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Unit circle

Take the circle of radius 1 centred at the origin, start at the point $(1, 0)$, and travel counterclockwise through an angle $t$. The unit circle definition of the trigonometric functions says the point you land on is $(\cos t,\ \sin t)$ - cosine is the horizontal coordinate and sine is the vertical one. OpenStax Precalculus builds section 5.2 on exactly this, and everything else in a first trigonometry course is a consequence of it.

Why the coordinates are the definition

In a right triangle, $\cos\theta$ is the adjacent side over the hypotenuse. Draw the radius to the point at angle $t$ and drop a vertical line to the $x$-axis. That is a right triangle whose hypotenuse is the radius, length 1. The adjacent side is the horizontal distance $x$, so

$$\cos t = \frac{x}{1} = x, \qquad \sin t = \frac{y}{1} = y$$

The hypotenuse being 1 is what removes the division. Ratios become coordinates.

The triangle picture only works for acute angles, though. There is no right triangle with an angle of $120^\circ$, and none at all for $-\pi/4$ or for $17\pi$. So the coordinate statement is promoted from consequence to definition: $\cos t$ and $\sin t$ are the coordinates, for every real $t$, and the triangle is the special case where $t$ lands in the first quadrant.

One identity comes free. The point is on the circle, so $x^2 + y^2 = 1$, which written in the new names is

$$\cos^2 t + \sin^2 t = 1$$

That is the Pythagorean identity, and it is not a separate theorem. It is the equation of the circle.

Radians are arc lengths

Arc length on a circle of radius $r$ is $s = r\theta$ when $\theta$ is in radians. On the unit circle $r = 1$, so $s = \theta$: the angle and the arc it cuts are the same number. Travel a distance of 1 along the rim and you have turned through 1 radian.

This makes radians the measurement the circle itself supplies. A full trip round is a distance of $2\pi$, so a full turn is $2\pi$ radians. Half is $\pi$, a quarter is $\pi/2$. Nothing was chosen. Degrees are the arbitrary unit here: 360 is a Babylonian convention with no geometric content, and the conversion $1$ radian $= 180/\pi = 57.2958^\circ$ is ugly precisely because one of the two units was invented rather than measured.

The payoff arrives in calculus, where $\frac{d}{dx}\sin x = \cos x$ holds only in radians. In degrees the derivative picks up a stray $\pi/180$, which is the conversion factor leaking into a formula that should not contain one. Our study guide for OpenStax Calculus 1.3 makes that the reason to convert once, at the start of a problem.

The two triangles behind every exact value

The exact values at $\pi/6$, $\pi/4$ and $\pi/3$ come from two triangles. Both are worth deriving once, because after that the table is reconstructible in a few seconds and there is nothing left to forget.

The 45-45-90 triangle. At $t = \pi/4$ the radius bisects the first quadrant, so the point sits on the line $y = x$. Substituting into $x^2 + y^2 = 1$ gives $2x^2 = 1$, so $x = 1/\sqrt{2} = \sqrt{2}/2 = 0.707107$. Both coordinates are that:

$$\cos\frac{\pi}{4} = \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}$$

The 30-60-90 triangle. Take an equilateral triangle with every side 1 and every angle $60^\circ$. Drop an altitude from the apex. It bisects the base, leaving a right triangle with hypotenuse 1, short leg $1/2$, and the angle at the apex cut down to $30^\circ$. The remaining leg is $\sqrt{1 - 1/4} = \sqrt{3}/2 = 0.866025$. The short leg is opposite the $30^\circ$ angle, so $\sin 30^\circ = 1/2$ and $\cos 30^\circ = \sqrt{3}/2$. In radians:

$$\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2},\quad \sin\frac{\pi}{6} = \frac{1}{2},\qquad \cos\frac{\pi}{3} = \frac{1}{2},\quad \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$$

The $\pi/3$ entries are the $\pi/6$ entries swapped, because those two angles are the two acute angles of the same triangle and swapping them swaps which leg is adjacent.

Now line up the sines at $0$, $\pi/6$, $\pi/4$, $\pi/3$, $\pi/2$:

$$\frac{\sqrt{0}}{2},\ \frac{\sqrt{1}}{2},\ \frac{\sqrt{2}}{2},\ \frac{\sqrt{3}}{2},\ \frac{\sqrt{4}}{2}$$

which is $0$, $1/2$, $\sqrt{2}/2$, $\sqrt{3}/2$, $1$. Cosine is the identical list read backwards, since the $x$-coordinate shrinks as the $y$-coordinate grows. That pattern is not a mnemonic pulled out of the air: the middle entry is the 45-45-90 result, the two either side are the 30-60-90 results, and the ends are the points $(1,0)$ and $(0,1)$ where the circle meets the axes. Check any column against $\cos^2 + \sin^2 = 1$ and it holds: at $\pi/6$, $3/4 + 1/4 = 1$.

Quadrants, reference angles, and the degree-mode trap

Because cosine is $x$ and sine is $y$, the sign of each function is just the sign of that coordinate. In quadrant I both are positive. In quadrant II, $x$ is negative and $y$ is positive, so cosine is negative and sine is positive. In quadrant III both are negative. In quadrant IV cosine is positive and sine is negative. There is nothing to memorise beyond which way the axes point.

The magnitudes never change. A reference angle is the acute angle between the radius and the nearest half of the $x$-axis, and the point at $t$ is a mirror image of the point at its reference angle, so the coordinates match except for signs.

Take $t = 5\pi/3$. It lies between $3\pi/2$ and $2\pi$, so quadrant IV, and the reference angle is $2\pi - 5\pi/3 = \pi/3$. From the table, $\cos\pi/3 = 1/2$ and $\sin\pi/3 = \sqrt{3}/2$. Quadrant IV makes $x$ positive and $y$ negative:

$$\cos\frac{5\pi}{3} = \frac{1}{2}, \qquad \sin\frac{5\pi}{3} = -\frac{\sqrt{3}}{2}$$

Check with the circle equation: $1/4 + 3/4 = 1$. Take $t = 7\pi/6$ instead. It is $\pi/6$ past $\pi$, so quadrant III with reference angle $\pi/6$, and both coordinates come out negative: $\cos = -\sqrt{3}/2$, $\sin = -1/2$. Same check, same 1. Reference and coterminal angles are where this reduction gets drilled; the unit circle values themselves are the table above.

The failure mode is the calculator. A problem posed in radians asks for $\sin(\pi/6)$, which is $0.5$ exactly. Type $\pi/6$ as the decimal $0.523599$ into a calculator left in degree mode and it returns $\sin(0.523599^\circ) = 0.009138$. That is off by a factor of 55, and the answer does not look like a botched sine - it looks like a sine of something near zero, which is what it is. The tell is that every radian input smaller than about 3 becomes a near-zero output, because in degree mode the calculator is computing roughly $t\pi/180$. If a batch of answers all come back tiny, check the mode before checking the algebra.

Tangent follows from the same picture as $y/x$, which makes it undefined exactly where $x = 0$, at $\pi/2$ and $3\pi/2$. The circle keeps paying out like that: define two functions as coordinates and the identities, the signs, the periodicity and the asymptotes all turn out to be geometry you can see.

References