The three Pythagorean identities

An equation that is always true

Most equations are true for only a few values. The equation $2x + 1 = 7$ holds only when $x = 3$. But $(x + 1)^2 = x^2 + 2x + 1$ holds for every $x$. An equation that is true for every value of its letters is an identity.

A trigonometric identity is an equation in the ratios of an angle that is true for every angle where both sides are defined.

The first identity

Take a right triangle $ABC$ with the right angle at $B$. Pythagoras says

Divide every term by $AC^2$:

For the angle $A$, $\dfrac{BC}{AC}$ is $\sin A$ and $\dfrac{AB}{AC}$ is $\cos A$. So

It holds for every angle from $0^\circ$ to $90^\circ$. Check it at $30^\circ$: $\dfrac{1}{4} + \dfrac{3}{4} = 1$.

Two more, by dividing

Divide Pythagoras by $AB^2$ instead:

This needs $AB \neq 0$, that is $\cos A \neq 0$. So it holds for $0^\circ \leq A < 90^\circ$.

Divide by $BC^2$:

This needs $\sin A \neq 0$. So it holds for $0^\circ < A \leq 90^\circ$.

Each identity can be turned round. For example, $\sec^2 A - \tan^2 A = 1$ and $\cos^2 A = 1 - \sin^2 A$.

Every ratio from one

Suppose we know only $\sin A$, and $A$ is acute. The first identity gives $\cos A = \sqrt{1 - \sin^2 A}$. The positive root is the one we want, since every side of a triangle is positive. Then

An example. If $\tan A = \dfrac{5}{12}$ for an acute angle, find $\sec A$ and $\cos A$.

1. Use the second identity. $\sec^2 A = 1 + \dfrac{25}{144} = \dfrac{169}{144}$.
2. Take the positive root. $\sec A = \dfrac{13}{12}$.
3. Flip it. $\cos A = \dfrac{12}{13}$.

No triangle had to be drawn.

Angles that add up to 90°

In the right triangle $ABC$, the two acute angles $A$ and $C$ add to $90^\circ$. The side opposite $A$ is the side adjacent to $C$. So $\sin A = \cos C$, that is

This is why $\sin 30^\circ = \cos 60^\circ$. It also gives quick answers: $\dfrac{\tan 35^\circ}{\cot 55^\circ} = 1$, since $\cot 55^\circ = \tan 35^\circ$.

Practise with new numbers

Try it in Python

The program tests the first two identities at every whole angle from $1^\circ$ to $89^\circ$, and prints the largest gap it finds.

Common mistakes

• Writing $\sin A + \cos A = 1$. It is the squares that add up to $1$. At $45^\circ$, $\sin A + \cos A = \sqrt{2}$.
• Mixing the pairs. $\tan$ goes with $\sec$, and $\cot$ goes with $\text{cosec}$. There is no identity $1 + \tan^2 A = \text{cosec}^2 A$.
• Taking the negative root when the angle is acute.
• Using an identity where it is not defined, such as $1 + \tan^2 A = \sec^2 A$ at $A = 90^\circ$.

Remember these

• $\sin^2 A + \cos^2 A = 1$.
• $1 + \tan^2 A = \sec^2 A$, for $A < 90^\circ$.
• $1 + \cot^2 A = \text{cosec}^2 A$, for $A > 0^\circ$.
• Any one ratio of an acute angle gives all the others.
• $\sin(90^\circ - A) = \cos A$ and $\tan(90^\circ - A) = \cot A$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Simplify $1 - \cos^2 A$.

2. If $\sec A = \dfrac{17}{8}$ for an acute angle, find $\tan A$.

3. What is $\text{cosec}^2 A - \cot^2 A$?

4. Write $\sin A$ in terms of $\cot A$.

5. Without a table, find $\dfrac{\cos 20^\circ}{\sin 70^\circ}$.