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11.4 Trigonometric Identities and Proofs

Updated March 2026Fiveable Content Team
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🍬Honors Algebra II Unit 11 Review

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11.4 Trigonometric Identities and Proofs

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Trigonometric Identities and Proofs

Trigonometric identities are equations that hold true for every value of the variable where both sides are defined. They let you rewrite trig expressions in equivalent forms, which is essential for simplifying complicated expressions, solving trig equations, and constructing proofs. This section covers the core identities you need to know, how to use them to simplify expressions, how to verify (prove) identities, and how to apply them when solving equations.

Trigonometric Identities

Fundamental Trigonometric Identities

Pythagorean Identities

The most important identity in trigonometry comes straight from the unit circle and the Pythagorean theorem:

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

From this single identity, you can derive two others by dividing through:

  • Divide everything by cos⁡2θ\cos^2\theta: tan⁡2θ+1=sec⁡2θ\tan^2\theta + 1 = \sec^2\theta
  • Divide everything by sin⁡2θ\sin^2\theta: 1+cot⁡2θ=csc⁡2θ1 + \cot^2\theta = \csc^2\theta

These are used constantly. If you know sin⁡θ\sin\theta, you can find cos⁡θ\cos\theta (up to a sign determined by the quadrant), and vice versa.

Reciprocal Identities

These define the three "secondary" trig functions:

  • sec⁡θ=1cos⁡θ\sec\theta = \frac{1}{\cos\theta}
  • csc⁡θ=1sin⁡θ\csc\theta = \frac{1}{\sin\theta}
  • cot⁡θ=1tan⁡θ\cot\theta = \frac{1}{\tan\theta}

Quotient Identities

These connect tangent and cotangent back to sine and cosine:

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

Converting everything to sine and cosine using the reciprocal and quotient identities is one of the most reliable strategies for simplifying expressions and proving identities.

Identities Involving Angle Relationships

Co-function Identities

These relate trig functions of complementary angles (angles that add to 90°90°). The idea is that the sine of an angle equals the cosine of its complement, and so on:

  • sin⁡(90°−θ)=cos⁡θ\sin(90° - \theta) = \cos\theta and cos⁡(90°−θ)=sin⁡θ\cos(90° - \theta) = \sin\theta
  • tan⁡(90°−θ)=cot⁡θ\tan(90° - \theta) = \cot\theta and cot⁡(90°−θ)=tan⁡θ\cot(90° - \theta) = \tan\theta
  • sec⁡(90°−θ)=csc⁡θ\sec(90° - \theta) = \csc\theta and csc⁡(90°−θ)=sec⁡θ\csc(90° - \theta) = \sec\theta

Odd-Even (Negative Angle) Identities

These tell you what happens when you negate the angle. Cosine and secant are even functions (the negative sign has no effect), while the other four are odd (the negative sign pulls out front):

  • Even: cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos\theta, sec⁡(−θ)=sec⁡θ\sec(-\theta) = \sec\theta
  • Odd: sin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin\theta, tan⁡(−θ)=−tan⁡θ\tan(-\theta) = -\tan\theta, csc⁡(−θ)=−csc⁡θ\csc(-\theta) = -\csc\theta, cot⁡(−θ)=−cot⁡θ\cot(-\theta) = -\cot\theta

A quick way to remember: cosine's graph is symmetric about the y-axis (even), while sine's graph has rotational symmetry about the origin (odd).

Sum and Difference Identities

These let you find exact values for angles like 75°75° or 15°15° by breaking them into angles you already know (like 45°+30°45° + 30°):

  • sin⁡(α±β)=sin⁡αcos⁡β±cos⁡αsin⁡β\sin(\alpha \pm \beta) = \sin\alpha\cos\beta \pm \cos\alpha\sin\beta
  • cos⁡(α±β)=cos⁡αcos⁡β∓sin⁡αsin⁡β\cos(\alpha \pm \beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta
  • tan⁡(α±β)=tan⁡α±tan⁡β1∓tan⁡αtan⁡β\tan(\alpha \pm \beta) = \frac{\tan\alpha \pm \tan\beta}{1 \mp \tan\alpha\tan\beta}

Notice the sign pattern: for sine, the ±\pm matches; for cosine, the signs flip (the ∓\mp). This is a common place to make errors, so pay close attention.

Simplifying Trigonometric Expressions

Fundamental Trigonometric Identities, TABLE OF TRIGONOMETRIC IDENTITIES - Engineering Mathematics 1 DBM10013 Politeknik

Applying Fundamental Identities

The general strategy for simplifying a trig expression:

  1. Convert to sine and cosine. Replace sec⁡\sec, csc⁡\csc, tan⁡\tan, and cot⁡\cot using the reciprocal and quotient identities.
  2. Combine fractions if there are multiple terms with different denominators.
  3. Look for Pythagorean identity substitutions. Whenever you see sin⁡2θ+cos⁡2θ\sin^2\theta + \cos^2\theta, replace it with 11. Similarly, tan⁡2θ+1=sec⁡2θ\tan^2\theta + 1 = \sec^2\theta and 1+cot⁡2θ=csc⁡2θ1 + \cot^2\theta = \csc^2\theta.
  4. Factor where possible, then cancel common factors.
  5. Simplify until you reach the target expression or the simplest form.

Example: Simplify sec⁡θ−sec⁡θsin⁡2θ\sec\theta - \sec\theta\sin^2\theta.

  1. Factor out sec⁡θ\sec\theta: sec⁡θ(1−sin⁡2θ)\sec\theta(1 - \sin^2\theta)

  2. Apply the Pythagorean identity (1−sin⁡2θ=cos⁡2θ1 - \sin^2\theta = \cos^2\theta): sec⁡θ⋅cos⁡2θ\sec\theta \cdot \cos^2\theta

  3. Replace sec⁡θ\sec\theta with 1cos⁡θ\frac{1}{\cos\theta}: cos⁡2θcos⁡θ=cos⁡θ\frac{\cos^2\theta}{\cos\theta} = \cos\theta

Simplifying Expressions with Double and Half Angles

Double-Angle Identities

These are derived from the sum identities by setting α=β=θ\alpha = \beta = \theta:

  • sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin\theta\cos\theta
  • cos⁡2θ=cos⁡2θ−sin⁡2θ=2cos⁡2θ−1=1−2sin⁡2θ\cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta
  • tan⁡2θ=2tan⁡θ1−tan⁡2θ\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}

The three forms of cos⁡2θ\cos 2\theta are all equivalent. Which one you use depends on what's in the problem. If the expression involves only sine, use 1−2sin⁡2θ1 - 2\sin^2\theta. If it involves only cosine, use 2cos⁡2θ−12\cos^2\theta - 1.

Half-Angle Identities

These come from solving the double-angle formulas for cos⁡2θ\cos 2\theta in terms of θ/2\theta/2:

  • sin⁡θ2=±1−cos⁡θ2\sin\frac{\theta}{2} = \pm\sqrt{\frac{1 - \cos\theta}{2}}
  • cos⁡θ2=±1+cos⁡θ2\cos\frac{\theta}{2} = \pm\sqrt{\frac{1 + \cos\theta}{2}}
  • tan⁡θ2=1−cos⁡θsin⁡θ=sin⁡θ1+cos⁡θ\tan\frac{\theta}{2} = \frac{1 - \cos\theta}{\sin\theta} = \frac{\sin\theta}{1 + \cos\theta}

The ±\pm sign is determined by the quadrant that θ/2\theta/2 falls in.

Power-Reducing Identities

These are rearrangements of the double-angle formulas, useful for rewriting squared trig functions as first-power expressions:

  • sin⁡2θ=1−cos⁡2θ2\sin^2\theta = \frac{1 - \cos 2\theta}{2}
  • cos⁡2θ=1+cos⁡2θ2\cos^2\theta = \frac{1 + \cos 2\theta}{2}

These show up frequently in calculus, so getting comfortable with them now is worthwhile.

Verifying Trigonometric Identities

Steps for Verifying Identities

Verifying (or "proving") an identity means showing that the left side and right side are equivalent using known identities. You are not solving an equation. You cannot perform the same operation on both sides. Instead, you transform one side until it matches the other.

Here's a reliable approach:

  1. Pick the more complex side to work with. Leave the simpler side as your target.
  2. Convert everything to sine and cosine if the path forward isn't obvious.
  3. Look for opportunities to factor, combine fractions, or apply Pythagorean identities.
  4. Work toward the other side. Each step should use a valid identity or algebraic manipulation.
  5. Make sure every step is reversible and that you haven't restricted the domain (for instance, don't multiply both sides by something that could be zero).

Example: Verify that sin⁡θ1+cos⁡θ=1−cos⁡θsin⁡θ\frac{\sin\theta}{1 + \cos\theta} = \frac{1 - \cos\theta}{\sin\theta}.

Start with the left side and multiply the numerator and denominator by the conjugate (1−cos⁡θ)(1 - \cos\theta):

sin⁡θ1+cos⁡θ⋅1−cos⁡θ1−cos⁡θ=sin⁡θ(1−cos⁡θ)1−cos⁡2θ\frac{\sin\theta}{1 + \cos\theta} \cdot \frac{1 - \cos\theta}{1 - \cos\theta} = \frac{\sin\theta(1 - \cos\theta)}{1 - \cos^2\theta}

Since 1−cos⁡2θ=sin⁡2θ1 - \cos^2\theta = \sin^2\theta:

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

This matches the right side, so the identity is verified.

Fundamental Trigonometric Identities, TrigCheatSheet.com: Identities

Graphical Verification of Identities

You can also check an identity by graphing both sides separately on a calculator or graphing tool. If the two graphs overlap completely across the entire domain, the identity is likely correct. This doesn't count as a formal proof, but it's a great way to confirm your algebraic work or to catch errors before you submit.

Solving Trigonometric Equations

Applying Identities to Solve Equations

Unlike identities (which are true for all valid angles), trig equations are true only for specific values of θ\theta. The goal is to use identities to rewrite the equation so you can isolate the trig function and then find the angle(s).

General strategy:

  1. Use identities to get the equation in terms of a single trig function. For example, if an equation has both sin⁡θ\sin\theta and cos⁡2θ\cos^2\theta, use cos⁡2θ=1−sin⁡2θ\cos^2\theta = 1 - \sin^2\theta to write everything in terms of sin⁡θ\sin\theta.

  2. Solve algebraically. The equation often becomes a quadratic (e.g., 2sin⁡2θ−sin⁡θ−1=02\sin^2\theta - \sin\theta - 1 = 0), which you can factor or use the quadratic formula on.

  3. Find all angles in the specified interval. Remember that most trig functions hit the same value twice per period. For example, sin⁡θ=12\sin\theta = \frac{1}{2} gives θ=30°\theta = 30° and θ=150°\theta = 150° in [0°,360°)[0°, 360°).

  4. Account for periodicity if no interval is specified. Add 360°n360°n (or 2πn2\pi n) for sine and cosine solutions, and 180°n180°n (or πn\pi n) for tangent solutions, where nn is any integer.

The double-angle, half-angle, and sum/difference identities are especially useful when the equation involves expressions like sin⁡2θ\sin 2\theta or cos⁡(θ+60°)\cos(\theta + 60°).

Checking Solutions

Always substitute your solutions back into the original equation. Trig equations can produce extraneous solutions, particularly when you square both sides or multiply by an expression that could equal zero. A solution that satisfies a transformed version of the equation might not satisfy the original.

Also watch for values that make any denominator zero. For instance, if the original equation contains csc⁡θ\csc\theta, then θ=0°\theta = 0° and θ=180°\theta = 180° are automatically excluded from the domain.