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Trigonometric Function Toolkit · Tutorial 133 of 1495

Deriving the Pythagorean Identity

Use the unit-circle equation to derive the three Pythagorean trigonometric identities and understand exactly where each one applies.

Beginner 9 min read

What You'll Learn

  • Derive the sine-cosine identity from the coordinates of a point on the unit circle
  • Explain why the identity holds for every real angle
  • Derive the tangent-secant identity and state its domain restriction
  • Derive the cotangent-cosecant identity and state its domain restriction
  • Use the identities to check exact values and determine a trigonometric value

Why These Identities Hold

The sine and cosine of an angle are coordinates of a point on the unit circle. Every point on that circle satisfies the same geometric equation, so its coordinates give a relationship between sine and cosine that is true for every angle.

That relationship is the starting point for three important trigonometric identities. The first applies to all real angles. The other two follow by division, so each has a restriction: the quantity being divided by must be nonzero.

The Unit-Circle Equation

The unit circle is the circle of radius \(1\) centered at the origin. A point \((u,v)\) on it has distance \(1\) from the origin. By the distance formula, its coordinates satisfy \(u^2+v^2=1\). For an angle \(x\), the corresponding unit-circle point has coordinates \((\cos x,\sin x)\).

Definition: For a real angle \(x\), the point on the unit circle corresponding to \(x\) has horizontal coordinate \(\cos x\) and vertical coordinate \(\sin x\). The unit-circle equation is \(u^2+v^2=1\).

Substitute \(u=\cos x\) and \(v=\sin x\) into the circle equation. This gives the Pythagorean identity for sine and cosine.

$$ \cos^2 x+\sin^2 x=1. $$

The conventional order writes the sine term first. The two terms can be reversed because addition is commutative.

Theorem: For every real angle \(x\), $$ \sin^2 x+\cos^2 x=1. $$ Here, \(\sin^2 x\) means \((\sin x)^2\), and \(\cos^2 x\) means \((\cos x)^2\).

The identity holds for every real angle because every angle selects a point on the unit circle, and every point on that circle satisfies its defining equation. The signs of sine and cosine may change in different quadrants, but squaring removes those signs; the sum of the squares remains \(1\).

Deriving the Tangent-Secant Identity

The earlier definitions of tangent and secant are \(\tan x=\frac{\sin x}{\cos x}\) and \(\sec x=\frac{1}{\cos x}\). Both expressions require \(\cos x\ne 0\). When that condition holds, divide every term of the sine-cosine identity by \(\cos^2 x\).

$$ \frac{\sin^2 x}{\cos^2 x} + \frac{\cos^2 x}{\cos^2 x} = \frac{1}{\cos^2 x}. $$

The first term is \(\tan^2 x\), the second is \(1\), and the right-hand side is \(\sec^2 x\). Therefore:

Theorem: If \(\cos x\ne 0\), then $$ \tan^2 x+1=\sec^2 x. $$

The condition matters: tangent and secant are undefined wherever \(\cos x=0\). The identity is not a claim that either side has a value at those angles.

Deriving the Cotangent-Cosecant Identity

The definitions of cotangent and cosecant are \(\cot x=\frac{\cos x}{\sin x}\) and \(\csc x=\frac{1}{\sin x}\). These expressions require \(\sin x\ne 0\). Divide the sine-cosine identity by \(\sin^2 x\) when that condition holds:

$$ \frac{\sin^2 x}{\sin^2 x} + \frac{\cos^2 x}{\sin^2 x} = \frac{1}{\sin^2 x}. $$

The terms become \(1\), \(\cot^2 x\), and \(\csc^2 x\), respectively. Hence:

Theorem: If \(\sin x\ne 0\), then $$ 1+\cot^2 x=\csc^2 x. $$

This identity is unavailable at angles where \(\sin x=0\), because cotangent and cosecant are undefined there.

Why the Derivations Work

Each derived identity comes from the same equation and a permitted division. The nonzero conditions are not optional technical details: division by zero is undefined. The steps below make the logic explicit.

1
Start with the unit-circle identity.
For every real \(x\), \(\sin^2 x+\cos^2 x=1\).
2
Choose a divisor that is nonzero.
To obtain the tangent-secant identity, require \(\cos x\ne 0\). To obtain the cotangent-cosecant identity, require \(\sin x\ne 0\).
3
Divide every term by its square.
Divide by \(\cos^2 x\) or \(\sin^2 x\), respectively. The condition ensures the divisor is nonzero.
4
Use the definitions.
Rewrite the resulting ratios as tangent, secant, cotangent, and cosecant.

This also explains the different domains. The sine-cosine identity itself requires no division, so it holds at every real angle. Each of the other identities uses a quotient with a denominator that must be nonzero.

Worked Examples

Worked Example: Check the Identities at \(x=\frac{\pi}{4}\)

At \(x=\frac{\pi}{4}\), the exact values are \(\sin x=\frac{\sqrt{2}}{2}\) and \(\cos x=\frac{\sqrt{2}}{2}\). First check the identity that applies to every angle:

$$ \sin^2 x+\cos^2 x = \left(\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{2}{4}+\frac{2}{4} = \frac{1}{2}+\frac{1}{2} = 1. $$

Since neither sine nor cosine is zero here, all the trigonometric functions in the other two identities are defined. From their definitions, \(\tan x=1\), \(\sec x=\sqrt{2}\), \(\cot x=1\), and \(\csc x=\sqrt{2}\). Check both:

$$ \tan^2 x+1=1^2+1=2=(\sqrt{2})^2=\sec^2 x, $$ $$ 1+\cot^2 x=1+1^2=2=(\sqrt{2})^2=\csc^2 x. $$

Both sides agree in each identity, as required.

Worked Example: Use the Identities When \(\sin x=\frac{3}{5}\) in Quadrant II

Suppose \(x\) lies in Quadrant II and \(\sin x=\frac{3}{5}\). The sine-cosine identity gives:

$$ \left(\frac{3}{5}\right)^2+\cos^2 x=1, \qquad \frac{9}{25}+\cos^2 x=1, $$ $$ \cos^2 x=1-\frac{9}{25}=\frac{16}{25}. $$

Thus \(\cos x\) is either \(\frac{4}{5}\) or \(-\frac{4}{5}\). In Quadrant II, cosine is negative, so \(\cos x=-\frac{4}{5}\). Now use the definitions to calculate and check the other functions:

$$ \tan x=\frac{3/5}{-4/5}=-\frac{3}{4}, \qquad \sec x=\frac{1}{-4/5}=-\frac{5}{4}, $$ $$ \cot x=\frac{-4/5}{3/5}=-\frac{4}{3}, \qquad \csc x=\frac{1}{3/5}=\frac{5}{3}. $$

Since both sine and cosine are nonzero, the two derived identities apply. Substitution verifies them:

$$ \sin^2 x+\cos^2 x = \frac{9}{25}+\frac{16}{25} =1, $$ $$ \tan^2 x+1 = \left(-\frac{3}{4}\right)^2+1 = \frac{9}{16}+\frac{16}{16} = \frac{25}{16} = \left(-\frac{5}{4}\right)^2 = \sec^2 x, $$ $$ 1+\cot^2 x = 1+\left(-\frac{4}{3}\right)^2 = \frac{9}{9}+\frac{16}{9} = \frac{25}{9} = \left(\frac{5}{3}\right)^2 = \csc^2 x. $$

The negative cosine, tangent, and cotangent agree with the quadrant signs, and all three identities check.

Worked Example: Find \(\sec x\) from \(\tan x=\frac{5}{12}\) in Quadrant I

The tangent-secant identity applies because \(x\) is in Quadrant I, where \(\cos x\ne 0\). Substitute the given tangent:

$$ \tan^2 x+1=\sec^2 x, \qquad \left(\frac{5}{12}\right)^2+1=\sec^2 x, $$ $$ \frac{25}{144}+\frac{144}{144}=\sec^2 x, \qquad \sec^2 x=\frac{169}{144}. $$

Taking square roots gives \(\sec x=\frac{13}{12}\) or \(\sec x=-\frac{13}{12}\). In Quadrant I, cosine is positive, so its reciprocal, secant, is also positive. Therefore \(\sec x=\frac{13}{12}\). Check by substitution:

$$ \tan^2 x+1 = \frac{25}{144}+1 = \frac{169}{144} = \left(\frac{13}{12}\right)^2 = \sec^2 x. $$

The positive value is selected by the quadrant, and its square satisfies the identity.

Common Pitfall: Dividing Without Checking

A frequent error is to divide the sine-cosine identity by \(\cos^2 x\) and then claim that \(\tan^2 x+1=\sec^2 x\) holds for every real \(x\). This overlooks the condition \(\cos x\ne 0\).

For example, at \(x=\frac{\pi}{2}\), \(\cos x=0\). The attempted division would include \(\frac{0}{0}\), which is undefined. In fact, \(\tan\frac{\pi}{2}\) and \(\sec\frac{\pi}{2}\) are undefined, so the tangent-secant identity cannot be evaluated there. The correction is to state its domain condition each time: it applies only when \(\cos x\ne 0\). The same care is needed when dividing by \(\sin^2 x\).

Using the Identities Carefully

The identities are useful for checking values and relating functions, but the squared equations do not determine signs on their own. For instance, \(\cos^2 x=\frac{16}{25}\) implies \(\cos x=\frac{4}{5}\) or \(\cos x=-\frac{4}{5}\). The quadrant or other information is needed to select the correct value.

Keep the three statements and their conditions distinct:

IdentityCondition
\(\sin^2 x+\cos^2 x=1\)All real \(x\)
\(\tan^2 x+1=\sec^2 x\)\(\cos x\ne 0\)
\(1+\cot^2 x=\csc^2 x\)\(\sin x\ne 0\)

The first identity is the geometric foundation. The other two are consequences obtained by division, and their restrictions follow directly from the definitions of the functions involved.

Check Your Understanding

Answer each question, paying attention to both the algebra and the conditions under which an identity applies.

  1. How does the unit-circle equation lead to \(\sin^2 x+\cos^2 x=1\)?
  2. What condition is required to divide the sine-cosine identity by \(\cos^2 x\), and what identity results?
  3. What condition is required to derive the cotangent-cosecant identity?
  4. If \(\cos^2 x=\frac{9}{25}\), what are the possible values of \(\cos x\), and what additional information could select one?
  5. Why can the tangent-secant identity not be evaluated at \(x=\frac{\pi}{2}\)?

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