Trigonometric Identities and Equations

⚡ Key Takeaways

Trigonometric identities are equations relating trig functions that hold for all valid inputs; trig equations ask for the specific inputs that make a relation true. Master the Pythagorean, sum/difference, double/half-angle identities and you can simplify any expression and solve any standard trig equation. Used everywhere: physics waves, engineering, navigation, computer graphics. Foundational work spans Hipparchus (c. 150 BC) through Euler (1748).

💡 The Core Insight

Sine and cosine are the coordinates of a point moving around the unit circle. Because the circle has radius 1, the Pythagorean theorem instantly gives $\sin^2+\cos^2=1$ . Because rotating by angle $A$ then by angle $B$ is the same as rotating by $A+B$ , the sum formulas follow. Every identity is a geometric or algebraic consequence of these two facts. An identity is a structural truth; an equation is a question about location.

DefinitionUnit-Circle Definitions

For an angle $\theta$ measured counterclockwise from the positive $x$ -axis, let $P=(x,y)$ be the point where the terminal ray meets the unit circle $x^2+y^2=1$ . Then:

1

\quad \cos\theta = x,\qquad \sin\theta = y,\qquad \tan\theta = \frac{y}{x}=\frac{\sin\theta}{\cos\theta}

Derivation hint: These are definitions, but note $\tan$ is the slope of the terminal ray, undefined when $x=\cos\theta=0$ . The reciprocal functions are:

2

\quad \csc\theta=\frac{1}{\sin\theta},\qquad \sec\theta=\frac{1}{\cos\theta},\qquad \cot\theta=\frac{1}{\tan\theta}=\frac{\cos\theta}{\sin\theta}

Derivation hint: Each is the multiplicative inverse of a primary function; domains exclude points where the denominator vanishes. Here $\theta$ is a real number (radians) or angle measure; $x,y$ are real coordinates with $x^2+y^2=1$ .

TheoremThe Pythagorean Identities

For all $\theta$ where the functions are defined:

3

\quad \sin^2\theta + \cos^2\theta = 1

Derivation hint: Apply $x^2+y^2=1$ with $x=\cos\theta$ , $y=\sin\theta$ .

4

\quad 1 + \tan^2\theta = \sec^2\theta

Derivation hint: Divide [3] by $\cos^2\theta$ .

5

\quad 1 + \cot^2\theta = \csc^2\theta

Derivation hint: Divide [3] by $\sin^2\theta$ .

TheoremSum and Difference Identities

For all angles $A,B$ :

6

\quad \cos(A-B) = \cos A\cos B + \sin A\sin B

Derivation hint: Compute the squared distance between unit-circle points at angles $A$ and $B$ two ways (law of cosines vs. coordinate distance).

7

\quad \cos(A+B) = \cos A\cos B - \sin A\sin B

Derivation hint: Replace $B$ with $-B$ in [6], using $\cos(-B)=\cos B$ , $\sin(-B)=-\sin B$ .

8

\quad \sin(A\pm B) = \sin A\cos B \pm \cos A\sin B

Derivation hint: Use the cofunction relation $\sin\phi=\cos(\tfrac{\pi}{2}-\phi)$ and apply [6]/[7].

9

\quad \tan(A\pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B}

Derivation hint: Divide [8] by [7] and simplify by dividing numerator and denominator by $\cos A\cos B$ .

TheoremDouble- and Half-Angle Identities

10

\quad \sin 2\theta = 2\sin\theta\cos\theta

Derivation hint: Set $A=B=\theta$ in [8] with the plus sign.

11

\quad \cos 2\theta = \cos^2\theta - \sin^2\theta = 1-2\sin^2\theta = 2\cos^2\theta - 1

Derivation hint: Set $A=B=\theta$ in [7]; obtain alternate forms using [3].

12

\quad \sin^2\theta = \frac{1-\cos 2\theta}{2},\qquad \cos^2\theta = \frac{1+\cos 2\theta}{2}

Derivation hint: Solve the second and third forms of [11] for $\sin^2$ and $\cos^2$ (the power-reduction formulas).

13

\quad \sin\frac{\theta}{2} = \pm\sqrt{\frac{1-\cos\theta}{2}},\qquad \cos\frac{\theta}{2} = \pm\sqrt{\frac{1+\cos\theta}{2}}

Derivation hint: Replace $\theta$ with $\theta/2$ in [12] and take square roots; sign depends on the quadrant of $\theta/2$ .

ProofDeriving the Difference Formula for Cosine — Equation 6

We prove $\cos(A-B)=\cos A\cos B+\sin A\sin B$ using the unit circle and the distance formula. 1. Place two points on the unit circle: $P=(\cos A,\sin A)$ at angle $A$ , and $Q=(\cos B,\sin B)$ at angle $B$ . 2. The angle between $OP$ and $OQ$ is $A-B$ . Compute the squared distance $PQ^2$ directly:

14

\quad PQ^2=(\cos A-\cos B)^2+(\sin A-\sin B)^2

Derivation hint: Distance formula between $P$ and $Q$ . 3. Expand: $PQ^2=\cos^2A-2\cos A\cos B+\cos^2B+\sin^2A-2\sin A\sin B+\sin^2B$ . Group using [3] twice:

15

\quad PQ^2 = 2 - 2(\cos A\cos B+\sin A\sin B)

Derivation hint: $\cos^2A+\sin^2A=1$ and $\cos^2B+\sin^2B=1$ . 4. Now compute $PQ^2$ using the law of cosines in triangle $OPQ$ where $|OP|=|OQ|=1$ and the included angle is $A-B$ :

16

\quad PQ^2 = 1^2+1^2-2(1)(1)\cos(A-B)=2-2\cos(A-B)

Derivation hint: Law of cosines $c^2=a^2+b^2-2ab\cos C$ . 5. Equate [15] and [16]: $2-2(\cos A\cos B+\sin A\sin B)=2-2\cos(A-B)$ . Cancel and divide by $-2$ :

17

\quad \cos(A-B)=\cos A\cos B+\sin A\sin B \qquad\blacksquare

Derivation hint: Subtract 2 from both sides, divide by $-2$ . Every other sum/difference and multiple-angle identity follows algebraically from this one and [3].

∎

mathematician · Leonhard Euler | 1707–1783

Euler unified trigonometry with the exponential function through his celebrated formula $e^{i\theta}=\cos\theta+i\sin\theta$ . From this single relation, all trig identities become consequences of the exponential law $e^{i(A+B)}=e^{iA}e^{iB}$ . Multiplying out the right side and matching real and imaginary parts instantly reproduces the sum formulas [7] and [8]. Euler's 1748 Introductio in analysin infinitorum standardized the modern function-based view of trigonometry — treating $\sin$ and $\cos$ as functions of a real variable rather than ratios in a particular triangle, which is exactly how we use them today.

🧠 Intuition

Think of $\cos\theta$ and $\sin\theta$ as the shadow of a rotating clock hand of length 1: the horizontal shadow is the cosine, the vertical shadow is the sine. As the hand sweeps, the two shadows trade size, but the Pythagorean identity [3] guarantees the combined shadow always reconstructs a full unit length — the hand never stretches. The sum formulas answer: "If I know the shadows at angle $A$ and at angle $B$ , what are the shadows at $A+B$ ?" The answer isn't simple addition — shadows interfere. Equation [7] says the new horizontal shadow is the product of the two horizontal shadows minus the product of the two vertical ones. That minus sign is the fingerprint of rotation, the same minus that appears in complex multiplication. For equations, switch viewpoints. Solving $\sin\theta=0.5$ isn't simplifying — it's asking "where on the circle is the height exactly $0.5$ ?" There are two such points per revolution and infinitely many revolutions, hence infinitely many solutions, which is why every trig equation answer carries a " $+2\pi n$ " tail.

Figure

The unit circle showing cosine and sine as coordinates of point P at angle θ

ExampleProving an Identity by Simplification

Given: Prove the identity $\dfrac{\sin\theta}{1+\cos\theta} + \dfrac{1+\cos\theta}{\sin\theta} = 2\csc\theta$ .

Steps: 1. Combine the left side over a common denominator $\sin\theta(1+\cos\theta)$ :

18

\quad \text{LHS}=\frac{\sin^2\theta + (1+\cos\theta)^2}{\sin\theta(1+\cos\theta)}

2. Expand the numerator: $\sin^2\theta + 1 + 2\cos\theta + \cos^2\theta$ . 3. Apply [3], $\sin^2\theta+\cos^2\theta=1$ , so the numerator becomes $1 + 1 + 2\cos\theta = 2 + 2\cos\theta = 2(1+\cos\theta)$ . 4. Substitute back:

19

\quad \text{LHS}=\frac{2(1+\cos\theta)}{\sin\theta(1+\cos\theta)} = \frac{2}{\sin\theta}

5. By definition [2], $\dfrac{2}{\sin\theta}=2\csc\theta$ .

\boxed{\dfrac{\sin\theta}{1+\cos\theta}+\dfrac{1+\cos\theta}{\sin\theta}=2\csc\theta \quad\checkmark}

ExampleSolving a Trigonometric Equation

Given: Solve $2\cos^2\theta + \cos\theta - 1 = 0$ for $0 \le \theta < 2\pi$ .

Steps: 1. Let $u=\cos\theta$ . The equation becomes the quadratic $2u^2+u-1=0$ . 2. Factor: $(2u-1)(u+1)=0$ , so $u=\tfrac{1}{2}$ or $u=-1$ . 3. Case $\cos\theta=\tfrac{1}{2}$ : On $[0,2\pi)$ the cosine equals $\tfrac12$ at $\theta=\tfrac{\pi}{3}$ (1st quadrant) and $\theta=\tfrac{5\pi}{3}$ (4th quadrant). 4. Case $\cos\theta=-1$ : This occurs only at $\theta=\pi$ . 5. Collect all solutions in the interval:

\boxed{\theta = \frac{\pi}{3},\ \pi,\ \frac{5\pi}{3}}

(If all real solutions were wanted, add $+2\pi n$ , $n\in\mathbb{Z}$ , to each.)

⚠️ Common mistake

An identity is not an equation to "solve." When asked to prove an identity, you may not move terms across the equals sign as if solving — that assumes the very thing you're proving. Instead, transform one side until it equals the other, or transform each side independently to a common third expression.

$\sin^2\theta$ means $(\sin\theta)^2$ , not $\sin(\theta^2)$ . The exponent applies to the function value. Also $\sin^{-1}\theta$ denotes the inverse function (arcsine), not $1/\sin\theta=\csc\theta$ . This notational clash is a frequent source of error.

Never cancel a function across a sum. $\sin(A+B)\ne \sin A+\sin B$ . The sum formula [8] is the correct expansion.

Counterexample

A student claims $\sqrt{\sin^2\theta}=\sin\theta$ . Test $\theta=\tfrac{3\pi}{2}$ : here $\sin\theta=-1$ , so $\sin^2\theta=1$ and $\sqrt{1}=1\ne -1$ . The square root returns the non-negative root, so correctly $\sqrt{\sin^2\theta}=|\sin\theta|$ . This same trap explains the $\pm$ sign in the half-angle formulas [13] — the sign must be chosen according to the quadrant, not assumed positive.

connection

Trig identities sit at a busy crossroads. They are the algebraic engine behind Trigonometric Functions and depend directly on the geometry of The Unit Circle. In analysis, the power-reduction forms [12] are essential for integration — every integral of $\sin^2$ or $\cos^2$ in Integral Calculus is computed by replacing them with $\cos 2\theta$ terms. The deepest link is to Complex Numbers via Euler's formula, which turns all of trigonometry into exponential algebra. The sum-to-product forms underpin Fourier Series, where any periodic signal decomposes into sines and cosines, and the double-angle relations connect to Rotation Matrices in Linear Algebra.

🔬 Where this appears

⏳ History

- c. 150 BC — Hipparchus of Nicaea compiles the first known table of chords, the ancestor of the sine table, to predict astronomical positions.

- c. 100 AD — Ptolemy states what is now equivalent to the sum and difference identities as "Ptolemy's theorem" on cyclic quadrilaterals in the Almagest.

- c. 500 AD — Aryabhata introduces the half-chord (the true sine), tabulating values and an early version of the half-angle relation in India.

- c. 1000 AD — Abu al-Wafa of Baghdad establishes the modern sine, tangent, secant relations and proves the double-angle and addition formulas geometrically.

- 1464 — Regiomontanus writes De triangulis omnimodis, the first European text treating trigonometry as a discipline independent of astronomy.

- 1593 — François Viète derives multiple-angle formulas and uses them to solve high-degree polynomial equations.

- 1722 — Abraham de Moivre publishes the formula

(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta

, linking powers to multiple angles.

- 1748 — Leonhard Euler unifies trigonometry with the exponential function via

e^{i\theta}=\cos\theta+i\sin\theta

in the Introductio, recasting all identities as exponential algebra.

- 1822 — Joseph Fourier shows arbitrary periodic functions decompose into sines and cosines, making trig identities central to modern physics and engineering.

open-problem

The "subtleties" here are domain and inverse-function issues rather than unsolved research problems. The chief one: trig equations have infinitely many solutions because $\sin$ and $\cos$ are periodic, so any complete answer requires the general form $\theta_0 + 2\pi n$ . Restricting to a principal interval (as in the inverse functions $\arcsin,\arccos,\arctan$ ) discards solutions, and students must consciously reintroduce them. A deeper subtlety: identities derived by squaring both sides may introduce extraneous solutions that satisfy the squared equation but not the original — every solution must be checked back in the unsquared form. Likewise, dividing by a trig function silently assumes it is nonzero and may lose solutions where that function equals zero. Finally, the half-angle sign ambiguity in [13] has no universal resolution; it can only be settled by knowing the quadrant of the half-angle, which is information outside the identity itself.

Trigonometric Identities and Equations

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also