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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.
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Right-angled triangle used for the base of Pythagoras theorem
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 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 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 \( 1 + \cot^2\theta = \csc^2\theta\).
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 \(x = 45^\circ\), \(105^\circ\), \(165^\circ\).
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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How do you derive Pythagorean identities?
The Pythagorean identities are derived from Pythagoras theorem and the unit circle.
What are Pythagorean identities?
They are expressions which are based on Pythagoras theorem and can be used to solve or simplify trigonometric equations.
What are the three Pythagorean identities?
sin^2(𝛉) +cos^2(𝛉) =1, tan^2(𝝷)+1=sec^2(𝝷) and 1+cot^2(𝝷)=csc^2(𝝷)
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