Trigonometry review

Section 8.3 Trigonometry review

There are certain trigonometric facts that should be in your mathematical toolkit. We review several here.

The law of cosines (Theorem 8.3.2)
The graphs of \(\sin(x)\) and \(\cos(x)\)
The parallelogram law (Theorem 8.3.13)
The addition laws for \(\sin(x)\) and \(\cos(x)\) (Theorem 8.3.18)
The law of sines (Theorem 8.3.20)

Subsection 8.3.1 Basic Definitions

Trigonometry was originally used for measuring triangles. Given a right triangle with hypotenuse \(c\text{,}\) legs \(a\) and \(b\) and angle \(\theta\) (as in the left illustration in Figure 8.3.1), the six trigonometric functions are defined by

\begin{equation*} \begin{array}{lll} \sin(\theta)=\frac bc \amp \cos(\theta)=\frac ac \amp \tan(\theta)=\frac {\sin(\theta)}{\cos(\theta)}=\frac ba \\ \cot(\theta)=\frac {\cos(\theta)}{\sin(\theta)}=\frac ab \amp \csc(\theta)=\frac 1{\sin(\theta)}=\frac cb \amp \sec(\theta)=\frac 1{\cos(\theta)}=\frac ca \end{array} \end{equation*}

The first three, the sine, the cosine and the tangent are by far the most important, even though the cotangent, cosecant and secant appear in many applications.

Figure 8.3.1. Trigonometric functions of angles

In any such right triangle, the angle \(\theta\) is necessarily acute. We extend the definition of the trigonometric functions to other angles by using the unit circle (centre at \(\vec 0=(0,0)\) and radius \(r=1\)). For any angle \(\theta\text{,}\) draw the radius of the unit circle that makes an angle \(\theta\) with the positive \(x\)-axis. The coordinates of the point where this radius meets the circle are \((\cos(\theta),\sin(\theta))\text{.}\) Note that when \(\theta\) is acute, the two definitions coincide.

Since the point \((\cos(\theta),\sin(\theta))\) is on the unit circle, its distance to the origin \((0,0)\) is \(1\text{.}\) This means \[ \cos^2(\theta) + \sin^2(\theta) =1 \textrm{ for any angle } \theta \]

Subsection 8.3.2 The law of cosines

In the right triangle above, the angle opposite the side labelled \(c\) is an right angle, and the Pythagorean theorem says \(c^2=a^2+b^2\text{.}\) The law of cosines extends this theorem to triangles with any angle opposite the side labelled \(c\text{.}\)

Theorem 8.3.2. The law of cosines.

If a triangle (as in the figure below) has sides of lengths \(a\text{,}\) \(b\) and \(c\text{,}\) and the angle at the vertex opposite \(c\) is \(\theta\text{,}\) then \[ c^2=a^2+b^2-2ab\,\cos\theta \]

Figure 8.3.3.

Proof.

Note that since \(a\) and \(b\) are interchangeable in the picture, there is no loss of generality to assume \(a \le b\text{.}\) We prove the law of cosines by dropping a perpendicular to the side \(b\text{:}\)

Figure 8.3.4. \(\theta\) acute

The perpendicular has length \(h\) and splits side \(b\) into two pieces of lengths \(b_1\) and \(b_2\text{.}\)

Now we have the following identities:

\(\displaystyle b=b_1+b_2\)
\(\displaystyle a^2=b_1^2 + h^2\)
\(\displaystyle c^2 = b_2^2 + h^2\)
\(\displaystyle b_1=a\cos(\theta)\)

which gives the equations

\begin{equation*} \begin{array}{rl} c^2 \amp=b_2^2+h^2\\ \amp=(b-b_1)^2+(a^2-b_1^2)\\ \amp=b^2 -2bb_1 + b_1^2 +a^2-b_1^2\\ \amp=a^2+b^2-2bb_1\\ \amp=a^2+b^2-2ab\cos(\theta) \end{array} \end{equation*}

We have cheated a little by assuming that the perpendicular divides \(b\) into two pieces. In fact if \(\theta\gt\frac\pi2\) then the picture looks like

Figure 8.3.5. \(\theta\) obtuse

Now we have

\(\displaystyle a^2=b_1^2+h^2\)
\(\displaystyle c^2=(b+b_1)^2+h^2\)
\(b_1=-a\cos(\theta)\) (remember that \(\cos(\theta)\lt0\) when \(\frac\pi 2\lt\theta\lt\pi\))

which gives

\begin{equation*} \begin{array}{rl} c^2 \amp= (b+b_1)^2+h^2\\ \amp= (b+b_1)^2+(a^2-b_1^2)\\ \amp= b^2+2bb_1+a^2\\ \amp= a^2+b^2-2ab\cos(\theta) \end{array} \end{equation*}

In principle there is another possibility:

Figure 8.3.6.

In this case we \(a \gt b\text{,}\) contrary to our assumption that \(a\leq b\text{.}\) Nonetheless we could use the argument:

\(\displaystyle a^2=(b+b_1)^2 + h^2\)
\(\displaystyle c^2 = b_1^2 + h^2\)
\(\displaystyle b+b_1=a\cos(\theta)\)

and can deduce \(c^2=a^2+b^2-2ab\cos(\theta)\) in a manner similar that used in the previous cases.

Subsection 8.3.3 The graphs of \(\sin(x)\) and \(\cos(x)\)

The graphs of the trigonometric functions are revealing. First, here is the graph of the function \(\sin(x)\text{:}\)

Figure 8.3.7.

Notice that \(\sin(x)\) is positive in this range precisely when \(0\lt x\lt\pi\text{.}\)

Here is the graph of \(\cos(x)\text{:}\)

Figure 8.3.8.

Notice that the function \(\cos(x)\) is decreasing for \(0\le x\le\pi\text{.}\) By decreasing we mean that if \(x_1 \lt x_2\text{,}\) then \(\cos(x_1) \gt \cos(x_2)\) (or, more intuitively, the function goes downhill as we go from left to right). There is an important implication from this: if \(y=\cos(x)\) for some \(x\) with \(0\le x\le\pi\text{,}\) then for any for any other \(x'\) in the same range, we have \(\cos(x') \gt y\) (when \(x'\lt x\)) or \(\cos(x') \lt y\) (when \(x'\gt x\)). In particular this means that for any given \(y_0\text{,}\) there is at most one \(x_0\) with \(0\le x_0\le\pi\) so that \(y_0=\cos(x_0)\text{.}\) In fact, the intermediate value theorem from the Calculus implies that for any \(y_0\) satisfying \(-1\leq y_0 \leq 1\) there is some \(x_0\) satisfying \(0\le x_0\le\pi\) so that \(y_0=\cos(x_0)\text{.}\)

This may be visualized as follows: pick any \(y_0\) between \(-1\) and \(1\text{.}\) From that value on the \(y\)-axis, go horizontally until you hit the graph of the function \(y=\cos(x)\) and then drop down to the \(x\)-axis at the point \(x_0\text{.}\) This value \(x_0\) is the unique one for which \(\cos(x_0)=y_0\text{.}\)

Figure 8.3.9.

Subsection 8.3.4 Parallelograms

A parallelogram is a quadrilateral with opposite sides parallel. Here is an example of one filled in yellow:

Figure 8.3.10.

Here are some important properties of parallelograms:

The angles on consecutive vertices are supplementary (sum to \(\pi\)).
Opposite angles are equal.
Opposite sides have the same length.

Here is why these properties hold:

Let one vertex have an angle of \(\theta\text{.}\) Extend one of the sides as in the figure. The lines being parallel implies that the angle outside the parallelogram is also \(\theta\text{,}\) which in turn makes the next angle within the parallelogram equal to \(\pi-\theta\text{.}\)

Figure 8.3.11.
If one vertex has an angle \(\theta\text{,}\) then next has angle \(\pi-\theta\text{.}\) The next vertex after that has angle \(\pi-(\pi-\theta)=\theta\text{,}\) and the final vertex has angle \(\pi-\theta\text{.}\) Hence two opposite vertices have an angle \(\theta\) while the other two opposite vertices have angle \(\pi-\theta\text{.}\)
Let \(b\) and \(b'\) be opposite sides of the parallelogram. Drop perpendiculars from two vertices to the opposite line as shown in the following figure. Then the length of both of the new lines is the distance, \(d\text{,}\) between the parallel lines, and so \(d=b\sin(\theta)=b'\sin(\theta)\) from which follows \(b=b'\text{.}\)

Figure 8.3.12.

The parallelogram law relates the lengths of sides of a parallelogram to the lengths of the diagonals.

Theorem 8.3.13. The parallelogram law.

If a parallelogram has sides of length \(a\) and \(b\text{,}\) and \(c\) and \(d\) are the lengths of the two diagonals, then \[ c^2+d^2=2a^2+2b^2 \] In other words, the sum of the squares of the sides is the sum of the squares of the diagonals.

Figure 8.3.14.

Proof.

The proof can be seen from the following two figures. Applying the law of cosines to the first figure gives \[ c^2=a^2+b^2-2ab\cos(\theta)\tag1 \] while applying it to the second figure gives \[ d^2=a^2+b^2-2ab\cos(\pi-\theta)= a^2+b^2+2ab\cos(\theta)\tag2 \]

Figure 8.3.15.

and

Figure 8.3.16.

Adding (1) and (2) gives \[ c^2+d^2=2a^2+2b^2 \]

Subsection 8.3.5 The addition formulas for \(\sin(x)\) and \(\cos(x)\)

Suppose we know \(\sin(\alpha)\text{,}\) \(\cos(\alpha)\text{,}\) \(\sin(\beta)\text{,}\) and \(\cos(\beta)\text{.}\) We wish to find formulas to compute \(\sin(\alpha+\beta)\text{,}\) and \(\cos(\alpha+\beta)\text{.}\) The following figure shows us how to do the computation.

The points \(A=(\cos(\alpha),\sin(\alpha))\text{,}\) \(B=\cos(\alpha+\beta),\sin(\alpha+\beta))\) and \((0,1)\) all lie on a circle with centre at the origin and radius \(1\text{.}\) These three points are joined by a line to the origin \(\vec 0\text{.}\) Drop perpendiculars from B to the line joining \(\vec0\) and \((1,0)\) and also to the line joining and \(\vec0\) and \(A\text{.}\) The points \(F\) and \(E\) are where the perpendiculars meet the respective lines. The line joining \(B\) and \(E\) is extended until it meets the line joining \(\vec 0\) and \((1,0)\) at \(D=(d,0)\text{.}\)

Figure 8.3.17.

Now we compute the area of the yellow triangle in two different ways and equate the results.

Consider the lengths of these lines:

\(\|BF\|=\sin(\alpha+\beta)\) (from triangle \(\vec0 B F\))
\(\displaystyle \|\vec 0 D\|=d\)
\(\|\vec 0 E\|=d\cos(\alpha)=\cos(\beta)\) (from triangles \(\vec0DE\) and \(\vec0BE\text{,}\) which implies that \(d=\frac{\cos(\beta)}{\cos(\alpha)}\))
\(\|BD\|=\|BE\|+\|ED\|=\sin(\beta)+d\sin(\alpha)\) (from triangles \(\vec0BE\) and \(\vec0DE\))

On the one hand, twice the area of the yellow triangle is \[ \|BF\|\,\|\vec0 D\| = d\sin(\alpha+\beta) \] On the other hand, twice the triangle area is also \[ \|\vec0 E\|\, \|BD\|=d\cos(\alpha)\left(\sin(\beta)+d\sin(\alpha)\right) \] Equating the two values we get \[ \sin(\alpha+\beta)=\cos(\alpha)\sin(\beta)+d\cos(\alpha)\sin(\alpha) =\cos(\alpha)\sin(\beta)+\cos(\beta)\sin(\alpha) \]

Note that, from the right triangle \(\vec0 E D\text{,}\) the angle at \(D\) is \(\frac\pi2-\alpha\text{,}\) and then from the right triangle \(B F D\) the angle at \(B\) is \(\alpha\text{.}\) We use this to compute some more lengths:

\(\displaystyle \|DF\| = \|BD\|\sin(\alpha)=(\sin(\beta)+d\sin(\alpha))\sin(\alpha)\)
\(\displaystyle \|\vec0 F\|=\cos(\alpha+\beta)\)

Now we have

\begin{equation*} \begin{array}{rl} \|\vec0 F\| \amp= \|\vec0 D\|- \|DF\|\\ \cos(\alpha+\beta) \amp=d-(\sin(\beta)+d\sin(\alpha))\sin(\alpha)\\ \amp=d-\sin(\beta)\sin(\alpha)-d\, \sin^2(\alpha)\\ \amp= d(1-\sin^2(\alpha))-\sin(\beta)\sin(\alpha)\\ \amp=d\,\cos^2(\alpha)-\sin(\beta)\sin(\alpha)\\ \amp= \cos(\beta)\cos(\alpha)-\sin(\beta)\sin(\alpha) \end{array} \end{equation*}

and so we have \[ \sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)\\ \cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta) \]

Theorem 8.3.18. Addition formulas for \(\sin(x)\) and \(\cos(x)\).

\begin{equation*} \sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)\\ \cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta) \end{equation*}

Subsection 8.3.6 The law of sines

Suppose a triangle has sides of lengths \(a\text{,}\) \(b\) and \(c\text{.}\) Let \(\alpha\text{,}\) \(\beta\text{,}\) and \(\gamma\) be the angles opposite \(a\text{,}\) \(b\) and \(c\) respectively. In addition, let the vertices of the triangle be labelled \(A\text{,}\) \(B\) and \(C\) at the angles \(\alpha\text{,}\) \(\beta\text{,}\) and \(\gamma\) respectively. Our picture looks like this:

Figure 8.3.19.

It is pretty obvious that the sides with longer length will have larger angles opposite. The law of sines makes this relationship precise.

Theorem 8.3.20. The law of sines.

Suppose a triangle has sides of lengths \(a\text{,}\) \(b\) and \(c\text{,}\) and \(\alpha\text{,}\) \(\beta\text{,}\) and \(\gamma\) are the angles opposite \(a\text{,}\) \(b\) and \(c\) respectively. Then \[ \frac{\sin(\alpha)}{a} =\frac{\sin(\beta)}{b} =\frac{\sin(\gamma)}{c} \]

Proof.

With no loss of generality, we may assume that \(\alpha\) is the largest angle. It may be acute or obtuse, but \(\beta\) and \(\gamma\) are definitely acute. That implies that we may drop a perpendicular from \(A\) that will divide the angle \(\alpha\) into \(\alpha_1\) and \(\alpha_2\) the side opposite into segments with lengths \(a_1\) and \(a_2\text{.}\)

Figure 8.3.21.

Then we have

\begin{equation*} \begin{array}{rl} \sin(\alpha) \amp = \sin(\alpha_1+\alpha_2)\\ \amp =\sin(\alpha_1)\cos(\alpha_2)+\cos(\alpha_1)\sin(\alpha_2)\\ \amp =\frac{a_1}c \frac hb + \frac hc \frac {a_2}b\\ \amp =\frac{ha}{bc} \end{array} \end{equation*}

and so \[ \frac{\sin(\alpha)}a = \frac h{bc} \] Using \(\sin(\beta)=\frac hc\) and \(\sin(\gamma)=\frac hb\text{,}\) it follows that \[ \frac{\sin(\beta)}b = \frac h{bc} \textrm{ and } \frac{\sin(\gamma)}c = \frac h{bc} \] and so \[ \frac h{bc}=\frac{\sin(\alpha)}a =\frac{\sin(\beta)}b =\frac{\sin(\gamma)}c \]