Pythagorean identities are equations based on Pythagoras' theorem \( a^2 + b^2 = c^2\). You can use this theorem to find the sides of a right-angled triangle. There are three Pythagorean identities.
Right-angled triangle used for the base of Pythagoras theorem
The first Pythagorean identity
The first Pythagorean identity is \( \sin^2 \theta + \cos^2 \theta = 1\). This can be derived using Pythagoras theorem and the unit circle.
Unit circle showing the derivation for the first Pythagorean identity
We know that \( a^2 + b^2 = c^2\) so \( \sin^2 \theta + \cos^2 \theta = 1\).
The second Pythagorean identity
The second Pythagorean identity is \( \tan^2\theta + 1 = \sec^2\theta \). This is derived by taking the first Pythagorean identity and dividing it by \(\cos^2\theta\):
\[ \frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta} {\cos^2\theta} = \frac{1}{\cos^2\theta} .\]
Remember that
\[ \frac{\sin\theta}{\cos\theta} = \tan\theta \mbox{ and } \frac{1}{\cos\theta} = \sec\theta.\]
Simplifying this expression we get \( \tan^2\theta + 1 = \sec^2\theta \).
The third Pythagorean identity
The third Pythagorean identity is \( 1 + \cot^2\theta = \csc^2\theta\). This is derived by taking the first Pythagorean identity and dividing it by \(\sin^2\theta\):
\[ \frac{\sin^2\theta}{\sin^2\theta} + \frac{\cos^2\theta} {\sin^2\theta} = \frac{1}{\sin^2\theta} .\]
Remember that
\[ \frac{\cos\theta}{\sin\theta} = \cot\theta \mbox{ and } \frac{1}{\sin\theta} = \csc\theta.\]
Now we can simplify this expression to \( 1 + \cot^2\theta = \csc^2\theta\).
How to use Pythagorean identities
We will now look at three examples of using each of the Pythagorean identities to answer questions.
Simplify \(\sin x \cos^2 x = \sin x -1\) and find the value of \(x\): \(0 ＜ x ＜ 2\pi\).
For this, we will need to use the first Pythagorean identity: \( \sin^2 \theta + \cos^2 \theta = 1\) and rearrange it:
\[ \cos^2 x = 1 - \sin^2 x .\]
We can now substitute \(1 - \sin^2 x \) into the expression:
\[ \sin x \cos^2 x = \sin x(1 - \sin^2 x ).\]
Simplifying this and setting it equal to the right hand side, we get
\[ \sin x - \sin^3 x = \sin x -1 \]
or
\[-\sin^3 x = -1. \]
So \( \sin x = 1 \) and \(x = \frac{\pi}{2}\).
If \(\cos x = 0.78\), what is the value of \(\tan x\)?
For this, we need to use the fact that \( \tan^2x + 1 = \sec^2x \). We also know that
\[ \sec x = \frac{1}{\cos x}\]
therefore
\[ \sec x = \frac{1}{0.78} = 1.282 .\]
We can now substitute this value into the equation and find \( \tan x\):
\[ \tan^2 x + 1 = (1.282)^2 \]
so
\[ \tan^2 x = (1.282)^2 -1 \]
and \( \tan x = 0.802\).
Solve for \(x\) between \(0^\circ\) and \(180^\circ\):
\[ \cot^2 (2x)+ \csc (2x) - 1 = 0.\]
In this case, we need to use the third Pythagorean identity, \( 1 + \cot^2\theta = \csc^2\theta\).
If we rearrange this identity, we get \( \cot^2\theta = \csc^2\theta - 1\). In this case \(\theta = 2x\) and we can plug in this rearranged identity into our equation:
\[ \left( \csc^2(2x) - 1 \right) + \csc 2x - 1 = 0 \]
so
\[ \csc^2 2x + \csc 2x - 2 = 0.\]
We can treat this as a quadratic that we can factorise into
\[(\csc 2x + 2)(\csc 2x - 1) = 0.\]
We can now solve this and get \( \csc 2x = -2\) or \( \csc 2x = 1\), so \( \sin 2x = -\frac{1}{2}\) or \(\sin x = 1\). Therefore \(2x = 210^\circ\), \(330^\circ\), \(90^\circ\). and \(x = 45^\circ\), \(105^\circ\), \(165^\circ\).
Pythagorean Identities - Key takeaways
The first Pythagorean identity is \( \sin^2 \theta + \cos^2 \theta = 1\). ![]()
The second Pythagorean identity is \( \tan^2\theta + 1 = \sec^2\theta \). ![]()
The third Pythagorean identity is \( 1 + \cot^2\theta = \csc^2\theta\). ![]()
The first identity is derived from the Pythagorean theorem \( a^2 + b^2 = c^2\) and the unit circle.
The second and third identities are derived from the first identity.
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