Baby Trigonometry

Now let’s discuss trigonometry, the bane of high school mathematics. In spite of its rather tragic reputation, the life goal of trigonometry is simple:

What is the relationship between line segments and angles?

An angle, at its heart, deals with circular motion. Angles take considerable effort to construct, but their main purpose is to quantify the “degree” (pun intended) of separation between two line segments.

Trigonometry aims to capture the precise effect angles have on line segments. The name trigono-metry itself suggests our starting point—the triangle. Furthermore, we should start with the simplest triangle—the right-angled triangle—then work our way up to more general triangles.

Definition 1. Consider the following right-angled triangle with acute angle $0^\circ < \theta < 90^\circ$ .

We abbreviate the words opposite, adjacent, and hypotenuse. We define the sine, cosine, and tangent of $\theta$ as follows:

$\begin{aligned} \sin(\theta) := \frac{\text{opp}}{\text{hyp}}, \quad \cos(\theta) := \frac{\text{adj}}{\text{hyp}}, \quad \tan(\theta) := \frac{\text{opp}}{\text{adj}}. \end{aligned}$

These definitions make sense thanks to similar triangles (i.e. right-angled triangles with the same shape will still give the same trigonometric ratios).

Example 1. Using a suitably-drawn triangle, evaluate the trigonometric ratios

$\sin(45^\circ), \quad \cos(45^\circ),\quad \tan(45^\circ).$

Solution. We draw a right-angled isosceles triangle with side length $1$ , whose base angles must be $(180^\circ - 90^\circ) /2 = 45^\circ$ :

Using Pythagoras’ theorem, the triangle has hypotenuse

$\sqrt{1^2 + 1^2} = \sqrt 2.$

By Definition 1,

$\displaystyle \sin(45^\circ) = \frac 1{\sqrt 2},\quad \cos(45^\circ) = \frac 1{\sqrt 2},\quad \tan(45^\circ) = \frac 1{1} = 1.$

Example 2. Using a suitably-drawn triangle, evaluate the trigonometric ratios

$\sin(60^\circ), \quad \cos(60^\circ),\quad \tan(60^\circ).$

Solution. We draw an equilateral triangle divided by its altitude:

We leave it as an exercise to check that the two right-angled triangles are congruent. Hence, relative to $60^\circ$ , the triangle has an adjacent side $2/2 = 1$ . Using Pythagoras’ theorem, the triangle has opposite side

$\sqrt{2^2 - 1^2} = \sqrt 3.$

By Definition 1,

$\displaystyle \sin(60^\circ) = \frac{\sqrt 3}2, \quad \cos(60^\circ) = \frac 12,\quad \tan(60^\circ) = \frac{\sqrt 3}{1} = \sqrt 3.$

Example 3. Use the same diagram in Example 2 to evaluate the trigonometric ratios

$\sin(30^\circ), \quad \cos(30^\circ),\quad \tan(30^\circ).$

Solution. Since angles in a triangle sum to $180^\circ$ , we can draw the $30^\circ$ angle as follows:

Relative to $30^\circ$ , the triangle has opposite side $1$ , adjacent side $\sqrt 3$ , and hypotenuse $2$ . By Definition 1,

$\displaystyle \sin(30^\circ) = \frac{1}2, \quad \cos(30^\circ) = \frac{\sqrt 3}2,\quad \tan(30^\circ) = \frac{1}{\sqrt 3}.$

Now, if we scrutinise these results more carefully, we will notice that

$\cos(60^\circ) = \sin(30^\circ),\quad \cos(30^\circ) = \sin(60^\circ), \quad \tan(60^\circ) \tan(30^\circ) = 1.$

These seeming coincidences are, in fact, not coincidences. We have several results that will always hold for any acute $\theta$ :

Theorem 1. For any acute angle $\theta$ ,

$\cos(\theta) = \sin(90^\circ - \theta),\quad \tan(\theta) \tan(90^\circ - \theta) = 1.$

Proof. Consider the triangle below:

Using Definition 1,

$\displaystyle \cos(\theta) = \frac{a}{c} = \sin(\alpha)$

and

$\displaystyle \tan(\theta) \tan(\alpha) = \frac{b}{a} \cdot \frac ab = 1.$

Since angles in a triangle sum to $180^\circ$ ,

$\alpha + \theta + 90^\circ = 180^\circ \quad \Rightarrow \quad \alpha = 90^\circ - \theta.$

Substituting $\alpha = 90^\circ - \theta$ ,

$\cos(\theta) = \sin(90^\circ - \theta),\quad \tan(\theta) \tan(90^\circ - \theta) = 1.$

We call the acute angles $\alpha, \beta$ complementary if $\alpha + \beta = 90^\circ$ . Hence, $90^\circ - \theta$ is complementary to $\theta$ . In fact, the “co-” in cosine really stands for “complement”, so that $\cos(\theta)$ is the sine of the complement of $\theta$ , i.e.

$\cos(\theta) = \sin(90^\circ - \theta).$

Remark 1. To reduce the clunkiness of the expression, we make the (tragically ambiguous) notations

$\sin^2(\theta):=(\sin(\theta))^2 , \quad \cos^2(\theta):=(\cos(\theta))^2.$

Theorem 2. For any acute angle $\theta$ ,

$\sin^2(\theta) + \cos^2(\theta) = 1.$

Proof. Consider the right-angled triangle below.

Using Definition 1 and Pythagoras’ theorem,

$\displaystyle \sin^2(\theta) + \cos^2(\theta) = \frac{b^2}{c^2} + \frac{a^2}{c^2} = \frac{b^2 + a^2}{c^2} = \frac{c^2}{c^2} = 1.$

This result is simply the Pythagoras’ theorem with the special case hypotenuse equals $1$ .

Strangely enough, together with geometry, we have everything we need to explore high school trigonometry. An alternate formulation of trigonometry, called rational trigonometry, captures more or less the same ideas but simplifies calculations.

Question 1. Given the following triangle with known side lengths $a , b$ and an acute angle $C$ between them, how do we calculate its area?

Solution. Let $h$ denote the height of the triangle.

Then the triangle has area $A = \frac 12 a h$ . By Definition 1,

$\displaystyle \sin(C) = \frac{h}{b} \quad \Rightarrow \quad h = b \sin(C)$ .

Therefore, $A = \frac 12 ab \sin (C)$ .

Remark 2. The proof still holds if $\angle ABC$ is right or obtuse.

Question 2. If instead the angle $C$ is obtuse, how would we calculate the area of the triangle?

Solution. Extend the base of the triangle by $a$ units as follows:

By Question 1, $\Delta AB'C$ has area $\frac 12 ab \sin(\alpha)$ . Since angles on a straight line sum to $180^\circ$ ,

$C + \alpha = 180^\circ \quad \Rightarrow \quad \alpha = 180^\circ - C.$

Since both triangles have the same base and height, they must have the same area $A$ , so that

$A = \frac 12 ab \sin(\alpha) = \frac 12 ab \sin(180^\circ - C).$

If we insist that the formula

$\text{(area of triangle)} = \frac 12 ab \sin(C)$

holds for obtuse $C$ , then we must have

$\sin(C) = \sin(180^\circ - C),$

just as we have explored.

Question 3. Using the same reasoning, how should we define $\sin(90^\circ)$ ?

Solution. Consider the right-angled triangle below.

Assuming the desired area formula holds, we have

$\frac 12 ab \sin(90^\circ) = \frac 12 ab \quad \Rightarrow \quad \sin(90^\circ) = 1.$

This result motivates our definition of $\sin(\theta)$ for $0^\circ < \theta < 180^\circ$ .

Definition 2. Define $\sin(90^\circ) := 1$ . For any acute $\theta$ , define

$\sin(180^\circ - \theta) := \sin(\theta) > 0.$

Using this result, we obtain the beautiful law of sines.

Theorem 3 (Law of Sines). Given the triangle below,

$\displaystyle \frac{\sin(A)}{a} = \frac{\sin(B)}{b} = \frac{\sin(C)}{c}.$

This result still holds even if some angle is right or obtuse.

Proof. Since the area formula holds in any type of angle (acute, right, obtuse),

$\frac 12 bc \sin(A) = \frac 12 ca \sin(B) = \frac 12 ab \sin(C).$

Dividing by $\frac 12 abc$ on all sides,

$\displaystyle \frac{\sin(A)}{a} = \frac{\sin(B)}{b} = \frac{\sin(C)}{c}.$

If there is a law of sines, would there be a law of cosines? Yes, and in fact we obtain it via Pythagoras’ theorem.

Lemma 1. Given the triangle below with all acute angles,

$c^2 = a^2 + b^2 - 2ab \cos(C).$

Proof. Sub-divide $a = s+t$ and draw the height of the triangle.

By Pythagoras’ theorem,

$\begin{aligned} c^2 &= t^2 + h^2 \\ &= (a-s)^2 + (b^2 - s^2) \\ &= a^2 - 2as + s^2 + b^2 - s^2 \\ &= a^2 + b^2 - 2as. \end{aligned}$

Using Definition 1,

$\displaystyle \cos(C) = \frac{s}{b} \quad \Rightarrow \quad s = b \cos(C).$

Therefore, $c^2 = a^2 + b^2 - 2ab \cos(C)$ .

Lemma 2. Given the triangle below with obtuse angle $C$ ,

$c^2 = a^2 + b^2 + 2ab \cos(180^\circ - C).$

Proof. Draw the height of the triangle and extend $BC$ as follows:

By Pythagoras’ theorem again,

$\begin{aligned} c^2 &= (a+s)^2 + h^2 \\ &= (a^2 + 2as + s^2) + (b^2-s^2) \\ &= a^2 + b^2 + 2as. \end{aligned}$

Using Definition 1,

$\displaystyle \cos(\alpha) = \frac sb \quad \Rightarrow \quad s = b \cos(\alpha).$

Therefore, $c^2 = a^2 + b^2 + 2ab \cos(\alpha)$ .

Since angles on a straight line sum to $180^\circ$ ,

$C + \alpha = 180^\circ \quad \Rightarrow \quad \alpha = 180^\circ - C.$

Hence,

$c^2 = a^2 + b^2 + 2ab \cos(180^\circ - C).$

Likewise, if we insist that the formula

$c^2 = a^2 + b^2 - 2ab \cos(C)$

holds for obtuse $C$ , then we must have

$\cos(C) = -{\cos(180^\circ - C)}.$

Question 4. Using the same reasoning, how should we define $\cos(90^\circ)$ ?

Solution. We draw a right-angled triangle:

By Pythagoras’ theorem,

$c^2 = a^2 + b^2.$

Therefore,

$c^2 = a^2 + b^2 - 2ab \cos(90^\circ) \quad \iff \quad \cos(90^\circ) = 0.$

Definition 3. Define $\cos(90^\circ) := 0$ . For any acute $\theta$ , define

$\cos(180^\circ - \theta) := -{\cos(\theta)} < 0.$

Using this result, we obtain the corresponding law of cosines.

Theorem 4 (Law of Cosines). Given the triangle below,

$\displaystyle c^2 = a^2 + b^2 - 2ab \cos(C).$

This result still holds even if $C$ is right or obtuse.

Proof. Apply Lemma 1, Lemma 2, and Question 4.

You would have noticed that we have neglected the tangent function. That is not too surprising, thanks to the following observation.

Theorem 5. For any acute $\theta$ , $\displaystyle \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$ .

Proof. We return to the very first triangle in this blog post.

By Definition 1,

$\displaystyle \frac{\sin(\theta)}{\cos(\theta)} = \frac{ \text{opp}/\text{hyp} }{ \text{adj}/\text{hyp} } = \frac{ \text{opp} }{ \text{adj} } = \tan(\theta).$

Since, $\sin(90^\circ) = 1$ and $\cos(90^\circ) = 0$ , superimposing the identity in Theorem 5 will lead to a mathematical error. Nevertheless, $\cos(\theta)$ is well-defined for obtuse $\theta$ . Hence, we define $\tan(\theta)$ according to Theorem 5.

Definition 4. For obtuse $\theta$ , define

$\displaystyle \tan(\theta) := \frac{\sin(\theta)}{\cos(\theta)}.$

In particular, if $\theta$ is acute, then Definitions 2–4 yield.

$\displaystyle \tan(180^\circ - \theta) = \frac{\sin(180^\circ - \theta)}{\cos(180^\circ - \theta)} = \frac{\sin(\theta)}{-{\cos(\theta)}} = -{\tan(\theta)}.$

Corollary 1 (Obtuse Angle Identities). For any $0^\circ < \theta < 180^\circ$ ,

$\begin{aligned} \sin(180^\circ - \theta) &= \sin(\theta), \\ \cos(180^\circ - \theta) &= -{\cos(\theta)}, \\ \tan(180^\circ - \theta) &= -{\tan(\theta)}, \end{aligned}$

whenever the right-hand side is well-defined.

You may explore more trigonometric identities in this exercise post, in which we used the extended definitions of $\sin(\theta)$ and $\cos(\theta)$ for $0^\circ \leq \theta \leq 360^\circ$ , made possible through a “doubling” trick. Here, the only geometric pre-requisite is a special case of the double angle formula.

Nevertheless, in the spirit of geometric reasoning, we will accomplish the same goal by revisiting an old friend—the unit circle.

—Joel Kindiak, 19 Dec 25, 1334H

KindiakMath

Baby Trigonometry

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Baby Trigonometry

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