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Pre-Calculus Trigonometric identities

Reciprocal, quotient, and Pythagorean identities

20 practice questions 0 video lessons Theory + worked examples

Reciprocal, Quotient, and Pythagorean Identities

California Pre-Calculus • Standard F-TF.8 • Trigonometric Identities

Reciprocal, Quotient, and Pythagorean Identities is the opening topic of Trigonometric Identities in the California Common Core State Standards. It is aligned to Standard F-TF.8, which requires students to prove and use the Pythagorean identity.

The fundamental identities — reciprocal, quotient, and Pythagorean \((\sin^2\theta+\cos^2\theta=1)\) — relate the trigonometric functions and are the basis for all the others.

California Pre-Calculus › Trigonometric Identities › Reciprocal, Quotient, and Pythagorean Identities  —  Standard F-TF.8

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Theory

An identity is an equation true for every value of the variable. The fundamental trig identities come in three families:

  • Reciprocal: \(\csc=\dfrac{1}{\sin}\), \(\sec=\dfrac{1}{\cos}\), \(\cot=\dfrac{1}{\tan}\).
  • Quotient: \(\tan=\dfrac{\sin}{\cos}\), \(\cot=\dfrac{\cos}{\sin}\).
  • Pythagorean: \(\sin^2\theta+\cos^2\theta=1\), and its relatives \(1+\tan^2\theta=\sec^2\theta\), \(1+\cot^2\theta=\csc^2\theta\).
The Pythagorean identity is the unit circle. Since \((\cos\theta,\sin\theta)\) lies on a circle of radius 1, \(\cos^2\theta+\sin^2\theta=1\) automatically.
Pythagorean identity from the unit circle On the unit circle the horizontal leg is cosine theta, the vertical leg is sine theta, and the hypotenuse is 1, so cosine squared plus sine squared equals one. cos θsin θ1
On the unit circle the legs are \(\cos\theta,\sin\theta\) and the hypotenuse is 1, giving \(\cos^2\theta+\sin^2\theta=1\).
Reciprocal and quotient identities Cosecant, secant, and cotangent are the reciprocals of sine, cosine, and tangent; tangent and cotangent are quotients of sine and cosine. Reciprocal & quotient reciprocal csc θ =1sin θ sec θ =1cos θ cot θ =1tan θ quotient tan θ =sin θcos θ cot θ =cos θsin θ
The reciprocal and quotient identities at a glance.

The three Pythagorean identities:

\[\sin^2\theta+\cos^2\theta=1,\quad 1+\tan^2\theta=\sec^2\theta,\quad 1+\cot^2\theta=\csc^2\theta\]
sine squared plus cosine squared equals one; one plus tangent squared equals secant squared; one plus cotangent squared equals cosecant squared
The last two come from the first by dividing through by \(\cos^2\theta\) or \(\sin^2\theta\).

How to use the fundamental identities

  1. To find another ratio: use \(\sin^2+\cos^2=1\) for the partner, then the quadrant for the sign.
  2. To simplify: rewrite \(\tan,\cot,\sec,\csc\) in terms of \(\sin\) and \(\cos\) and cancel.
  3. To switch identities: divide the Pythagorean identity by \(\cos^2\) or \(\sin^2\) to reach the version you need.
Example 1 — Find cosine from sine
If \(\sin\theta=\dfrac{3}{5}\) and \(\theta\) is in Quadrant II, find \(\cos\theta\).
Solution

Use the Pythagorean identity, then choose the sign for the quadrant.

\(\cos^2\theta\)\(=\)\(1-\sin^2\theta=1-\dfrac{9}{25}\)
\(=\)\(\dfrac{16}{25}\)
\(\cos\theta\)\(=\)\(-\dfrac{4}{5}\)

Cosine is negative in Quadrant II.

cosine theta is negative four fifths
Example 2 — Simplify with the quotient identity
Simplify \(\dfrac{\sin\theta}{\cos\theta}\cdot\cos\theta\).
Solution

The quotient identity gives \(\dfrac{\sin\theta}{\cos\theta}=\tan\theta\), but here the \(\cos\theta\) cancels directly.

\(\dfrac{\sin\theta}{\cos\theta}\cdot\cos\theta\)\(=\)\(\sin\theta\)
expression simplifies to sine theta
Example 3 — Derive a Pythagorean identity
Show that \(1+\tan^2\theta=\sec^2\theta\).
Solution

Start from \(\sin^2\theta+\cos^2\theta=1\) and divide every term by \(\cos^2\theta\).

\(\dfrac{\sin^2\theta}{\cos^2\theta}+\dfrac{\cos^2\theta}{\cos^2\theta}\)\(=\)\(\dfrac{1}{\cos^2\theta}\)
\(\tan^2\theta+1\)\(=\)\(\sec^2\theta\)
dividing the Pythagorean identity by cosine squared gives 1 plus tan squared equals secant squared
Example 4 — All six from one
Given \(\cos\theta=\dfrac{5}{13}\) in Quadrant I, find \(\sin\theta\) and \(\tan\theta\).
Solution

Pythagorean identity for sine, then quotient identity for tangent.

\(\sin\theta\)\(=\)\(\sqrt{1-\dfrac{25}{169}}=\dfrac{12}{13}\)
\(\tan\theta\)\(=\)\(\dfrac{\sin\theta}{\cos\theta}=\dfrac{12}{5}\)
sine is twelve thirteenths, tangent is twelve fifths

Common pitfalls

Choose the sign by quadrant. \(\cos\theta=\pm\sqrt{1-\sin^2\theta}\); the quadrant fixes which.
\(\sin^2\theta\) means \((\sin\theta)^2\). Square the value, not the angle.
Reciprocal \(\neq\) inverse. \(\csc\theta=\dfrac{1}{\sin\theta}\), not \(\arcsin\theta\).

Frequently asked questions

What are the Pythagorean identities?

\(\sin^2\theta+\cos^2\theta=1\), and dividing by \(\cos^2\) or \(\sin^2\) gives \(1+\tan^2\theta=\sec^2\theta\) and \(1+\cot^2\theta=\csc^2\theta\).

What are the reciprocal and quotient identities?

Reciprocal: \(\csc=1/\sin\), \(\sec=1/\cos\), \(\cot=1/\tan\). Quotient: \(\tan=\sin/\cos\) and \(\cot=\cos/\sin\).

How do you find cosine if you know sine?

Use \(\cos\theta=\pm\sqrt{1-\sin^2\theta}\) and pick the sign from the quadrant of \(\theta\).

Why is sin squared plus cos squared equal to 1?

Because \((\cos\theta,\sin\theta)\) is a point on the unit circle, whose radius is 1; the Pythagorean theorem then gives the identity.

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