Every trig value is the coordinate of a point moving around the unit circle.
Points at related angles, such as an angle and its mirror image, share coordinates up to a sign.
This means the sine or cosine of a large or negative angle is always the sine or cosine of a first-quadrant reference angle, give or take a minus sign.
Knowing these relationships lets you evaluate any angle, find every solution of a trig equation in an interval (not just the one your calculator gives), and simplify expressions quickly.
This article reads the relationships straight off the unit circle, cross-checks them with the compound angle identities, and links each to the symmetry of the graphs.
Note
The point on the unit circle at angle $\theta$ has coordinates $(\cos\theta, \sin\theta)$, and $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ is the gradient of the radius.
The Unit Circle Is The Source Of Every Relationship
Move a point to a related angle and watch which coordinate changes sign.
That single observation gives every identity in this topic.
A reflection in the $y$-axis flips the $x$-coordinate; a reflection in the $x$-axis flips the $y$-coordinate; a half-turn flips both.
Since $x = \cos\theta$ and $y = \sin\theta$, a sign change in a coordinate is a sign change in that ratio: nothing more is needed.
The Four Key Angle Relationships
Reflection In The y-axis: θ And π − θ
The point at $\pi - \theta$ is the mirror image of the point at $\theta$ in the $y$-axis, with coordinates $(-\cos\theta, \sin\theta)$.
Reading off the coordinates: $$\sin(\pi - \theta) = \sin\theta, \qquad \cos(\pi - \theta) = -\cos\theta$$
Cross-check with a compound identity: $$\sin(\pi-\theta) = \sin\pi\cos\theta - \cos\pi\sin\theta = 0 - (-1)\sin\theta = \sin\theta$$
Half Turn: θ And π + θ
Adding $\pi$ rotates the point by half a turn to $(-\cos\theta, -\sin\theta)$, so both coordinates flip: $$\sin(\pi + \theta) = -\sin\theta, \qquad \cos(\pi + \theta) = -\cos\theta$$
The two sign changes cancel in the ratio, which is exactly why tangent repeats every $\pi$: $$\tan(\pi + \theta) = \tan\theta$$
Reflection In The x-axis: θ And −θ
The point at $-\theta$ is the mirror image in the $x$-axis, $(\cos\theta, -\sin\theta)$, which gives the odd and even rules: $$\cos(-\theta) = \cos\theta, \qquad \sin(-\theta) = -\sin\theta, \qquad \tan(-\theta) = -\tan\theta$$
So cosine is an even function, while sine and tangent are odd.
Complementary Angles: θ And π/2 − θ
The complementary angle swaps the roles of the two coordinates, which is where the "co" in cosine comes from:
These symmetry relationships are not printed in the formula booklet, so learn to regenerate them.
The fast way is to sketch the unit-circle point; the reliable back-up is a compound angle identity (AHL 3.10), which is in the booklet.
Reducing Any Angle To A Reference Angle
Put the relationships to work: to evaluate an awkward angle, rewrite it in terms of a first-quadrant reference angle, then attach the sign from its quadrant.
Example question
Find all solutions of $\sin x = \tfrac12$ for $0 \le x \le 2\pi$.
Solution
Find the reference angle from the first quadrant: $\sin\tfrac{\pi}{6} = \tfrac12$, so $x = \tfrac{\pi}{6}$.
Sine is also positive in the second quadrant, where the related angle is $\pi - \tfrac{\pi}{6}$.
Apply $\sin(\pi - \theta) = \sin\theta$: the second solution is $x = \dfrac{5\pi}{6}$.
Solutions: $x = \dfrac{\pi}{6}$ and $x = \dfrac{5\pi}{6}$.
Symmetry Of The Graphs
Each relationship above shows up as a visible symmetry of the graph.
The graph and the identity are two views of the same fact.
Sine: Odd, Period 2π
Because $\sin(-x) = -\sin x$, the sine graph has rotational symmetry about the origin.
Because $\sin(\pi - x) = \sin x$, each hump is also symmetric about the vertical line through its peak, such as $x = \tfrac{\pi}{2}$.
Cosine: Even, Period 2π
Because $\cos(-x) = \cos x$, the cosine graph is symmetric in the $y$-axis.
It is the sine graph shifted left by $\tfrac{\pi}{2}$, matching the complementary-angle relationship $\cos\theta = \sin\!\left(\tfrac{\pi}{2}-\theta\right)$.
Tangent: Odd, Period π
Because $\tan(-x) = -\tan x$, the tangent graph has rotational symmetry about the origin.
Because $\tan(x + \pi) = \tan x$, it repeats every $\pi$, not every $2\pi$, with vertical asymptotes wherever $\cos x = 0$.
Exam technique
Quick simplifier when memory fails: for $\sin$ and $\cos$, adding or subtracting $\pi$ flips the sign, so $\cos\theta = -\cos(\theta + \pi)$.
For $\tan$ and $\cot$, adding or subtracting $\pi$ leaves the value unchanged, because their period is $\pi$.
Putting Relationships Together
Many exam simplifications collapse to a single value once each term is rewritten with these rules.
