Trig Identities

The Pythagorean identity: one line

The unit circle has equation $x^2 + y^2 = 1$. The point on the unit circle at angle $\theta$ is $(\cos\theta, \sin\theta)$. Substitute:

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

That's it. The Pythagorean identity is the Pythagorean theorem applied to the unit circle. The unit circle has radius $1$; its points satisfy $x^2 + y^2 = 1$; since $x = \cos\theta$ and $y = \sin\theta$, we get $\cos^2\theta + \sin^2\theta = 1$. One sentence, full proof. Anyone who made you memorize this without that sentence owes you an apology.

The identity works for every real $\theta$ — no angle excluded. From it, we get two companions by dividing through by $\cos^2\theta$ and $\sin^2\theta$ respectively:

$1 + \tan^2\theta = \sec^2\theta, \qquad \cot^2\theta + 1 = \csc^2\theta.$

(Where $\sec\theta = 1/\cos\theta$ and $\csc\theta = 1/\sin\theta$ and $\cot\theta = 1/\tan\theta$. These reciprocal functions appear rarely but you should recognize them when they do.)

The sum formulas: derivation via rotation

We want to prove:

$\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta.$

Proof. Consider two points on the unit circle: $A = (\cos\alpha, \sin\alpha)$ and $B = (\cos\beta, \sin\beta)$. The distance $|AB|^2$ between them:

$|AB|^2 = (\cos\alpha - \cos\beta)^2 + (\sin\alpha - \sin\beta)^2.$

Expand: $= \cos^2\alpha - 2\cos\alpha\cos\beta + \cos^2\beta + \sin^2\alpha - 2\sin\alpha\sin\beta + \sin^2\beta.$

Using $\cos^2\theta + \sin^2\theta = 1$ twice: $= 1 + 1 - 2(\cos\alpha\cos\beta + \sin\alpha\sin\beta) = 2 - 2(\cos\alpha\cos\beta + \sin\alpha\sin\beta).$

Now rotate both points by $-\beta$: $A$ goes to $(\cos(\alpha-\beta), \sin(\alpha-\beta))$ and $B$ goes to $(1, 0)$. Rotation preserves distances (it's an isometry), so the distance is unchanged. But now:

$|AB|^2 = (\cos(\alpha-\beta) - 1)^2 + (\sin(\alpha-\beta) - 0)^2$ $= \cos^2(\alpha-\beta) - 2\cos(\alpha-\beta) + 1 + \sin^2(\alpha-\beta)$ $= 1 - 2\cos(\alpha-\beta) + 1 = 2 - 2\cos(\alpha-\beta).$

Setting the two expressions equal: $2 - 2\cos(\alpha-\beta) = 2 - 2(\cos\alpha\cos\beta + \sin\alpha\sin\beta).$ $\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta. \qquad \square$

Two distances, one rotation, zero hand-waving. That is a proof, damn it.

Deriving the rest

From $\cos(\alpha - \beta)$ we get everything else by substitution and symmetry:

$\cos(\alpha + \beta)$: replace $\beta$ with $-\beta$ and use $\cos(-\beta) = \cos\beta$, $\sin(-\beta) = -\sin\beta$: $\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta.$

$\sin(\alpha + \beta)$: use the identity $\sin\theta = \cos(\pi/2 - \theta)$: $\sin(\alpha + \beta) = \cos\!\left(\frac{\pi}{2} - (\alpha+\beta)\right) = \cos\!\left(\left(\frac{\pi}{2} - \alpha\right) - \beta\right).$

Apply the $\cos(\cdot - \beta)$ formula: $= \cos\!\left(\frac{\pi}{2}-\alpha\right)\cos\beta + \sin\!\left(\frac{\pi}{2}-\alpha\right)\sin\beta = \sin\alpha\cos\beta + \cos\alpha\sin\beta.$

$\sin(\alpha - \beta)$: replace $\beta$ with $-\beta$: $\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta.$

The four sum/difference formulas:

$\boxed{\cos(\alpha \pm \beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta}$ $\boxed{\sin(\alpha \pm \beta) = \sin\alpha\cos\beta \pm \cos\alpha\sin\beta}$

The $\mp$ in the cosine formula: the upper sign on the left goes with the lower sign on the right. Many students mix these up. The easiest fix: remember the cosine formula has mixed signs ($+\beta$ on left gives $-$ between terms on right), while the sine formula has same signs.

Drag the widget and watch the sum formula in action:

Double-angle formulas: just set α = β

Set $\alpha = \beta$ in the sum formulas:

$\sin(2\alpha) = 2\sin\alpha\cos\alpha.$

$\cos(2\alpha) = \cos^2\alpha - \sin^2\alpha.$

The cosine double-angle has two equivalent alternate forms, obtained by substituting $\cos^2\alpha = 1 - \sin^2\alpha$ or $\sin^2\alpha = 1 - \cos^2\alpha$:

$\cos(2\alpha) = 1 - 2\sin^2\alpha = 2\cos^2\alpha - 1.$

The last two forms are the power-reduction formulas in disguise: $\sin^2\alpha = (1 - \cos 2\alpha)/2$ and $\cos^2\alpha = (1 + \cos 2\alpha)/2$. These appear constantly in calculus integrals — you don't need calculus to see the identity; you need calculus to care about it. But file it the hell away. It'll save your ass later.

Proving identities: the etiquette

An identity is a statement that two trig expressions are equal for all valid angles. Proving one means transforming one side into the other using known identities. The rules:

1. Work one side only. Pick the more complicated side and manipulate it into the other. Never "cross" the equals sign. Writing "$\sin^2\theta + \cos^2\theta = 1$, therefore LHS $=$ RHS" is circular when the identity you're proving IS the substitution you're using.
2. Never add the same thing to both sides (that's solving an equation, not proving an identity).
3. Conjugates and $1 = \cos^2+\sin^2$ are your two main tricks. When stuck, try substituting $1$ for $\cos^2\theta + \sin^2\theta$, or multiply by a clever form of $1$.

The payoff: cos/sin sum formulas are the 2D rotation matrix

Here is the reason the Matrices stratum is collecting on a debt from this lesson. Consider the 2D rotation by angle $\theta$: it sends

$(1, 0) \mapsto (\cos\theta, \sin\theta)$ $(0, 1) \mapsto (-\sin\theta, \cos\theta).$

(The second mapping: $(0,1)$ is at angle $\pi/2$. Rotating by $\theta$ gives angle $\pi/2 + \theta$, with coordinates $(\cos(\pi/2+\theta), \sin(\pi/2+\theta)) = (-\sin\theta, \cos\theta)$.)

The matrix whose columns are the images of the basis vectors is:

$R_\theta = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}.$

Now apply this to a general vector $(x, y)$:

$R_\theta \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x\cos\theta - y\sin\theta \\ x\sin\theta + y\cos\theta \end{pmatrix}.$

The entries of the rotated vector are exactly the sum formulas for $\cos(\theta - \phi)$ and $\sin(\theta + \phi)$ (where $x = \cos\phi$, $y = \sin\phi$ for a unit vector). The rotation matrix IS the sum formula, packaged in a grid.

In the Matrices stratum, we will see that $R_\alpha R_\beta = R_{\alpha+\beta}$ — multiplying two rotation matrices is the addition formula for angles. The trig and the linear algebra are the same thing, viewed from different windows. File this away. It detonates in the eigenvalue lesson.