Trigonometry

Toolkit

Sample Problems

1. Determine the values of $x$ such that $2{\sin }^{3}x-5{\sin }^{2}x+2\sin x=0$ given that $0\le x\le 2\pi$.

   Solution

   We factor the given equation to obtain $$\begin{array}{rl}\sin x(2{\sin }^{2}x-5\sin x+2)& =0\\ \sin x(2\sin x-1)(\sin x-2)& =0\end{array}$$ So $\sin x=0$, ${\displaystyle \frac{1}{2}}$ or $2$. But $|\sin x|\le 1$. So $\sin x\ne 2$. Therefore, in the interval $0\le x\le 2\pi$, we have $x=0$, $\pi$, $2\pi$, ${\displaystyle \frac{\pi }{6}}$ or ${\displaystyle \frac{5\pi }{6}}$.

2. An airplane leaves an aircraft carrier and flies due south at $400$ km/hr. The carrier proceeds at a heading of $60\mathrm{°}$ west of north at $32$ km/hr. If the plane has $5$ hours of fuel, what is the maximum distance south the plane can travel so that the fuel remaining will allow a safe return to the carrier at $400$ km/hr?

   Solution

   The first step in solving this problem is to draw a diagram (as shown). If we let $x$ be the number of hours that the plane flies south, then the distance that the plane flies south is $400x$. The plane then flies a distance $400(5-x)$ in the remaining time, while the total distance the carrier travels is $5(32)=160$.

   

   Using these distances, the cosine law gives us $$(400(5-x){)}^2={160}^2+(400x{)}^2-2\cdot 160\cdot 400x\cdot \cos 120\mathrm{°}.$$ Simplifying we obtain $$4000000-1600000x+160000x^2=25600+160000x^2+64000x$$ which we can solve to get $x={\displaystyle \frac{621}{260}}$. Thus, the maximum distance the plane should travel south is $400\left({\displaystyle \frac{621}{260}}\right)={\displaystyle \frac{12420}{13}}$ km, which is approximately $955.4$ km.

3. In triangle $ABC$, the point $D$ is on $BC$ such that $AD$ bisects $\mathrm{\angle }A$.

   

   Show that ${\displaystyle \frac{AB}{BD}}={\displaystyle \frac{AC}{CD}}$.

   Solution

   We call $\mathrm{\angle }ADC=\theta$ and $\mathrm{\angle }BAC=\alpha$. We use the sine law in triangles $ADC$ and $ADB$ to obtain ${\displaystyle \frac{\sin {\displaystyle \frac{\alpha }{2}}}{\sin \theta }}={\displaystyle \frac{CD}{AC}}$ and ${\displaystyle \frac{\sin {\displaystyle \frac{\alpha }{2}}}{\sin (180\mathrm{°}-\theta )}}={\displaystyle \frac{BD}{AB}}$. But $\sin \theta =\sin (180\mathrm{°}-\theta )$ and so ${\displaystyle \frac{AB}{BD}}={\displaystyle \frac{AC}{CD}}$. This result is known as the angle bisector theorem.

4. For the given triangle $ABC$, $\mathrm{\angle }C=\mathrm{\angle }A+60\mathrm{°}$. If $BC=1$, $AC=r$ and $AB=r^2$, where $r>1$, prove that $r<\sqrt{2}.$

   

   Solution

   We represent the angles of the triangle as: $\mathrm{\angle }A=\alpha$, $\mathrm{\angle }C=\alpha +60\mathrm{°}$, and $\mathrm{\angle }B=120\mathrm{°}-2\alpha$. So the sine law states $$\begin{array}{rl}{\displaystyle \frac{r^2}{1}}& ={\displaystyle \frac{\sin (\alpha +60\mathrm{°})}{\sin \alpha }}\\ & ={\displaystyle \frac{\sin \alpha \cos 60\mathrm{°}+\cos \alpha \sin 60\mathrm{°}}{\sin \alpha }}\\ & ={\displaystyle \frac{1}{2}}+{\displaystyle \frac{\sqrt{3}}{2}}\cot \alpha \end{array}$$ Since all three angles in the triangle are positive, we can see that $0\mathrm{°}<\alpha <60\mathrm{°}$. In this range, the tangent function is increasing, and its reciprocal, the cotangent function, is decreasing.
   The cosine law gives $$r^2=1+r^4-2r^2\cos (120\mathrm{°}-2\alpha ).$$ But $r>1$ and so $$(r^2-1{)}^2>0\text{ or }{r}^4+1>2r^2.$$ Substituting the second inequality into the equation gives $r^2>2r^2-2r^2\cos (120\mathrm{°}-2\alpha )$ which implies $\cos (120\mathrm{°}-2\alpha )>{\displaystyle \frac{1}{2}}$. Thus, $\alpha >30\mathrm{°}$ and

   $$r^2={\displaystyle \frac{1}{2}}+{\displaystyle \frac{\sqrt{3}}{2}}\cot \alpha <{\displaystyle \frac{1}{2}}+{\displaystyle \frac{\sqrt{3}}{2}}\cdot {\displaystyle \frac{\sqrt{3}}{1}}=2$$ Thus, $r^2<2$ and so $r<\sqrt{2}.$

Problem Set

1. 1. If $2\sin (2\theta )+1=0$, find the smallest positive value of $\theta$ (in degrees).

   2. For $-\pi \le \theta \le \pi$, find all solutions to the equation $2({\sin }^2\theta -{\cos }^2\theta )=8\sin \theta -5$.

2. In $\mathrm{△}ABC$, $M$ is a point on $BC$ such that $BM=5$ and $MC=6$. If $AM=3$ and $AB=7$, determine the exact value of $AC$.

   

3. In determining the height, $MN$, of a tower on an island, two points $A$ and $B$, $100$ m apart, are chosen on the same horizontal plane as $N$ (the base of the tower). If $\mathrm{\angle }NAB=108\mathrm{°}$, $\mathrm{\angle }ABN=47\mathrm{°}$ and $\mathrm{\angle }MBN=32\mathrm{°}$, determine the height of the tower to the nearest metre.

   

4. A rectangle $PQRS$ has side $PQ$ on the $x$-axis and touches the graph of $y=k\cos x$ at the points $S$ and $R$ as shown.

   

   If the length of $PQ$ is ${\displaystyle \frac{\pi }{3}}$ and the area of the rectangle is ${\displaystyle \frac{5\pi }{3}}$, what is the value of $k$?

5. The graph of the equation $y=a\sin kx$ is shown in the diagram, and the point $({\displaystyle \frac{3\pi }{4}}$,$-2)$ is the minimum point indicated. The line $y=1$ intersects the graph at point $D$. What are the coordinates of $D$?

   

6. A square with an area of $9{\text{ cm}}^2$ is surrounded by four congruent triangles, forming a larger square with an area of $89{\text{ cm}}^2$. If each of the triangles has an angle $\theta$ as shown, find the value of $\tan \theta$.

   

7. A rectangular box has a square base of length $1$ cm, and height $\sqrt{3}$ cm as shown in the diagram.

   

   What is the cosine of angle $FAC$?

8. In the grid, each small equilateral triangle has side length $1$.

   

   If the vertices of $\mathrm{△}WAT$ are themselves vertices of small equilateral triangles, what is the area of $\mathrm{△}WAT$?

9. In $\mathrm{△}ABC$, $AB=8$, and $\mathrm{\angle }CAB=60\mathrm{°}$. Sides $BC$ and $AC$ have integer lengths $a$ and $b$, respectively. Find all possible values of $a$ and $b$.