Trigonometry

Trigonometric Identities

Last updated: March 2026 · Advanced
Before you start

You should be comfortable with:

Real-world applications
📐
Carpentry

Measurements, material estimation, cutting calculations

⚡
Electrical

Voltage drop, wire sizing, load balancing

A trigonometric identity is an equation that is true for all values of the variable where both sides are defined. Unlike regular equations (which are true only for specific values), identities are always true. They are the tools you use to simplify expressions, verify equations, and solve advanced trig problems.

The Pythagorean Identity

The most important trig identity comes directly from the Pythagorean theorem applied to the unit circle. For any point (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta) on the unit circle, x2+y2=1x^2 + y^2 = 1:

sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1

Why sin²θ + cos²θ = 1 — The Unit Circle Connection

θcos θsin θ1(cos θ, sin θ)cos²θ + sin²θ = 1² → the Pythagorean theorem on this triangle

This identity is used constantly. Two rearranged forms are equally useful:

sin⁡2θ=1−cos⁡2θ\sin^2\theta = 1 - \cos^2\theta

cos⁡2θ=1−sin⁡2θ\cos^2\theta = 1 - \sin^2\theta

Derived Pythagorean Identities

Dividing the Pythagorean identity by cos⁡2θ\cos^2\theta:

sin⁡2θcos⁡2θ+cos⁡2θcos⁡2θ=1cos⁡2θ\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta}

tan⁡2θ+1=sec⁡2θ\tan^2\theta + 1 = \sec^2\theta

Dividing by sin⁡2θ\sin^2\theta:

1+cot⁡2θ=csc⁡2θ1 + \cot^2\theta = \csc^2\theta

Reciprocal Identities

Three additional trig functions are defined as the reciprocals of sine, cosine, and tangent:

csc⁡θ=1sin⁡θsec⁡θ=1cos⁡θcot⁡θ=1tan⁡θ\csc\theta = \frac{1}{\sin\theta} \qquad \sec\theta = \frac{1}{\cos\theta} \qquad \cot\theta = \frac{1}{\tan\theta}

These are read as cosecant, secant, and cotangent. They appear less frequently than sin, cos, and tan, but they show up in calculus, physics, and advanced test questions.

Memory aid: each reciprocal pairs with the function that does not share its first letter:

  • sin and csc (not sec)
  • cos and sec (not csc)
  • tan and cot

Quotient Identities

The tangent and cotangent can be expressed as quotients of sine and cosine:

tan⁡θ=sin⁡θcos⁡θcot⁡θ=cos⁡θsin⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta} \qquad \cot\theta = \frac{\cos\theta}{\sin\theta}

Complete Identity Reference Table

CategoryIdentity
Pythagoreansin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1
tan⁡2θ+1=sec⁡2θ\tan^2\theta + 1 = \sec^2\theta
1+cot⁡2θ=csc⁡2θ1 + \cot^2\theta = \csc^2\theta
Reciprocalcsc⁡θ=1sin⁡θ\csc\theta = \dfrac{1}{\sin\theta}
sec⁡θ=1cos⁡θ\sec\theta = \dfrac{1}{\cos\theta}
cot⁡θ=1tan⁡θ\cot\theta = \dfrac{1}{\tan\theta}
Quotienttan⁡θ=sin⁡θcos⁡θ\tan\theta = \dfrac{\sin\theta}{\cos\theta}
cot⁡θ=cos⁡θsin⁡θ\cot\theta = \dfrac{\cos\theta}{\sin\theta}
Co-functionsin⁡θ=cos⁡(90°−θ)\sin\theta = \cos(90° - \theta)
cos⁡θ=sin⁡(90°−θ)\cos\theta = \sin(90° - \theta)
tan⁡θ=cot⁡(90°−θ)\tan\theta = \cot(90° - \theta)
Even/Oddsin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin\theta (odd)
cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos\theta (even)
tan⁡(−θ)=−tan⁡θ\tan(-\theta) = -\tan\theta (odd)

Using Identities to Simplify Expressions

The general strategy for simplifying trig expressions:

  1. Convert everything to sin and cos. This is almost always a good first step.
  2. Look for Pythagorean identity patterns. If you see sin⁡2θ+cos⁡2θ\sin^2\theta + \cos^2\theta, replace it with 1.
  3. Factor when possible. Difference of squares, common factors, etc.
  4. Simplify fractions. Combine terms, cancel common factors.

Worked Examples

Example 1: Simplify sin⁡2θ+cos⁡2θ+tan⁡2θ\sin^2\theta + \cos^2\theta + \tan^2\theta

Start with the Pythagorean identity: sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1.

1+tan⁡2θ1 + \tan^2\theta

By the derived Pythagorean identity, 1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta.

Answer: sec⁡2θ\sec^2\theta

Example 2: Simplify sin⁡θcsc⁡θ+cos⁡θsec⁡θ\frac{\sin\theta}{\csc\theta} + \frac{\cos\theta}{\sec\theta}

Convert using reciprocal identities:

sin⁡θ1/sin⁡θ+cos⁡θ1/cos⁡θ=sin⁡2θ+cos⁡2θ=1\frac{\sin\theta}{1/\sin\theta} + \frac{\cos\theta}{1/\cos\theta} = \sin^2\theta + \cos^2\theta = 1

Answer: 11

Example 3: Verify that 1−cos⁡2θsin⁡θ=sin⁡θ\frac{1 - \cos^2\theta}{\sin\theta} = \sin\theta

The numerator 1−cos⁡2θ=sin⁡2θ1 - \cos^2\theta = \sin^2\theta (Pythagorean identity rearranged).

sin⁡2θsin⁡θ=sin⁡θ\frac{\sin^2\theta}{\sin\theta} = \sin\theta

Answer: Verified. Both sides equal sin⁡θ\sin\theta.

Example 4: Simplify cot⁡θ⋅sin⁡θ\cot\theta \cdot \sin\theta

Convert cotangent using the quotient identity:

cot⁡θ⋅sin⁡θ=cos⁡θsin⁡θ⋅sin⁡θ=cos⁡θ\cot\theta \cdot \sin\theta = \frac{\cos\theta}{\sin\theta} \cdot \sin\theta = \cos\theta

The sin⁡θ\sin\theta terms cancel.

Answer: cos⁡θ\cos\theta

Example 5: Simplify sec⁡2θ−1sec⁡2θ\frac{\sec^2\theta - 1}{\sec^2\theta}

Split the fraction:

sec⁡2θ−1sec⁡2θ=1−1sec⁡2θ=1−cos⁡2θ\frac{\sec^2\theta - 1}{\sec^2\theta} = 1 - \frac{1}{\sec^2\theta} = 1 - \cos^2\theta

By the Pythagorean identity, 1−cos⁡2θ=sin⁡2θ1 - \cos^2\theta = \sin^2\theta.

Answer: sin⁡2θ\sin^2\theta

Example 6: Given cos⁡θ=35\cos\theta = \frac{3}{5} in Quadrant IV, find sin⁡θ\sin\theta and tan⁡θ\tan\theta

Using the Pythagorean identity:

sin⁡2θ=1−cos⁡2θ=1−925=1625\sin^2\theta = 1 - \cos^2\theta = 1 - \frac{9}{25} = \frac{16}{25}

sin⁡θ=±45\sin\theta = \pm\frac{4}{5}

In Quadrant IV, sine is negative, so sin⁡θ=−45\sin\theta = -\frac{4}{5}.

tan⁡θ=sin⁡θcos⁡θ=−4/53/5=−43\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{-4/5}{3/5} = -\frac{4}{3}

Answer: sin⁡θ=−45\sin\theta = -\frac{4}{5} and tan⁡θ=−43\tan\theta = -\frac{4}{3}.

Why Identities Matter

Trig identities are not just academic exercises. They are essential for:

  • Simplifying complex expressions before solving equations — a messy equation with mixed trig functions often reduces to a simple one after applying identities
  • Calculus — integration techniques like substitution and partial fractions rely heavily on trig identities
  • Physics and engineering — wave interference, AC power calculations, and signal processing all use identities to combine or decompose sinusoidal functions
  • Proving new results — more advanced identities (sum and difference formulas, double-angle formulas) are all derived from the basic identities on this page

The identities covered here — Pythagorean, reciprocal, and quotient — form the foundation. Once you are fluent with these, the more advanced identity families build on them naturally.

Real-World Connection: AC Circuits

In electrical work, alternating current is described by sinusoidal functions. The power in an AC circuit involves the identity:

sin⁡2(ωt)=1−cos⁡(2ωt)2\sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2}

This identity (derived from the Pythagorean identity and double-angle formulas) is how electricians and engineers calculate RMS (root mean square) power — the effective power delivered by AC current. Without this identity, computing average power over a full cycle would require advanced calculus.

The result: the RMS voltage of standard 170V-peak household current is:

Vrms=1702≈120 VV_{\text{rms}} = \frac{170}{\sqrt{2}} \approx 120 \text{ V}

This is why household outlets are labeled 120V even though the peak voltage is 170V.

Common Mistakes

  1. Confusing sin⁡2θ\sin^2\theta with sin⁡(θ2)\sin(\theta^2). The notation sin⁡2θ\sin^2\theta means (sin⁡θ)2(\sin\theta)^2 — you square the result of the sine function, not the angle.
  2. Forgetting that sin⁡−1\sin^{-1} is not csc⁡\csc. sin⁡−1\sin^{-1} is the inverse sine (arcsin). csc⁡θ=1sin⁡θ\csc\theta = \frac{1}{\sin\theta} is the reciprocal.
  3. Trying to add fractions without a common denominator. When simplifying, always find a common denominator before combining fractions.
  4. Not converting to sin and cos first. Most simplification problems become straightforward once everything is in terms of sine and cosine.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Problem 1: Simplify sec⁡2θ−tan⁡2θ\sec^2\theta - \tan^2\theta.

By the Pythagorean identity: tan⁡2θ+1=sec⁡2θ\tan^2\theta + 1 = \sec^2\theta, so sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1.

Answer: 11

Problem 2: Simplify sin⁡θ⋅cos⁡θcos⁡θ\frac{\sin\theta \cdot \cos\theta}{\cos\theta}.

Cancel the common factor of cos⁡θ\cos\theta:

sin⁡θ⋅cos⁡θcos⁡θ=sin⁡θ\frac{\sin\theta \cdot \cos\theta}{\cos\theta} = \sin\theta

Answer: sin⁡θ\sin\theta

Problem 3: Verify that tan⁡θ⋅cos⁡θ=sin⁡θ\tan\theta \cdot \cos\theta = \sin\theta.

Convert tangent to sin⁡θcos⁡θ\frac{\sin\theta}{\cos\theta}:

sin⁡θcos⁡θ⋅cos⁡θ=sin⁡θ\frac{\sin\theta}{\cos\theta} \cdot \cos\theta = \sin\theta

The cos⁡θ\cos\theta cancels.

Answer: Verified. tan⁡θ⋅cos⁡θ=sin⁡θ\tan\theta \cdot \cos\theta = \sin\theta.

Problem 4: If sin⁡θ=0.8\sin\theta = 0.8 and θ\theta is in Quadrant I, find cos⁡θ\cos\theta and tan⁡θ\tan\theta.

Using the Pythagorean identity:

cos⁡2θ=1−sin⁡2θ=1−0.64=0.36\cos^2\theta = 1 - \sin^2\theta = 1 - 0.64 = 0.36

cos⁡θ=0.6(positive in Q I)\cos\theta = 0.6 \quad (\text{positive in Q I})

tan⁡θ=sin⁡θcos⁡θ=0.80.6=43≈1.333\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{0.8}{0.6} = \frac{4}{3} \approx 1.333

Answer: cos⁡θ=0.6\cos\theta = 0.6 and tan⁡θ=43\tan\theta = \frac{4}{3}.

Problem 5: Simplify 1sin⁡2θ−cos⁡2θsin⁡2θ\frac{1}{\sin^2\theta} - \frac{\cos^2\theta}{\sin^2\theta}.

Combine over the common denominator sin⁡2θ\sin^2\theta:

1−cos⁡2θsin⁡2θ=sin⁡2θsin⁡2θ=1\frac{1 - \cos^2\theta}{\sin^2\theta} = \frac{\sin^2\theta}{\sin^2\theta} = 1

Answer: 11

Key Takeaways

  • The Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 is the most-used identity in all of trigonometry
  • Reciprocal identities define csc, sec, and cot as 1/sin⁡1/\sin, 1/cos⁡1/\cos, and 1/tan⁡1/\tan
  • Quotient identities express tangent and cotangent as ratios of sine and cosine
  • To simplify trig expressions: convert to sin and cos first, then look for Pythagorean identity patterns
  • These identities are not arbitrary — they follow from the geometry of the unit circle and the Pythagorean theorem

Return to Trigonometry for more topics in this section.

Last updated: March 28, 2026