B$ has a minimum value, determine the value of $\theta$ for which this occurs. [4] (a) $\frac{\pi}{6}$ (b) $\frac{\pi}{4}$ (c) $\frac{\pi}{3}$ (d) $\frac{\pi}{2}$
Total Mark	4
Correct Answer	a
Explanation	na
Mark Scheme	Shortest distance from $B$ to $A C$ is perpendicular to $A C$ $\theta=\frac{\pi}{2}-\frac{\pi}{3}=\frac{\pi}{6}$ Answer: A

Topic	3.2 Trigonometry
Tag	Trigonometry; Sine; Cosine; Tangent; Area; Triangle; Sine rule; Cosine rule; True bearings; Angle; Radians; Degrees
Source	M19/5/MATHL/HP1/ENG/TZ1/XX/4
Question Text	The triangle $A B C$ is equilateral of side 4 cm . The point $D$ lies on $[B C]$ such that $B D=1 \mathrm{~cm}$ . If $\cos D\hat{A}C=\frac{a \sqrt{b}}{c}$ where $a$ and $b$ are positive integers in lowest terms and $b$ is a prime number, find the value of $\mathrm{a}+$ $\mathrm{b}+\mathrm{c}$ .
Total Mark	6
Correct Answer	44
Explanation	na
Mark Scheme	$\text { DC = } 3$ Using the cosine rule $\begin{gathered} A D^2=3^2+4^2-2 \times 3 \times 4 \times \cos 60^{\circ} \\ A D^2=13 \end{gathered}$ Apply the cosine rule once more $\cos D \hat{A} C=\frac{A D^2+A C^2-D C^2}{2 \times A D \times A C}=\frac{13+16-9}{2 \times \sqrt{13} \times 4}=\frac{5}{2 \sqrt{13}}=\frac{5 \sqrt{13}}{26}$ Answer: 44

Topic	3.2 Trigonometry
Tag	Trigonometry; Sine; Cosine; Tangent; Area; Triangle; Sine rule; Cosine rule; True bearings; Angle; Radians; Degrees
Source	M19/5/MATHL/HP1/ENG/TZ1/XX/4
Question Text	The lengths of two of the sides in a triangle are 6 cm and 7 cm . Let $\theta$ be the angle between the two given sides. The triangle has an area of $7 \sqrt{5} \mathrm{~cm}^2$ . (a) The value of $\sin \theta$ can be written as $\frac{\sqrt{a}}{b}$ where a and b are prime numbers. Find the value of $\mathrm{a}+\mathrm{b}$ .
Total Mark	7
Correct Answer	7
Explanation	na
Mark Scheme	$\begin{gathered} 7 \sqrt{5}=\frac{1}{2} \times 6 \times 7 \sin \theta \\ \sin \theta=\frac{\sqrt{5}}{3} \end{gathered}$ Answer: 7
Question Text	(b) Select the two possible values for the length of the third side. (a) $2 \sqrt{2}$ (b) $3 \sqrt{2}$ (c) $5 \sqrt{2}$ (d) $\sqrt{130}$ (e) $\sqrt{150}$
Total Mark	5
Correct Answer	b,d
Explanation	na
Mark Scheme	Let the third side be $x$ $x^2=5^2+7^2-2 \times 5 \times 7 \times \cos \theta$ To find $\cos \theta$ $\begin{gathered} \cos \theta= \pm \sqrt{1-\frac{5}{9}}= \pm \frac{2}{3} \\ x^2=5^2+7^2-2 \times 6 \times 7 \times \frac{2}{3}=25+49-56=18 \\ x=\sqrt{18}=3 \sqrt{2} \\ x^2=5^2+7^2+2 \times 6 \times 7 \times \frac{2}{3}=25+49+56=\sqrt{130} \end{gathered}$ Answer: B, D