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| Primary source | OpenStax Precalculus 2e, Section 7.1: "Simplifying and Verifying Trigonometric Identities" |
| Direct link | https://openstax.org/books/precalculus-2e/pages/7-1-simplifying-and-verifying-trigonometric-identities |
| Sum/difference | OpenStax Precalculus 2e, Section 7.2: "Sum and Difference Identities" |
| Sum/difference link | https://openstax.org/books/precalculus-2e/pages/7-2-sum-and-difference-identities |
| Double/half angle | OpenStax Precalculus 2e, Section 7.3: "Double-Angle, Half-Angle, and Reduction Formulas" |
| Double/half angle link | https://openstax.org/books/precalculus-2e/pages/7-3-double-angle-half-angle-and-reduction-formulas |
Precalculus 2e Chapter 7 begins on page 591 of the PDF.
Every trigonometric identity derives from two facts.
Fact 1 (Pythagorean identity). For any angle $\theta$, the point $(\cos\theta, \sin\theta)$ lies on the unit circle $x^2 + y^2 = 1$. Therefore: \[ \cos^2\theta + \sin^2\theta = 1 \]
Fact 2 (angle addition formula for cosine). This can be derived from the distance formula on the unit circle: \[ \cos(A - B) = \cos A \cos B + \sin A \sin B \]
From Fact 1 alone: the other two Pythagorean identities, the reciprocal and quotient identities, and the even/odd symmetry properties.
From Fact 2 alone: the sine and cosine sum/difference formulas (by substitution and algebra), the double angle formulas (by setting $A = B$), the half-angle formulas (by solving the double angle formula for $\sin^2$ and $\cos^2$), and the power-reduction formulas (rearrangement).
If you forget an identity during a problem, you can derive it from these two starting points. The identities are not a list to memorize; they are a family you can rebuild.
Pythagorean identities: \[ \sin^2\theta + \cos^2\theta = 1 \] \[ 1 + \tan^2\theta = \sec^2\theta \qquad 1 + \cot^2\theta = \csc^2\theta \]
Reciprocal identities: \[ \csc\theta = \frac{1}{\sin\theta} \qquad \sec\theta = \frac{1}{\cos\theta} \qquad \cot\theta = \frac{1}{\tan\theta} \]
Quotient identities: \[ \tan\theta = \frac{\sin\theta}{\cos\theta} \qquad \cot\theta = \frac{\cos\theta}{\sin\theta} \]
Even/odd: \[ \cos(-\theta) = \cos\theta \quad \text{(even)} \qquad \sin(-\theta) = -\sin\theta \quad \text{(odd)} \]
\[ \sin(A \pm B) = \sin A \cos B \pm \cos A \sin B \] \[ \cos(A \pm B) = \cos A \cos B \mp \sin A \sin B \] \[ \tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B} \]
\[ \sin 2\theta = 2\sin\theta\cos\theta \] \[ \cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta \] \[ \tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta} \]
\[ \sin^2\theta = \frac{1 - \cos 2\theta}{2} \qquad \cos^2\theta = \frac{1 + \cos 2\theta}{2} \] \[ \sin\frac{\theta}{2} = \pm\sqrt{\frac{1 - \cos\theta}{2}} \qquad \cos\frac{\theta}{2} = \pm\sqrt{\frac{1 + \cos\theta}{2}} \]
(Sign determined by the quadrant of $\theta/2$.)
The unit circle is the circle of radius 1 centered at the origin: $x^2 + y^2 = 1$.
For any angle $\theta$ (measured counterclockwise from the positive $x$-axis), the point on the unit circle at angle $\theta$ is defined to be $(\cos\theta, \sin\theta)$.
Since this point lies on the unit circle: \[ \cos^2\theta + \sin^2\theta = 1 \]
This is the master identity. Everything else follows from it.
Deriving the other two Pythagorean identities.
Divide $\cos^2\theta + \sin^2\theta = 1$ by $\cos^2\theta$ (assuming $\cos\theta \neq 0$): \[ 1 + \tan^2\theta = \sec^2\theta \]
Divide by $\sin^2\theta$ (assuming $\sin\theta \neq 0$): \[ \cot^2\theta + 1 = \csc^2\theta \]
The three Pythagorean identities are not three separate facts. They are the same fact written three different ways.
These follow directly from definitions. They are NOT identities to derive; they are definitions expressed as equations.
| Definition | Identity form |
|---|---|
| $\csc\theta = 1/\sin\theta$ | $\sin\theta \cdot \csc\theta = 1$ |
| $\sec\theta = 1/\cos\theta$ | $\cos\theta \cdot \sec\theta = 1$ |
| $\cot\theta = 1/\tan\theta$ | $\tan\theta \cdot \cot\theta = 1$ |
| $\tan\theta = \sin\theta/\cos\theta$ | $\tan\theta\cos\theta = \sin\theta$ |
Example 1. Simplify $\dfrac{\sin\theta}{\cos\theta} + \cos\theta\sec\theta$.
\[ \tan\theta + \cos\theta \cdot \frac{1}{\cos\theta} = \tan\theta + 1 \]
The unit circle has bilateral symmetry. For any angle $\theta$:
\[ \cos(-\theta) = \cos\theta \qquad \sin(-\theta) = -\sin\theta \]
Consequences:
Even/odd is why you see $\cos^2$ appearing frequently in integrals (cosine is even, so it pairs symmetrically), while $\sin^2$ and $\cos^2$ are related by the Pythagorean identity.
The derivation of $\cos(A - B) = \cos A \cos B + \sin A \sin B$ comes from the distance formula applied to two points on the unit circle at angles $A$ and $B$. The full derivation is in OpenStax Precalculus 2e Section 7.2. Here we state the formulas and show how to use them.
Applying the sum formulas to find exact values.
Example 2. Find $\cos 75°$ exactly.
Write $75° = 45° + 30°$: \[ \cos 75° = \cos(45° + 30°) = \cos 45°\cos 30° - \sin 45°\sin 30° \] \[ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4} \]
Example 3. Find $\sin(A + B)$ given $\sin A = 3/5$ (with $A$ in Quadrant I) and $\cos B = 5/13$ (with $B$ in Quadrant I).
First find the missing trig values.
From $\sin A = 3/5$ and $\sin^2 A + \cos^2 A = 1$: $\cos A = 4/5$ (positive, Q1).
From $\cos B = 5/13$: $\sin B = 12/13$ (positive, Q1).
\[ \sin(A + B) = \sin A \cos B + \cos A \sin B = \frac{3}{5} \cdot \frac{5}{13} + \frac{4}{5} \cdot \frac{12}{13} = \frac{15}{65} + \frac{48}{65} = \frac{63}{65} \]
Deriving identities from the sum formulas.
Set $B = A$ in $\sin(A + B)$: \[ \sin(A + A) = \sin A \cos A + \cos A \sin A = 2\sin A\cos A \]
That is the double angle formula for sine. The double angle formula for cosine comes from $\cos(A + A)$.
Derivation (for the student watching): Set $A = B = \theta$ in the sum formulas.
$\sin(\theta + \theta) = \sin\theta\cos\theta + \cos\theta\sin\theta = 2\sin\theta\cos\theta$
$\cos(\theta + \theta) = \cos\theta\cos\theta - \sin\theta\sin\theta = \cos^2\theta - \sin^2\theta$
The two additional forms of $\cos 2\theta$ come from substituting $\cos^2\theta = 1 - \sin^2\theta$ or $\sin^2\theta = 1 - \cos^2\theta$:
\[ \cos 2\theta = \cos^2\theta - \sin^2\theta = (1 - \sin^2\theta) - \sin^2\theta = 1 - 2\sin^2\theta \] \[ \cos 2\theta = \cos^2\theta - (1 - \cos^2\theta) = 2\cos^2\theta - 1 \]
Example 4. If $\sin\theta = -\frac{3}{5}$ and $\theta$ is in Quadrant III, find $\sin 2\theta$ and $\cos 2\theta$.
In Q3, $\cos\theta < 0$. From Pythagorean: $\cos\theta = -\frac{4}{5}$.
$\sin 2\theta = 2(-\frac{3}{5})(-\frac{4}{5}) = \frac{24}{25}$.
$\cos 2\theta = \cos^2\theta - \sin^2\theta = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}$.
Derivation. Solve $\cos 2\theta = 1 - 2\sin^2\theta$ for $\sin^2\theta$: \[ \sin^2\theta = \frac{1 - \cos 2\theta}{2} \]
Solve $\cos 2\theta = 2\cos^2\theta - 1$ for $\cos^2\theta$: \[ \cos^2\theta = \frac{1 + \cos 2\theta}{2} \]
These are the power reduction formulas: they convert $\sin^2$ and $\cos^2$ into expressions linear in $\cos 2\theta$. This is essential for integrating even powers of sine and cosine.
For the half angle formulas, substitute $\theta = \alpha/2$ (so $2\theta = \alpha$): \[ \sin\frac{\alpha}{2} = \pm\sqrt{\frac{1 - \cos\alpha}{2}} \qquad \cos\frac{\alpha}{2} = \pm\sqrt{\frac{1 + \cos\alpha}{2}} \]
The sign is $+$ or $-$ depending on which quadrant $\alpha/2$ lies in.
Example 5. Find $\cos(22.5°)$ exactly.
$22.5° = 45°/2$. Use the half-angle formula for cosine with $\alpha = 45°$: \[ \cos 22.5° = +\sqrt{\frac{1 + \cos 45°}{2}} = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} = \sqrt{\frac{2 + \sqrt{2}}{4}} = \frac{\sqrt{2 + \sqrt{2}}}{2} \]
(Positive because $22.5°$ is in Q1.)
To verify a trigonometric identity, start with one side and transform it into the other using known identities and algebra. Do NOT move terms from one side to the other (that would assume what you are trying to prove).
Strategy.
Example 6. Verify $\tan\theta + \cot\theta = \sec\theta\csc\theta$.
Start with the left side: \[ \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} = \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta} \]
Right side: $\sec\theta\csc\theta = \dfrac{1}{\cos\theta} \cdot \dfrac{1}{\sin\theta} = \dfrac{1}{\sin\theta\cos\theta}$.
Both sides equal $\dfrac{1}{\sin\theta\cos\theta}$. Identity verified.
Example 7. Verify $\dfrac{1 - \cos\theta}{\sin\theta} = \dfrac{\sin\theta}{1 + \cos\theta}$.
Cross-multiply to see both sides are equivalent (multiply left side's denominator and right side's numerator): $(1 - \cos\theta)(1 + \cos\theta) = 1 - \cos^2\theta = \sin^2\theta = \sin\theta \cdot \sin\theta$.
But the proper verification method works from one side. Start with the left:
\[ \frac{1 - \cos\theta}{\sin\theta} \cdot \frac{1 + \cos\theta}{1 + \cos\theta} = \frac{1 - \cos^2\theta}{\sin\theta(1+\cos\theta)} = \frac{\sin^2\theta}{\sin\theta(1+\cos\theta)} = \frac{\sin\theta}{1 + \cos\theta} \]
This equals the right side. Identity verified.
You may not assume the identity while proving it. Do not add the same expression to both sides, cross-multiply, or perform any other step that assumes the equality holds. Each step must follow from a known identity or algebraic rule.
| Error | Example | Correction |
|---|---|---|
| $\sin(A + B) = \sin A + \sin B$ | "distributing" the sine | The sum formula has cross terms: $\sin A \cos B + \cos A \sin B$ |
| $\cos 2\theta = 2\cos\theta$ | "distributing" the 2 | $\cos 2\theta = \cos^2\theta - \sin^2\theta$; there is no such simplification |
| Wrong sign in $\cos(A - B)$ | $\cos A \cos B - \sin A \sin B$ | Correct for $\cos(A + B)$; for $\cos(A - B)$: $\cos A \cos B + \sin A \sin B$ |
| Forgetting to find missing trig value | Using $\sin A$ and $\sin B$ in $\sin(A+B)$ without finding $\cos A$, $\cos B$ | The sum formulas need all four values: $\sin A$, $\cos A$, $\sin B$, $\cos B$ |
| Moving terms across the equal sign in a verification | Rewriting $\tan + \cot = \sec\csc$ by subtracting $\cot$ from both sides | Verification requires transforming one side into the other; no cross-side moves |
Problem 1. Given $\cos\theta = -\frac{5}{13}$ with $\theta$ in Quadrant II, find $\sin\theta$, $\tan\theta$, $\csc\theta$, and $\sec\theta$.
Pythagorean: $\sin^2\theta = 1 - \frac{25}{169} = \frac{144}{169} \implies \sin\theta = \frac{12}{13}$ (positive in Q2).
$\tan\theta = \frac{12/13}{-5/13} = -\frac{12}{5}$.
$\csc\theta = \frac{13}{12}$, $\sec\theta = -\frac{13}{5}$.
Problem 2. Find $\sin 15°$ exactly using a sum or difference formula.
$\sin 15° = \sin(45° - 30°) = \sin 45°\cos 30° - \cos 45°\sin 30°$
$= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} - \sqrt{2}}{4}$
Problem 3. Use the power reduction formula to write $\sin^2(3x)$ in terms of $\cos(6x)$.
$\sin^2(3x) = \dfrac{1 - \cos(2 \cdot 3x)}{2} = \dfrac{1 - \cos 6x}{2}$
Problem 4. Verify: $\sec^2\theta - \tan^2\theta = 1$.
From the Pythagorean identity $1 + \tan^2\theta = \sec^2\theta$:
$\sec^2\theta - \tan^2\theta = 1$. $\checkmark$
This is the same identity rearranged; verification is immediate.
Problem 5. Verify: $\dfrac{\sin^2\theta}{1 - \cos\theta} = 1 + \cos\theta$.
Left side: substitute $\sin^2\theta = 1 - \cos^2\theta = (1 - \cos\theta)(1 + \cos\theta)$:
$\dfrac{(1 - \cos\theta)(1 + \cos\theta)}{1 - \cos\theta} = 1 + \cos\theta$ (for $\cos\theta \neq 1$). $\checkmark$
Problem 6. Verify: $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$.
$\cos 3\theta = \cos(2\theta + \theta) = \cos 2\theta \cos\theta - \sin 2\theta \sin\theta$
$= (2\cos^2\theta - 1)\cos\theta - (2\sin\theta\cos\theta)\sin\theta$
$= 2\cos^3\theta - \cos\theta - 2\sin^2\theta\cos\theta$
Substitute $\sin^2\theta = 1 - \cos^2\theta$:
$= 2\cos^3\theta - \cos\theta - 2(1 - \cos^2\theta)\cos\theta$
$= 2\cos^3\theta - \cos\theta - 2\cos\theta + 2\cos^3\theta$
$= 4\cos^3\theta - 3\cos\theta$. $\checkmark$
Problem 7. Write $\sin^4 x$ in terms of $\cos 2x$ and $\cos 4x$ using power reduction. (This form appears in integration.)
$\sin^4 x = (\sin^2 x)^2 = \left(\frac{1 - \cos 2x}{2}\right)^2 = \frac{1 - 2\cos 2x + \cos^2 2x}{4}$
Apply power reduction to $\cos^2 2x = \dfrac{1 + \cos 4x}{2}$:
$= \frac{1 - 2\cos 2x + \frac{1 + \cos 4x}{2}}{4} = \frac{\frac{3}{2} - 2\cos 2x + \frac{\cos 4x}{2}}{4} = \frac{3}{8} - \frac{\cos 2x}{2} + \frac{\cos 4x}{8}$
This form integrates immediately: $\displaystyle\int \sin^4 x \, dx = \frac{3x}{8} - \frac{\sin 2x}{4} + \frac{\sin 4x}{32} + C$.
Problem 8. Use the identity $1 + \tan^2\theta = \sec^2\theta$ to simplify $\sqrt{1 + \tan^2\theta}$.
$\sqrt{1 + \tan^2\theta} = \sqrt{\sec^2\theta} = |\sec\theta|$.
If $\theta$ is in Q1 or Q4 (where $\sec\theta > 0$): $|\sec\theta| = \sec\theta$.
In trigonometric substitution (MATH162), you set $x = a\tan\theta$ to handle integrals involving $\sqrt{a^2 + x^2}$. The substitution converts: $\sqrt{a^2 + a^2\tan^2\theta} = a\sqrt{1 + \tan^2\theta} = a|\sec\theta|$. This Pythagorean identity is the key step.
Problem 9. Show that $\displaystyle\int_0^{\pi/2} \sin^2 x \, dx = \frac{\pi}{4}$ using the power reduction formula.
$\sin^2 x = \dfrac{1 - \cos 2x}{2}$.
$\displaystyle\int_0^{\pi/2} \sin^2 x \, dx = \int_0^{\pi/2} \frac{1 - \cos 2x}{2}\,dx = \left[\frac{x}{2} - \frac{\sin 2x}{4}\right]_0^{\pi/2}$
$= \left(\frac{\pi/2}{2} - \frac{\sin\pi}{4}\right) - \left(0 - 0\right) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$. $\square$
This computation appears in MATH162 when integrating even powers of sine. Without the power reduction formula, the integral is not straightforward.
All of trigonometry lives on the unit circle.
The unit circle is the set of points $(x, y)$ with $x^2 + y^2 = 1$. Every angle $\theta$ labels a point: the point you reach after traveling arc length $\theta$ counterclockwise from $(1, 0)$. That point is $(\cos\theta, \sin\theta)$.
The Pythagorean identity $\cos^2\theta + \sin^2\theta = 1$ is just $x^2 + y^2 = 1$ for that point. It is always true; it is the equation of the circle.
The sum formula for cosine comes from measuring the distance between two points on the unit circle in two ways: once using the distance formula for points, once using the angle between them. Both calculations give the same answer, and that equality is $\cos(A - B) = \cos A \cos B + \sin A \sin B$.
Everything else -- double angles, half angles, power reduction -- is algebra done to those two starting points. When you learn a new identity, ask: "Which of the two starting facts does this come from, and which algebraic step takes me there?" That question converts a list of memorizations into a network of derived facts.
For students who find math intimidating: The most important single identity is $\sin^2\theta + \cos^2\theta = 1$. If you know this one and know how to divide it by $\cos^2\theta$ and $\sin^2\theta$, you have the three Pythagorean identities. Everything else can be looked up during the course. Focus first on recognizing when a Pythagorean substitution helps (e.g., replace $1 - \sin^2\theta$ with $\cos^2\theta$) -- that skill appears in almost every problem.
For students interested in proof: The cosine subtraction formula $\cos(A - B) = \cos A \cos B + \sin A \sin B$ can be proved rigorously using the distance formula. The distance between the unit circle points at angles $A$ and $B$ can be computed two ways: as $\sqrt{(\cos A - \cos B)^2 + (\sin A - \sin B)^2}$ (coordinate distance), and also as the distance between the points at angle $A - B$ and angle $0 = (1, 0)$. Setting these equal and simplifying yields the formula. This is a short and elegant proof accessible with only coordinate geometry.
For students interested in careers: Trigonometric identities are the foundation of Fourier analysis: decomposing signals into sums of sines and cosines. The power reduction formulas explain how different frequency components interact when multiplied. In signal processing, audio compression, and medical imaging (MRI), Fourier analysis is the underlying mathematical structure.