Trigonometric Identities

A trigonometric identity is an equation involving trig functions that is true for every value of the angle where both sides are defined. That "for every angle" part is what makes identities so powerful: once you know one, you can swap one expression for an equal one anywhere it appears, no matter what the angle is. Identities are the grammar of trigonometry — they let you rewrite messy expressions into simple ones, solve equations, collapse complicated integrals, and describe how waves add and interfere.

If earlier trig felt like memorizing a pile of unrelated facts, this page is where the pile turns into a connected system. Almost everything below flows from one picture — a point on the unit circle — and a single theorem you already know: the Pythagorean theorem. Learn to derive the identities rather than memorize them, and you will never be stuck when your memory blanks in an exam.

Learning Objectives

By the end of this page, you should be able to:

• State and derive the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ and its two companions.
• Use the reciprocal and quotient identities fluently to convert between the six trig functions.
• Apply the angle-sum, angle-difference, and double-angle formulas to compute exact values and rewrite expressions.
• Prove an identity by transforming one side into the other using a reliable strategy.
• Simplify trigonometric expressions and recognize which identity a problem is inviting you to use.

Quick Answer

The three Pythagorean identities are $\sin^2\theta + \cos^2\theta = 1$, $1 + \tan^2\theta = \sec^2\theta$, and $1 + \cot^2\theta = \csc^2\theta$; the last two come from dividing the first by $\cos^2\theta$ and $\sin^2\theta$. The reciprocal identities define $\csc\theta = 1/\sin\theta$, $\sec\theta = 1/\cos\theta$, $\cot\theta = 1/\tan\theta$, and the quotient identities give $\tan\theta = \sin\theta/\cos\theta$ and $\cot\theta = \cos\theta/\sin\theta$. The angle-sum formulas — for example $\sin(A+B) = \sin A\cos B + \cos A\sin B$ — let you break apart or combine angles, and setting $A=B$ produces the double-angle formulas like $\sin 2\theta = 2\sin\theta\cos\theta$. To prove an identity, start from the more complicated side and rewrite it, usually by converting everything to sines and cosines, until it matches the other side. These tools simplify expressions, solve equations, and are indispensable in calculus and wave physics.

Where It Came From

Trigonometric identities were born from astronomy, not abstract mathematics. Ancient astronomers needed to predict the positions of the Sun, Moon, and planets, and that meant doing arithmetic with angles and chords on a sphere. Around 150 CE the Greek astronomer Ptolemy, building on Hipparchus (often called the founder of trigonometry, c. 150 BCE), compiled a table of chord lengths in his great work the Almagest. To build that table he needed a way to find the chord of the sum or difference of two angles from chords he already knew — this is exactly what we now call Ptolemy's theorem, and it is essentially the angle-sum formula for sine in disguise. Without such an identity, every new angle would require fresh, brutal measurement; with it, a whole table could be generated from a handful of known values.

The identities matured in the medieval Islamic world and in India, where mathematicians such as Aryabhata, Al-Battani, and later Nasir al-Din al-Tusi worked directly with sine and cosine (concepts inherited from Indian astronomy) and treated trigonometry as a subject in its own right. The compact algebraic notation we use today arrived with Leonhard Euler in the 18th century, whose formula $e^{i\theta} = \cos\theta + i\sin\theta$ revealed why all these identities hang together: they are shadows of the simple rules for multiplying exponentials.

The modern reason every calculus and physics student meets identities is different from the astronomical one. In the 1700s and 1800s, identities became the essential trick for simplifying integrals (you cannot integrate $\sin^2\theta$ until you rewrite it using a double-angle formula) and for modeling waves — sound, light, alternating current, and quantum states are all built from sines and cosines, and combining them requires exactly these identities. The need that forced them into existence keeps renewing itself.

The Foundation: Pythagorean, Reciprocal, and Quotient Identities

Everything starts on the unit circle — a circle of radius $1$ centered at the origin. If you rotate by an angle $\theta$ from the positive $x$-axis, the point you land on has coordinates $(\cos\theta, \sin\theta)$. That is the definition of cosine and sine, and it is the single most useful fact in trigonometry.

Because the point $(\cos\theta, \sin\theta)$ lies on a circle of radius $1$, the Pythagorean theorem applied to the little right triangle with legs $\cos\theta$ and $\sin\theta$ and hypotenuse $1$ gives immediately:

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

This is the Pythagorean identity, and it is true for every angle — no exceptions. Notice $\sin^2\theta$ is shorthand for $(\sin\theta)^2$.

The other two Pythagorean identities are not new facts; they are the same identity wearing different clothes. Divide both sides by $\cos^2\theta$:

$\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta} \quad\Longrightarrow\quad \tan^2\theta + 1 = \sec^2\theta$

Divide the original instead by $\sin^2\theta$ and you get $1 + \cot^2\theta = \csc^2\theta$. Rederiving these on scratch paper takes ten seconds and never fails you.

The reciprocal identities just name the flips of the three basic functions:

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

and the quotient identities connect tangent and cotangent to sine and cosine:

$\tan\theta = \frac{\sin\theta}{\cos\theta}, \qquad \cot\theta = \frac{\cos\theta}{\sin\theta}$

Worked Example: Finding all six functions from one

Suppose $\sin\theta = \tfrac{3}{5}$ and $\theta$ is in the second quadrant (between $90°$ and $180°$). Find $\cos\theta$ and $\tan\theta$.

From the Pythagorean identity, $\cos^2\theta = 1 - \sin^2\theta = 1 - \tfrac{9}{25} = \tfrac{16}{25}$, so $\cos\theta = \pm\tfrac{4}{5}$. The quadrant decides the sign: in the second quadrant cosine is negative, so $\cos\theta = -\tfrac{4}{5}$.

Then by the quotient identity,

$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{3/5}{-4/5} = -\frac{3}{4}.$

The lesson: the identity narrows the answer to two possibilities, and the geometry (which quadrant) picks the right one.

Angle-Sum and Angle-Difference Formulas

The next layer answers a question tables of values cannot: what is $\sin$ or $\cos$ of a sum of two angles? You cannot simply distribute — $\sin(A+B)$ is not $\sin A + \sin B$. The correct formulas are:

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

For differences, flip the signs (replace $B$ with $-B$, using that cosine is even and sine is odd):

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

A handy memory hook: the cosine formula flips the sign (plus becomes minus) and keeps functions matched ($\cos\cos$, $\sin\sin$), while the sine formula keeps the sign and mixes functions ($\sin\cos$, $\cos\sin$). For tangent,

$\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A\tan B}.$

Worked Example: An exact value

Find the exact value of $\cos 75°$.

Write $75° = 45° + 30°$, two angles whose values we know exactly. Then

$\cos 75° = \cos(45° + 30°) = \cos 45°\cos 30° - \sin 45°\sin 30°.$

Substitute the known values $\cos 45° = \sin 45° = \tfrac{\sqrt2}{2}$, $\cos 30° = \tfrac{\sqrt3}{2}$, $\sin 30° = \tfrac12$:

$\cos 75° = \frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2} - \frac{\sqrt2}{2}\cdot\frac12 = \frac{\sqrt6}{4} - \frac{\sqrt2}{4} = \frac{\sqrt6 - \sqrt2}{4}.$

Numerically that is about $0.2588$, which matches a calculator's $\cos 75°$. This is exactly how the ancient chord tables were extended: known angles combined to reach new ones.

Double-Angle and Half-Angle Formulas

Set $A = B = \theta$ in the sum formulas and something beautifully useful drops out — the double-angle formulas:

$\sin 2\theta = 2\sin\theta\cos\theta$ $\cos 2\theta = \cos^2\theta - \sin^2\theta$

Using $\sin^2\theta + \cos^2\theta = 1$, the cosine version has two equally valid alternate forms:

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

Those two rearrangements are the ones calculus loves, because solving them gives the power-reduction formulas $\cos^2\theta = \tfrac{1 + \cos 2\theta}{2}$ and $\sin^2\theta = \tfrac{1 - \cos 2\theta}{2}$ — the standard trick for integrating $\sin^2$ or $\cos^2$.

Worked Example: Simplifying an integral setup

Rewrite $\sin^2\theta$ so it can be integrated, then integrate $\int \sin^2\theta\, d\theta$.

Using the power-reduction formula,

$\sin^2\theta = \frac{1 - \cos 2\theta}{2}.$

Now the integral is elementary:

$\int \sin^2\theta\, d\theta = \int \frac{1 - \cos 2\theta}{2}\, d\theta = \frac{\theta}{2} - \frac{\sin 2\theta}{4} + C.$

Without the double-angle identity, this integral has no obvious antiderivative; with it, the problem dissolves. This single move appears constantly in physics when computing average power of an oscillating signal.

Proving and Simplifying with Identities

"Prove that the left side equals the right side" is a staple exam task. Unlike solving an equation, you may not move terms across the equals sign as if it were already true — that would assume what you are proving. Instead you transform one side into the other. A reliable strategy:

1. Start with the more complicated side; there is more to simplify.
2. Convert everything to sines and cosines when stuck — it exposes cancellations.
3. Look for a Pythagorean opportunity, especially $1 - \sin^2$ or $\sec^2 - 1$.
4. Combine fractions over a common denominator, or factor.
5. Keep the target side in view so you know what you are aiming at.

Worked Example: A full proof

Prove that $\dfrac{\cos\theta}{1 - \sin\theta} = \sec\theta + \tan\theta$.

The left side looks messier, so start there. Multiply top and bottom by the conjugate $1 + \sin\theta$:

$\frac{\cos\theta}{1 - \sin\theta}\cdot\frac{1 + \sin\theta}{1 + \sin\theta} = \frac{\cos\theta(1 + \sin\theta)}{1 - \sin^2\theta}.$

The denominator is a Pythagorean opportunity: $1 - \sin^2\theta = \cos^2\theta$. So

$\frac{\cos\theta(1 + \sin\theta)}{\cos^2\theta} = \frac{1 + \sin\theta}{\cos\theta} = \frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta} = \sec\theta + \tan\theta.$

That equals the right side, so the identity is proved. Notice how the conjugate trick manufactured a $1 - \sin^2\theta$ precisely so the Pythagorean identity could fire.

Real-World Applications

• Physics of waves and sound. When two sound waves overlap, the product-to-sum identities turn $\cos A\cos B$ into a sum, explaining the "beats" you hear when two nearly-tuned strings play together. Musicians and audio engineers rely on this directly.
• Electrical engineering. Alternating current is a sine wave. Computing average power involves $\sin^2$, which engineers immediately reduce with $\sin^2\theta = \tfrac{1-\cos2\theta}{2}$ to find that average power is half the peak.
• Calculus and integration. Trigonometric substitution and the power-reduction formulas are how integrals containing $\sqrt{1 - x^2}$ or $\sin^2 x$ get solved — the backbone of much of second-semester calculus.
• Computer graphics and robotics. Rotating an object by a combined angle uses angle-sum formulas; a rotation by $A$ then $B$ composes into a rotation by $A+B$ exactly because of $\sin(A+B)$ and $\cos(A+B)$.
• Astronomy and GPS. The spherical-trigonometry descendants of these identities still position satellites and convert between celestial coordinate systems, just as they once tracked planets.

Common Mistakes

Mistake 1: Thinking $\sin(A+B) = \sin A + \sin B$. Trig functions are not linear, so you cannot distribute them over a sum. A quick test kills this instantly: $\sin(90° + 90°) = \sin 180° = 0$, but $\sin 90° + \sin 90° = 2$. The correct rule is $\sin(A+B) = \sin A\cos B + \cos A\sin B$.

Mistake 2: Writing $\sin^2\theta$ as $\sin\theta^2$. The notation $\sin^2\theta$ means $(\sin\theta)^2$ — square the value of the sine. Writing $\sin\theta^2$ suggests $\sin(\theta^2)$, squaring the angle, which is completely different. Keep the exponent right after the function name and apply it to the whole output.

Mistake 3: Treating a "prove" problem like a "solve" problem. When proving an identity you may not perform the same operation on both sides or cross-multiply as if equality is established — that assumes the conclusion. Work one side down to match the other. Only equations (where you solve for $\theta$) allow two-sided manipulation.

Mistake 4: Losing the sign when taking a square root. From $\cos^2\theta = \tfrac{16}{25}$ it does not follow that $\cos\theta = \tfrac{4}{5}$; it could be $-\tfrac{4}{5}$. The Pythagorean identity only gives the magnitude. You must use the quadrant (or given information) to choose the correct sign.

Comparison and Connections

The word "identity" is easily confused with the word "equation." Both use an equals sign, but they demand different things of you.

Feature	Identity	Equation
True for...	every valid angle	only specific angles (solutions)
Your goal	prove both sides always agree	find the angles that make it true
Allowed move	transform one side into the other	operate on both sides equally
Example	$\sin^2\theta + \cos^2\theta = 1$	$\sin\theta = \tfrac12$

Identities also connect tightly to Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$: multiplying $e^{iA}\cdot e^{iB} = e^{i(A+B)}$ and comparing real and imaginary parts reproduces the angle-sum formulas exactly. If you ever forget them, you can regenerate them from this one exponential rule. They are also the trigonometric cousins of algebraic identities like $(a+b)^2 = a^2 + 2ab + b^2$ — permanent truths you exploit rather than solve.

Practice Questions

Recall

State the three Pythagorean identities.

Answer: $\sin^2\theta + \cos^2\theta = 1$, $\;1 + \tan^2\theta = \sec^2\theta$, $\;1 + \cot^2\theta = \csc^2\theta$.

Understanding

Explain why $\cos 2\theta$ can be written in three different forms, and derive $1 - 2\sin^2\theta$ from $\cos^2\theta - \sin^2\theta$.

Answer: Starting from $\cos 2\theta = \cos^2\theta - \sin^2\theta$, replace $\cos^2\theta$ with $1 - \sin^2\theta$: this gives $(1 - \sin^2\theta) - \sin^2\theta = 1 - 2\sin^2\theta$. Replacing $\sin^2\theta$ with $1 - \cos^2\theta$ instead gives $2\cos^2\theta - 1$. All three are equal because they differ only by substitutions of the Pythagorean identity.

Application

Find the exact value of $\sin 15°$.

Answer: $\sin 15° = \sin(45° - 30°) = \sin45°\cos30° - \cos45°\sin30° = \tfrac{\sqrt2}{2}\cdot\tfrac{\sqrt3}{2} - \tfrac{\sqrt2}{2}\cdot\tfrac12 = \tfrac{\sqrt6 - \sqrt2}{4} \approx 0.2588.$

Analysis

Prove that $\dfrac{1 - \cos 2\theta}{\sin 2\theta} = \tan\theta$.

Answer: Use $1 - \cos 2\theta = 2\sin^2\theta$ (from the double-angle form) and $\sin 2\theta = 2\sin\theta\cos\theta$. Then the left side is $\dfrac{2\sin^2\theta}{2\sin\theta\cos\theta} = \dfrac{\sin\theta}{\cos\theta} = \tan\theta$. Proved.

FAQ

Do I really have to memorize all of these? Memorize a small core: the Pythagorean identity, the reciprocal/quotient definitions, and the two angle-sum formulas. Everything else — the other Pythagorean identities, the double-angle formulas, the difference formulas — you can derive on the spot in seconds. Deriving is faster and safer than trusting a shaky memory.

How do I know which identity to use on a problem? Look at what is present and what you want. Seeing $\sin^2$ or $\cos^2$ suggests the Pythagorean identity or power reduction. Seeing a doubled angle like $2\theta$ suggests double-angle formulas. Seeing a sum inside the function, like $\sin(x + 30°)$, points to angle-sum. Fractions of trig functions usually want everything converted to sine and cosine.

Why can't I just move terms across the equals sign when proving an identity? Because doing so assumes the two sides are already equal — the very thing you are asked to demonstrate. A valid proof transforms one side using known identities until it becomes the other, never relying on the conclusion.

Is $\sin^{-1}\theta$ the same as $\csc\theta$? No, and this is a notorious trap. $\csc\theta = 1/\sin\theta$ is the reciprocal, but $\sin^{-1}\theta$ (also written $\arcsin$) is the inverse function that returns an angle. Because of this clash, many people avoid the $\sin^{-1}$ notation for reciprocals entirely.

Where will I actually use these after this course? In calculus you will use power-reduction and double-angle formulas constantly for integration. In physics and engineering you will use angle-sum and product-to-sum formulas for waves, oscillations, and AC circuits. In computer graphics, rotations rely on the angle-sum formulas. They are among the most reused tools in all of quantitative science.

Quick Revision

• Pythagorean: $\sin^2\theta + \cos^2\theta = 1$; divide by $\cos^2$ → $1 + \tan^2\theta = \sec^2\theta$; divide by $\sin^2$ → $1 + \cot^2\theta = \csc^2\theta$.
• Reciprocal: $\csc = 1/\sin$, $\sec = 1/\cos$, $\cot = 1/\tan$. Quotient: $\tan = \sin/\cos$, $\cot = \cos/\sin$.
• Angle-sum: $\sin(A\pm B) = \sin A\cos B \pm \cos A\sin B$; $\cos(A\pm B) = \cos A\cos B \mp \sin A\sin B$.
• Double-angle: $\sin 2\theta = 2\sin\theta\cos\theta$; $\cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$.
• Power reduction: $\sin^2\theta = \tfrac{1 - \cos2\theta}{2}$, $\cos^2\theta = \tfrac{1 + \cos2\theta}{2}$.
• Prove, don't solve: transform the harder side; convert to sine/cosine; hunt for a Pythagorean cancellation.
• $\sin(A+B) \ne \sin A + \sin B$. Watch quadrant signs when taking square roots.

Related Topics

Prerequisites

• The Unit Circle
• Right-Triangle Trigonometry

Related Topics

• Graphs and Periodic Motion
• Trigonometry Overview

Next Topics

• Integrals
• Applications of Calculus