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Section 4 Trigonometry

Throughout this discussion we will use the following right triangle as a notational reference.
Figure 4.1. Right triangle
The side labeled \(a\) is the side opposite the angle \(\theta\) since it is on the other side of the triangle. For this reason it is sometimes labeled opp. The side labeled \(b\) is the side adjacent to the angle \(\theta\) since it is on the same side of the triangle. For this reason it is sometimes labeled adj. The side labeled \(c\) is the hypotenuse and is sometimes labeled hyp. The sine of \(\theta\) is denoted as \(\sin(\theta)\) and it is the length of the opposite side \(a\) divided by the hypotenuse \(c,\) that is,
\begin{equation*} \sin(\theta)=\frac{a}{c}=\frac{opp}{hyp}. \end{equation*}
The cosine of \(\theta\) is denoted as \(\cos(\theta)\) and it is the length of the adjacent side \(b\) divided by the hypotenuse \(c,\) that is,
\begin{equation*} \cos(\theta)=\frac{b}{c}=\frac{adj}{hyp}. \end{equation*}
In some cases we may apply the theorem of Pythagoras to compute values of these trigonometric functions.
Figure 4.2. 45-45-90 right triangle
Here we have that \(\sin(45^{\circ})=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\) and \(\cos(45^{\circ})=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}.\)
Figure 4.3. 30-60-90 right triangle
In this case we obtain that \(\sin(30^{\circ})=\frac{a}{c}=\frac{1}{2}\) and \(\cos(30^{\circ})=\frac{b}{c}=\frac{\sqrt{3}}{2},\) similarly we have that \(\sin(60^{\circ})=\frac{a}{c}=\frac{\sqrt{3}}{2}\) and \(\cos(60^{\circ})=\frac{b}{c}=\frac{1}{2}.\) From the definition we also get a nice and important identity, namely
\begin{equation*} \sin(\theta)^2+\cos(\theta)^2=1. \end{equation*}
It easily follows from the theorem of Pythagoras,
\begin{equation*} \sin(\theta)^2+\cos(\theta)^2=\left(\frac{a}{c}\right)^2+\left(\frac{b}{c}\right)^2=\left(\frac{a^2+b^2}{c^2}\right)=1. \end{equation*}
In what follows we will measure angles in degrees and radians. Radians are simply another unit for measuring the size of an angle. It turns out that there is an easy way to convert from degrees to radians and back:
\begin{equation*} x^{\circ}=x\left(\frac{\pi}{180}\right)\mbox{ radians and } x\mbox{ radians}=x\left(\frac{180}{\pi}\right)\mbox{ degrees.} \end{equation*}
Figure 4.4. Degree-Radian examples
Now consider the right triangle inside the unit circle.
Figure 4.5. Unit circle I.
Figure 4.6. Unit circle II.
So every point on the unit circle has the coordinates
\begin{equation*} (\cos(\theta), \sin(\theta)). \end{equation*}
We may extend our earlier computations based on the following figure:
Figure 4.7. Known values of Sine and Cosine
With some further ideas we may provide exact values of the trigonometric functions. Let us see an example:
Figure 4.8. The golden triangle subdivided
Comparing ratios of side lengths gives
\begin{equation*} \frac{x}{1-x}=\frac{1}{x}, \end{equation*}
from which we conclude
\begin{equation*} x=\frac{\sqrt{5}-1}{2}. \end{equation*}
We obtain from the original figure that
\begin{equation*} \sin(18^{\circ})=\frac{\sqrt{5}-1}{4}. \end{equation*}
Using a cute geometric trick we may show that
\begin{equation*} \sin(\alpha-\beta)=\sin(\alpha)\cos(\beta)-\cos(\alpha)\sin(\beta)\mbox{ and }\cos(\alpha-\beta)=\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta). \end{equation*}
The idea is based on the figure
Figure 4.9. Verifying the angle-subtraction formulas
Using the fact that \(\sin(-\theta) = - \sin(\theta)\) and \(\cos(-\theta) = \cos(\theta),\) both of which can be seen from the unit circle constructions above. We obtain that
\begin{equation*} \sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)\mbox{ and }\cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta). \end{equation*}
We may apply these formulas to compute \(\sin(2\alpha),\cos(2\alpha):\)
\begin{equation*} \sin(2\alpha)=\sin(\alpha+\alpha)=2\sin(\alpha)\cos(\alpha)\mbox{ and }\cos(2\alpha)=\cos(\alpha+\alpha)=\cos(\alpha)^2-\sin(\alpha)^2. \end{equation*}

Subsection 4.1 Exercises

Convert the following angles given in degrees, to radians. Do not use a calculator and give your answers as multiples of \(\pi.\)
\begin{equation*} a) 15^{\circ}\quad b) 75^{\circ}\quad c) 120^{\circ}\quad b) 150^{\circ}. \end{equation*}
Convert each of the following angles given in radians, to degrees. Do not use a calculator.
\begin{equation*} a) \frac{\pi}{6}\quad b) \frac{3\pi}{4}\quad c) \frac{2\pi}{5}\quad a) \frac{7\pi}{10}. \end{equation*}
In the Figure, \(CB\) is a diameter of a circle with a radius of 2 cm and center \(O, ABC\) is a right triangle, and \(CD\) has length \(\sqrt{3}\) cm. Find \(\sin A.\) Find the length of \(AC.\) Find the length of \(AD.\)
Figure 4.13. Triangles
Represent the following sine and cosine values in exact forms (in terms of square-root radicals, and the four common operations of arithmetic).
\begin{equation*} \cos(18^{\circ}),\quad \sin(15^{\circ}),\quad \cos(15^{\circ}). \end{equation*}
Provide formulas for \(\sin(3\alpha)\) and \(\cos(3\alpha)\) in terms of \(\sin(\alpha)\) and \(\cos(\alpha).\)
Use the formulas of the previous exercise to give a polynomial with integral coefficients having \(\sin(10^{\circ})\) as a root.

Subsection 4.2 Sine and cosine rules

The sine and cosine rules are particularly useful when dealing with triangles that are not right angled. The area of a triangle is 1/2(base)(perpendicular height).
Figure 4.17. Triangle
In the above triangle we may compute the area as
\begin{equation*} \frac{1}{2}bc\sin A =\frac{1}{2}ac\sin B = \frac{1}{2}ab \sin C. \end{equation*}
If each expression is then divided by \(1/2abc\) then what results is called the sine rule:
\begin{equation*} \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}. \end{equation*}
Consider the triangle below with perpendicular height \(h.\)
Figure 4.18. Triangle
Theorem of Pythagoras implies that
\begin{equation*} a^2-x^2=h^2=b^2-(c-x)^2\longrightarrow b^2=a^2+c^2-2cx=a^2+c^2-2ac\cos B. \end{equation*}
In a similar way we obtain formulas called the cosine rule:
\begin{align*} a^2= & b^2+c^2-2bc\cos A,\\ b^2= & a^2+c^2-2ac\cos B,\\ c^2= & a^2+b^2-2ab\cos C. \end{align*}

Subsection 4.3 Exercises

Find the value of the unknown length \(x,\) correct to 2 decimal places.
Figure 4.20. Triangle
Find the two possible values of angle \(A\) for triangle \(ABC,\) given \(a = 8, c = 6\) and \(C = 43^{\circ}.\)
Find all the side lengths, correct to 2 decimal places, and angle values for the triangle \(ABC,\) given \(a = 10.5, B = 60^{\circ}\) and \(C = 72^{\circ}.\)

Find the value of the unknown length \(x,\) correct to 2 decimal places.

Figure 4.24. Triangle

For triangle \(ABC,\) find the magnitude of angle \(A,\) expressed in decimal form, given \(a = 5, b = 7 \mbox{ and } c = 4.\)

Subsection 4.4 Applications I.

We consider the internal case only, the proof of the other one is similar.
Figure 4.28. Internal case
We apply the sine rule to the triangles \(ABP, ACP\) and we use the fact that \(\sin(\theta)=\sin(180^{\circ}-\theta):\)
\begin{equation*} \frac{\sin(\theta)}{AB}=\frac{\sin(\alpha)}{BP}\quad\mbox{ and }\quad \frac{\sin(180^{\circ}-\theta)}{AC}=\frac{\sin(\alpha)}{PC}. \end{equation*}
Thus we get that
\begin{equation*} \sin(\theta)=\frac{AB\sin(\alpha)}{BP}=\frac{AC\sin(\alpha)}{PC}. \end{equation*}
It follows that
\begin{equation*} \frac{AB}{BP}=\frac{AC}{PC}. \end{equation*}
Figure 4.30. Apollonius' circle
We draw the internal and external bisectors of angle \(APB.\) Notice that \(2\alpha+2\beta=180^{\circ}\) (angles on a line). Hence for all positions of \(P\) we have \(\alpha+\beta=90^{\circ}.\) The angle bisector theorem implies that
\begin{equation*} \frac{AP_1}{BP_1}=\frac{AP}{BP}=k, \end{equation*}
that is \(P_1\) is a fixed point. Similarly,
\begin{equation*} \frac{AP_2}{BP_2}=\frac{AP}{BP}=k, \end{equation*}
that is \(P_2\) is a fixed point. Therefore \(P_1,P_2\) are fixed points, \(P_1PP_2\) is a right angle, consequently, \(P\) traces out a circle, with centre the midpoint of \(P_1P_2.\)
Figure 4.32. Apollonius' median
We apply the theorem of Pythagoras a couple of times:
\begin{align*} AB^2+AC^2 & = (BE^2+AE^2) + (CE^2+AE^2) =\\ & = (BD-ED)^2 + (CD + ED)^2 + 2AE^2=\\ & =(BD-ED)^2 + (BD + ED)^2 + 2AE^2 \mbox{ since } CD=BD\\ & =2BD^2 + 2(ED^2 + AE^2)=2BD^2 + 2AD^2. \end{align*}

Subsection 4.5 Exercises

The sides of a triangle are 8, 12, and 15. An angle bisector meets the side of length 15. Find the lengths \(x\) and \(y.\)
Figure 4.34. Exercise
Given \(π‘ͺ𝑫 = πŸ‘, 𝑫𝑩 = πŸ’, 𝑩𝑭 = πŸ’, 𝑭𝑬 = πŸ“, 𝑨𝑩 = πŸ”,\) and \(\angle π‘ͺ𝑨𝑫 = \angle 𝑫𝑨𝑩 = \angle 𝑩𝑨𝑭 = \angle 𝑭𝑨𝑬,\) find the perimeter of quadrilateral \(𝑨𝑬𝑩π‘ͺ.\)
Figure 4.36. Exercise
In a quadrilateral \(ABCD,\) the bisectors of \(\angle B\) and \(\angle D\) intersect on \(AC\) at \(E.\) Prove that
\begin{equation*} \frac{AB}{BC} = \frac{AD}{DC}. \end{equation*}
Figure 4.38. Exercise
Find the length of the shortest median of a triangle with sides 8 cm, 6 cm and 5 cm.
Find the length of the longest median of a triangle with sides 7 cm, 9 cm and 10 cm.
Two sides of a triangle have lengths 12 cm and 9 cm. The median to the third side has length 7 cm. Find the length of the third side.

Subsection 4.6 Applications II.

Figure 4.43. Stewart's theorem
Let us apply the cosine rule two times:
\begin{align*} AB^2 & = AD^2 + BD^2 - 2AD\cdot BD\cos(\theta),\\ AC^2 & = AD^2 + DC^2 - 2AD\cdot DC\cos(180^{\circ}-\theta) = AD^2 + DC^2 + 2AD\cdot DC\cos(\theta). \end{align*}
Based on these equations let us compute \(nAB^2+mAC^2:\)
\begin{equation*} nAB^2+mAC^2= (m + n)AD^2 + mDC^2 + nBD^2 + 2AD\cos(\theta)(mDC-nBD). \end{equation*}
We also have that \(\frac{BD}{DC}=\frac{m}{n}.\) Hence we obtain
\begin{equation*} nAB^2+mAC^2= (m + n)AD^2 + mDC^2 + nBD^2. \end{equation*}
Figure 4.46. Ptolemy's theorem - proof
We construct similar triangles. We draw \(AH,\) where \(H\) lies on \(BD\) such that \(\theta_1=\theta_2.\) In the triangles \(ABH, ACD\) we have \(\theta_1=\theta_2\) and \(\alpha_1=\alpha_2\) (angles in same segment). Therefore we get that \(\frac{AB}{AC}=\frac{BH}{CD}.\) We also have that \(\alpha+\theta=\alpha_1+\theta_2=\alpha_1+\theta_3\) and \(\beta_1=\beta_2.\) Hence \(ADH\) is similar to \(ABC,\) thus we obtain \(\frac{HD}{BC}=\frac{AD}{AC}.\) Combining these equations yields \(BD=BH+HD=\frac{AB\cdot CD+BC\cdot DA}{AC}\) and the statement follows.

Subsection 4.7 Exercises

\(Q\) lies on \(AC\) such that \(AQ : QC = 2 : 1.\) If \(AC = 6 cm, BC = 7 cm\) and \(AB = 8 cm,\) find the length of \(BQ.\)
We have that \(PS : SR = 5 : 3, QS = QP = 9 cm\) and \(PR = 8 cm.\) Find the length of \(QR.\)
Figure 4.49. Exercise
Find \(AC\) given that \(BD\) has length 12 cm.
Figure 4.51. Exercise
The sides of a cyclic quadrilateral given in clockwise order are 6 cm, 7 cm, 9 cm and 10 cm. If one diagonal is 8 cm, find the length of the other diagonal.
Three consecutive sides of a cyclic quadrilateral have lengths 6 cm, 7 cm and 11 cm. Its diagonals have lengths 8 cm and 12 cm. Find the length of the fourth side of the cyclic quadrilateral.

Subsection 4.8 Concurrency and Collinearity

Definition 4.54.
Three or more lines are called concurrent if they meet at a single point. Three or more points are said to be collinear if they lie on a single straight line.
Let \(W\) be the point on \(AB\) such that \(CW\parallel XY.\) Then,
\begin{equation*} \frac{BX}{XC}=\frac{BZ}{ZW}\mbox{ and }\frac{CY}{YA}=\frac{WZ}{ZA}. \end{equation*}
It follows that
\begin{equation*} \frac{BX}{XC}\cdot \frac{CY}{YA}\cdot \frac{AZ}{ZB}=\frac{BZ}{ZW}\cdot \frac{WZ}{ZA}\cdot \frac{AZ}{ZB}=\frac{BZ}{ZB}\cdot \frac{WZ}{ZW}\cdot \frac{AZ}{ZA}=(-1)(-1)(-1)=-1. \end{equation*}
Let us now prove the opposite direction. Suppose the line joining \(X\) and \(Z\) intersects \(AC\) at \(Y'.\) From above,
\begin{equation*} \frac{BX}{XC}\cdot \frac{CY'}{Y'A}\cdot \frac{AZ}{ZB}=-1=\frac{BX}{XC}\cdot \frac{CY}{YA}\cdot \frac{AZ}{ZB}. \end{equation*}
It follows that
\begin{equation*} \frac{CY'}{Y'A}=\frac{CY}{YA}. \end{equation*}
The points \(Y'\) and \(Y\) divide the segment \(CA\) in the same ratio. These must be the same point, and \(X, Y , Z\) are collinear.
Suppose the lines \(AX, BY , CZ\) intersect at a point \(P.\) Consider the line \(BPY\) cutting the sides of \(\triangle CAX.\) By Menelaus' theorem,
\begin{equation*} \frac{CY}{YA}\cdot \frac{AP}{PX}\cdot \frac{XB}{BC}=-1\mbox{ or }\frac{CY}{YA}\cdot \frac{PA}{XP}\cdot \frac{BX}{BC}=+1 \end{equation*}
Also, consider the line \(CPZ\) cutting the sides of \(\triangle ABX.\) By Menelaus’ theorem again,
\begin{equation*} \frac{AZ}{ZB}\cdot \frac{BC}{CX}\cdot \frac{XP}{PA}=-1\mbox{ or }\frac{AZ}{ZB}\cdot \frac{BC}{XC}\cdot \frac{XP}{PA}=+1 \end{equation*}
Multiplying the two equations together, we have
\begin{equation*} \frac{BX}{XC}\cdot \frac{CY}{YA}\cdot \frac{AZ}{ZB}=1. \end{equation*}
In the Figure, \(ABC\) is a triangle with \(LB = 90^{\circ}, BC = 3cm\) and \(AB = 4cm.\)\(D\) is a point on \(AC\) such that \(AD= 1cm,\) and \(E\) is the mid-point of \(AB.\) Join \(D\) and \(E,\) and extend \(DE\) to meet \(CB\) extended at \(F.\) Find \(BF.\)
Figure 4.60. Figure
Solution
Figure 4.61. .
Apply Ceva’s theorem to prove that the medians of a triangle are concurrent. Solution
Let the medians of \(\triangle ABC\) be \(AP, BQ\) and \(CR\) respectively. Now
\begin{equation*} \frac{AR}{RB}\cdot \frac{BP}{PC}\cdot \frac{CQ}{QA}=1\times 1\times 1=1, \end{equation*}
hence \(AP, BQ\) and \(CR\) are concurrent.