Unit Circle and Trigonometric Identities

The circle has radius \(1\), so one full turn has length \(2\pi\). Travelling an extra \(2\pi\) therefore brings the point back exactly where it was, and \(k\) extra turns bring it back \(k\) times: \(M(x + 2k\pi) = M(x)\). Conversely, suppose \(M(x) = M(y)\) with \(x \leq y\). Going from \(M(x)\) to \(M(y)\) one travels a distance \(y - x\) and comes back to the starting point: that distance is a whole number of turns, so \(y - x = 2k\pi\) for some integer \(k\).

In a right-angled triangle with angle \(\theta\), opposite side \(o\), adjacent side \(a\) and hypotenuse \(h\):$$\begin{aligned}[t] \sin^2\theta + \cos^2\theta &= \left(\dfrac{o}{h}\right)^2 + \left(\dfrac{a}{h}\right)^2 && \text{(definitions of sine and cosine)} \\ &= \dfrac{o^2 + a^2}{h^2} && \text{(same denominator)} \\ &= \dfrac{h^2}{h^2} && \text{(Pythagoras' theorem: } o^2 + a^2 = h^2\text{)} \\ &= 1. \end{aligned}$$

Let \(M(\cos \theta,\sin\theta)\) be the point on the unit circle at the angle \(\theta\), and \(H\) the foot of the perpendicular from \(M\) to the \(x\)-axis. The \(x\)-coordinate of \(H\) is \(\cos\theta\), and the length \(OH\) is its absolute value: \(OH = |\cos\theta|\); in the same way, \(HM = |\sin\theta|\). By Pythagoras' theorem in the triangle \(OHM\), right-angled at \(H\): $$\begin{aligned}[t] OH^2+HM^2&=OM^2 && \text{(Pythagoras' theorem)}\\ |\cos \theta|^2+|\sin \theta|^2&=1^2 && (OM = 1)\\ \cos ^2 \theta+\sin ^2 \theta&=1 && \text{(a square does not depend on the sign)} \end{aligned}$$ When \(M\) lies on an axis, the triangle is flat: one coordinate is \(0\), the other is \(1\) or \(-1\), and the equality still holds. On the figure \(\theta\) is acute, and the two coordinates are the lengths \(OH\) and \(HM\) themselves.

• Geometric proof: Let \(H\) be the foot of the perpendicular from \(M(\cos\theta, \sin\theta)\) to the \(x\)-axis. The coordinate \(\cos\theta\) is the \(x\)-coordinate of \(H\), negative when \(M\) lies to the left of the \(y\)-axis, whereas the length \(OH\) is its absolute value: \(OH = |\cos\theta|\). In the right-angled triangle \(OHM\), the side \(OH\) is at most the hypotenuse \(OM = 1\) (when \(M\) lies on an axis, the triangle is flat and the inequality is immediate), so \(|\cos\theta| \leq 1\), that is \(-1 \leq \cos \theta \leq 1\). In the same way, the length \(HM = |\sin\theta|\) is at most \(OM = 1\), so \(-1 \leq \sin \theta \leq 1\). On the figure \(\theta\) is acute, and the two coordinates are the lengths \(OH\) and \(HM\) themselves.
• Analytical proof (for cosine; the proof for sine is analogous):$$\begin{aligned} 0 &\leq \sin ^2 \theta &&\text{(a square is always non-negative)} \\ \cos ^2 \theta &\leq \cos ^2 \theta +\sin ^2 \theta &&\text{(add } \cos ^2 \theta\text{ to both sides)} \\ \cos ^2 \theta &\leq 1 &&(\cos ^2 \theta+\sin ^2 \theta=1) \\ |\cos \theta| &\leq 1 &&\text{(taking square roots)} \\ -1 &\leq \cos \theta \leq 1 && \end{aligned}$$ The same reasoning applied to \(\sin \theta\) gives \(-1 \leq \sin \theta \leq 1\).

Let \(M(\cos \theta,\sin\theta)\) be the point on the unit circle at the angle \(\theta\).
Let \(M'(\cos(\theta+2\pi),\sin(\theta+2\pi))\) be the point on the unit circle at the angle \(\theta+2\pi\). By the proposition on the reals having the same image, \(M(\theta+2\pi)=M(\theta)\): the two points coincide on the circle, \(M'=M\).
Thus, \(\cos (\theta+2 \pi)=\cos \theta\) and \(\sin (\theta+2 \pi)=\sin \theta\).
The same argument holds for any multiple of \(2\pi\), i.e. for any integer \(k\).

Let \(\theta\) be an angle.
Let \(M(\cos \theta,\sin\theta)\) be the point on the unit circle at the angle \(\theta\).
Let \(M'(\cos(\pi+\theta),\sin(\pi+\theta))\) be the point on the unit circle at the angle \(\pi+\theta\).
A rotation by an angle \(\pi\) is a half-turn about the origin \(O\), which is the same as a point reflection through \(O\). Therefore, the coordinates of \(M'\) are the opposites of the coordinates of \(M\).
So \(\begin{aligned}[t] \sin (\pi+\theta)&=-\sin \theta \\ \cos (\pi+\theta)&=-\cos \theta \end{aligned}\)

Let \(M(\cos \theta,\sin\theta)\) be the point on the unit circle at the angle \(\theta\).
Let \(M'(\cos(-\theta),\sin(-\theta))\) be the point on the unit circle at the angle \(-\theta\).
The point \(M'\) is the reflection of \(M\) in the \(x\)-axis. Therefore, the \(x\)-coordinate of \(M'\) is the same as that of \(M\), and the \(y\)-coordinate of \(M'\) is the opposite of the \(y\)-coordinate of \(M\).
So \(\begin{aligned}[t] \sin (-\theta)&=-\sin \theta \\ \cos (-\theta)&=\cos \theta \end{aligned}\)

Write \(\pi - \theta = \pi + (-\theta)\) and apply the two preceding propositions, first the one for the opposite, then the one for adding \(\pi\):$$\begin{aligned}[t] \cos(\pi - \theta) &= \cos\big(\pi + (-\theta)\big) \\ &= -\cos(-\theta) && \scriptscriptstyle\text{(adding } \pi \scriptscriptstyle\text{)} \\ &= -\cos\theta && \scriptscriptstyle\text{(cosine of the opposite)} \end{aligned}$$and, in the same way,$$\begin{aligned}[t] \sin(\pi - \theta) &= \sin\big(\pi + (-\theta)\big) \\ &= -\sin(-\theta) && \scriptscriptstyle\text{(adding } \pi \scriptscriptstyle\text{)} \\ &= -(-\sin\theta) = \sin\theta. && \scriptscriptstyle\text{(sine of the opposite)} \end{aligned}$$

Let \(\theta\) be an angle.
Let \(M(\cos \theta,\sin\theta)\) be the point on the unit circle at the angle \(\theta\).
Let \(M'(\cos(\frac \pi 2 -\theta),\sin(\frac \pi 2 -\theta))\) be the point on the unit circle at the angle \(\frac \pi 2 -\theta\).
The points \(M\) and \(M'\) are symmetric with respect to the line \(y=x\), so the \(x\)-coordinate of \(M\) is the \(y\)-coordinate of \(M'\), and the \(y\)-coordinate of \(M\) is the \(x\)-coordinate of \(M'\).
So \(\begin{aligned}[t] \cos\left(\frac \pi 2 -\theta\right) &= \sin \theta \\ \sin\left(\frac \pi 2 -\theta\right) &= \cos \theta \\ \end{aligned}\)

Write \(\dfrac{\pi}{2} + \theta = \pi - \left(\dfrac{\pi}{2} - \theta\right)\), and use \(\cos(\pi - x) = -\cos x\), \(\sin(\pi - x) = \sin x\) with \(x = \dfrac{\pi}{2} - \theta\), then the identities with \(\dfrac{\pi}{2} - \theta\): $$\begin{aligned}[t]\cos\left(\frac{\pi}{2} + \theta\right) &= \cos\left(\pi - \left(\frac{\pi}{2} - \theta\right)\right) = -\cos\left(\frac{\pi}{2} - \theta\right) = -\sin\theta,\\ \sin\left(\frac{\pi}{2} + \theta\right) &= \sin\left(\pi - \left(\frac{\pi}{2} - \theta\right)\right) = \sin\left(\frac{\pi}{2} - \theta\right) = \cos\theta.\end{aligned}$$

The sum of the angles in a triangle is \(\pi\), so \(\angle{OMH}=\pi - \dfrac \pi 4 - \dfrac \pi 2 = \dfrac \pi 4\).
Since \(\angle{OMH}=\angle{MOH}\), the triangle \(OHM\) is isosceles.
Let \(a=OH=HM\). $$\begin{aligned} a^2+a^2 &=1^2 \quad \text { (Pythagoras' theorem in the right-angled triangle } OHM)\\ 2 a^2 &=1 \\ a^2 &=\frac{1}{2} \\ a &=\frac{\sqrt{2}}{2} \quad \text { since } a\geq 0 \end{aligned}$$ So \(M\left(\dfrac{\sqrt{2}}{2},\dfrac{\sqrt{2}}{2}\right)\).
As \(\cos \theta\) is the \(x\)-coordinate of \(M\) and \(\sin \theta\) is the \(y\)-coordinate of \(M\), $$\cos \frac{\pi}{4}=\frac{\sqrt{2}}{2} \quad \text { and } \quad \sin \frac{\pi}{4}=\frac{\sqrt{2}}{2}$$

The coordinates of each point are found by using the symmetries about the axes and the half-turn about \(O\).

Let \(\angle{MON}=\dfrac \pi 3\).
As \(ON=OM=1\), the triangle \(OMN\) is isosceles at \(O\): its base angles are equal, \(\angle{OMN}=\angle{ONM}\).
Since the sum of the angles in a triangle is \(\pi\), \(2\,\angle{OMN}=\pi-\dfrac \pi 3=\dfrac{2\pi}{3}\), so \(\angle{OMN}=\angle{ONM}=\dfrac \pi 3\).
So the triangle \(OMN\) is equilateral.
The altitude \(MH\) bisects the base \(ON\).
Thus \(OH=\dfrac{1}{2}\). $$\begin{aligned} OH^2+HM^2 &=OM^2 \quad \text { (Pythagoras' theorem in the right-angled triangle } OHM)\\ \left(\frac{1}{2}\right)^2 + HM^2 &=1 \\ HM^2 &=\frac{3}{4} \\ HM &=\frac{\sqrt{3}}{2} \quad \text { since } HM\geq 0 \end{aligned}$$ As \(\cos \theta\) is the \(x\)-coordinate of \(M\) and \(\sin \theta\) is the \(y\)-coordinate of \(M\): $$\cos \frac{\pi}{3}=\frac{1}{2} \quad \text { and } \quad \sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$$ By the symmetry about the line \(y=x\) (the identities with \(\frac{\pi}{2}-\theta\)), the twelfth of a turn, \(\frac{\pi}{6}\), follows at once:$$\cos \frac{\pi}{6}=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2} \quad \text { and } \quad \sin \frac{\pi}{6}=\cos \frac{\pi}{3}=\frac{1}{2}.$$

The coordinates of each point are found by applying the symmetries about the axes and the half-turn about \(O\).

Consider the two unit vectors whose tips are the points of the unit circle associated with \(A\) and \(B\): $$\vec{u} = \begin{pmatrix}\cos A \\ \sin A \end{pmatrix} \quad\text{and}\quad \vec{v} = \begin{pmatrix}\cos B \\ \sin B \end{pmatrix}.$$ The geometric angle \(\theta \in [0, \pi]\) between \(\vec{u}\) and \(\vec{v}\) is equal to \(A - B\) or to \(B - A\), up to a multiple of \(2\pi\). Since cosine is even and \(2\pi\)-periodic, \(\cos\theta = \cos(A - B)\).
The scalar product \(\vec{u} \cdot \vec{v}\) can be computed in two ways:

1. From the angle between the vectors: $$\vec{u} \cdot \vec{v} = \|\vec{u}\| \,\|\vec{v}\| \cos\theta = 1 \cdot 1 \cdot \cos(A - B) = \cos(A - B).$$
2. From the components: $$\vec{u} \cdot \vec{v} = \begin{pmatrix}\cos A \\ \sin A \end{pmatrix} \cdot \begin{pmatrix}\cos B \\ \sin B \end{pmatrix} = (\cos A)(\cos B) + (\sin A)(\sin B).$$

Equating the two expressions gives the identity $$\cos(A - B) = \cos A \cos B + \sin A \sin B.$$ Another proof, with distances. Let \(P\) and \(Q\) be the points of the unit circle associated with \(A\) and \(B\): \(P(\cos A, \sin A)\) and \(Q(\cos B, \sin B)\). Then $$\begin{aligned}[t] PQ^2 &= (\cos A - \cos B)^2 + (\sin A - \sin B)^2 && \text{(distance between two points)}\\ &= \cos^2 A + \sin^2 A + \cos^2 B + \sin^2 B\\ &\quad - 2(\cos A\cos B + \sin A\sin B) && \text{(expanding)}\\ &= 2 - 2(\cos A\cos B + \sin A\sin B). && (\cos^2 + \sin^2 = 1) \end{aligned}$$ The rotation about \(O\) through the angle \(-B\) keeps distances. It sends \(Q\) to the point \(Q'(1, 0)\), of angle \(0\), and \(P\) to the point \(P'\big(\cos(A - B), \sin(A - B)\big)\), of angle \(A - B\). So $$\begin{aligned}[t] PQ^2 = P'Q'^2 &= \big(\cos(A - B) - 1\big)^2 + \sin^2(A - B) && \text{(distance between two points)}\\ &= \cos^2(A - B) + \sin^2(A - B) - 2\cos(A - B) + 1 && \text{(expanding)}\\ &= 2 - 2\cos(A - B). && (\cos^2 + \sin^2 = 1) \end{aligned}$$ The two expressions of \(PQ^2\) are equal: \(2 - 2\cos(A - B) = 2 - 2(\cos A\cos B + \sin A\sin B)\), which gives the identity again, without the scalar product.

Substitute \(B\) by \(-B\) in the cosine difference formula: $$\begin{aligned} \cos(A + B) &= \cos\bigl(A - (-B)\bigr)\\ &= \cos A \cos(-B) + \sin A \sin(-B)\\ &= \cos A \cos B + \sin A (-\sin B)\quad(\text{since } \cos(-\theta) = \cos \theta\text{ and }\sin(-\theta) = -\sin \theta)\\ &= \cos A \cos B - \sin A \sin B. \end{aligned}$$

Using the complementary angle identity \(\sin \theta = \cos\left(\dfrac{\pi}{2} - \theta\right)\): $$\begin{aligned} \sin(A + B) &= \cos\left(\frac{\pi}{2} - (A + B)\right)\\ &= \cos\left( \left(\frac{\pi}{2} - A\right) - B \right)\\ &= \cos\left(\frac{\pi}{2} - A\right) \cos B + \sin\left(\frac{\pi}{2} - A\right) \sin B\\ &= \sin A \cos B + \cos A \sin B. \end{aligned}$$ For the difference, write $$\begin{aligned} \sin(A - B) &= \sin\bigl(A + (-B)\bigr)\\ &= \sin A \cos(-B) + \cos A \sin(-B)\\ &= \sin A \cos B - \cos A \sin B. \end{aligned}$$

Assume \(\cos A \cos B \neq 0\) and \(\cos(A + B) \neq 0\) so that all ratios below are defined: $$\begin{aligned} \tan(A + B) &= \frac{\sin(A + B)}{\cos(A + B)} \\ &= \frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B - \sin A \sin B} \\ &= \frac{\dfrac{\sin A \cos B + \cos A \sin B}{\cos A \cos B}}{\dfrac{\cos A \cos B - \sin A \sin B}{\cos A \cos B}} \quad \text{(divide numerator and denominator by } \cos A \cos B\text{)} \\ &= \frac{\dfrac{\sin A}{\cos A} + \dfrac{\sin B}{\cos B}}{1 - \dfrac{\sin A \sin B}{\cos A \cos B}}\\ &= \frac{\tan A + \tan B}{1 - \tan A \tan B}. \end{aligned}$$ The formula for \(\tan(A-B)\) is obtained by replacing \(B\) with \(-B\) and using \(\tan(-B) = -\tan B\).