DWG NO. 5.1 — Lesson 1 of 5

Fundamental Identities

Unit 5: Trigonometric Identities and Equations · ~20 min

Objective Use the reciprocal, quotient, and Pythagorean identities to simplify trigonometric expressions and find unknown ratios.

An identity is different from an equation you solve. A trig equation like sinx = ½ is only true for specific values of x. An identity like sin²x + cos²x = 1 is true for every value of x — it's not a statement to solve, it's a rule that always holds. Unit 4 defined sine, cosine, and the other four trig functions directly from the unit circle; this lesson collects the identities that follow immediately from those definitions, and uses them to rewrite trig expressions in simpler forms.

Reciprocal identities

Each of the three "extra" trig functions is just the reciprocal of one you already know:

cscθ = \(\frac{1}{sin\theta}\) secθ = \(\frac{1}{cos\theta}\) cotθ = \(\frac{1}{tan\theta}\)

Quotient identities

Tangent and cotangent can always be rewritten in terms of sine and cosine — this single move is often the fastest way into a simplification:

tanθ = \(\frac{sin\theta}{cos\theta}\) cotθ = \(\frac{cos\theta}{sin\theta}\)

Pythagorean identities

These come straight from the unit circle: any point (cosθ, sinθ) sits on a circle of radius 1, so its coordinates must satisfy x² + y² = 1.

sin²θ + cos²θ = 1
1 + tan²θ = sec²θ (divide the identity above by cos²θ)
1 + cot²θ = csc²θ (divide the identity above by sin²θ)

The point (cosθ, sinθ) sits on the unit circle, so its legs and radius satisfy sin²θ + cos²θ = 1

Worked Example 1 · Simplify using quotient identity

ProblemSimplify sinθ · cotθ.

1Rewrite cotθ using the quotient identity: cotθ = \(\frac{cos\theta}{sin\theta.}\)

2sinθ · \(\frac{cos\theta}{sin\theta}\) — the sinθ factors cancel, leaving cosθ.

sinθ · cotθ = cosθ

Worked Example 2 · Simplify using a Pythagorean identity

ProblemSimplify \(\frac{1 - cos^{2}\theta}{sin\theta.}\)

1From sin²θ + cos²θ = 1, rearranged: 1 − cos²θ = sin²θ.

2Substitute: \(\frac{sin^{2}\theta}{sin\theta}\) = sinθ.

\(\frac{1 - cos^{2}\theta}{sin\theta}\) = sinθ

Worked Example 3 · Find missing ratios

ProblemGiven sinθ = \(\frac{3}{5,}\) with θ in Quadrant II, find cosθ and tanθ.

1Use sin²θ + cos²θ = 1: cos²θ = 1 − (\(\frac{3}{5}\))² = 1 − \(\frac{9}{25}\) = \(\frac{16}{25.}\)

2cosθ = \(\frac{\pm4}{5.}\) In Quadrant II, cosine is negative, so cosθ = \(\frac{-4}{5.}\)

3tanθ = \(\frac{sin\theta}{cos\theta}\) = \(\frac{\frac{3}{5}}{\frac{-4}{5}}\) = \(\frac{-3}{4}\).

cosθ = \(\frac{-4}{5}\) and tanθ = \(\frac{-3}{4}\)

Guided practice

1.Simplify cosθ · tanθ.
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cosθ · (\(\frac{sin\theta}{cos\theta}\)) = sinθ.
2.Simplify secθ · sinθ · cotθ.
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(\(\frac{1}{cos\theta}\)) · sinθ · (\(\frac{cos\theta}{sin\theta}\)) — the sinθ and cosθ factors both cancel, leaving 1.
3.Given cosθ = \(\frac{5}{13,}\) with θ in Quadrant I, find sinθ and tanθ.
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sin²θ = 1 − \(\frac{25}{169}\) = \(\frac{144}{169}\), so sinθ = \(\frac{12}{13}\) (positive in QI). tanθ = \(\frac{\frac{12}{13}}{\frac{5}{13}}\) = \(\frac{12}{5}\).
4.Simplify \(\frac{sec^{2}\theta - 1}{tan^{2}\theta.}\)
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Since 1 + tan²θ = sec²θ, sec²θ − 1 = tan²θ. So the expression is \(\frac{tan^{2}\theta}{tan^{2}\theta}\) = 1.
5.Explain how 1 + cot²θ = csc²θ follows from sin²θ + cos²θ = 1.
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Divide every term of sin²θ + cos²θ = 1 by sin²θ: 1 + \(\frac{cos^{2}\theta}{sin^{2}\theta}\) = \(\frac{1}{sin^{2}\theta,}\) which is exactly 1 + cot²θ = csc²θ using the quotient and reciprocal identities.

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