Pythagoras Theorem

Level 1 - Math I (Physics) topic page in Trigonometry.

Principle

Pythagoras' theorem connects the side lengths of a right triangle. In trigonometry it becomes the reason that sine and cosine are linked: a point on the unit circle always has distance 1 from the origin.

The same idea appears in physics whenever perpendicular components combine, such as horizontal and vertical displacement, rectangular vector components, or orthogonal oscillations.

Notation

\(a,b\)

the two perpendicular side lengths of a right triangle

\(c\)

the hypotenuse, opposite the right angle

\(\theta\)

an angle measured in radians unless stated otherwise

\(\sin\theta\)

opposite side divided by hypotenuse, or y-coordinate on the unit circle

\(\cos\theta\)

adjacent side divided by hypotenuse, or x-coordinate on the unit circle

\(\tan\theta\)

\sin\theta/\cos\theta where \cos\theta\ne0

\(\sec\theta\)

1/\cos\theta where \cos\theta\ne0

\(\csc\theta\)

1/\sin\theta where \sin\theta\ne0

\(\cot\theta\)

\cos\theta/\sin\theta where \sin\theta\ne0

The Core Method

Start with a right triangle:

Pythagoras theorem

\[a^2+b^2=c^2\]

On the unit circle, the point at angle \(\theta\) has coordinates \((\cos\theta,\sin\theta)\) and radius 1. Substituting the horizontal leg \(\cos\theta\), the vertical leg \(\sin\theta\), and hypotenuse \(1\) into Pythagoras gives the fundamental identity:

Unit-circle identity

\[\sin^2\theta+\cos^2\theta=1\]

Divide this identity by \(\cos^2\theta\) when \(\cos\theta\ne0\):

Tangent identity

\[1+\tan^2\theta=\sec^2\theta\]

Divide it by \(\sin^2\theta\) when \(\sin\theta\ne0\):

Cotangent identity

\[1+\cot^2\theta=\csc^2\theta\]

Worked Cases

Question

A right triangle has perpendicular sides \(5\) and \(12\). Find the hypotenuse.

Answer

Use Pythagoras: \(a^2+b^2=c^2\). Substitute \(a=5\) and \(b=12\): \(5^2+12^2=c^2\). This gives \(25+144=169=c^2\), so \(c=13\). The hypotenuse is \(13\).

Examples

Question

Simplify \(\sin^2\theta+\cos^2\theta+\tan^2\theta\).

Answer

Use \(\sin^2\theta+\cos^2\theta=1\). The expression becomes \(1+\tan^2\theta\). Then use \(1+\tan^2\theta=\sec^2\theta\). So the simplified expression is \(\sec^2\theta\).

Mistake Filter

Mistake: Treating \(a^2+b^2=c^2\) as valid for every triangle. Correction: it only applies to right triangles, where \(c\) is the hypotenuse.
Mistake: Dropping the square on \(\sin^2\theta+\cos^2\theta=1\). Correction: the identity is about squared values, not \(\sin\theta+\cos\theta\).
Mistake: Using \(1+\tan^2\theta=\sec^2\theta\) when \(\cos\theta=0\). Correction: tangent and secant are undefined there.
Mistake: Forgetting the \(\pm\) sign after solving \(\cos^2\theta=k\). Correction: use quadrant information to decide the sign.

Fast Summary

Pythagoras gives \(a^2+b^2=c^2\) for right triangles. On the unit circle this becomes \(\sin^2\theta+\cos^2\theta=1\), and dividing by \(\cos^2\theta\) or \(\sin^2\theta\) gives the tangent/secant and cotangent/cosecant identities.

Inverse Functions

Trig Functions