# Pythagorean Identities

**Pythagorean identities** are identities in trigonometry that are extensions of the Pythagorean theorem. The fundamental identity states that for any angle \(\theta,\)

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

Pythagorean identities are useful in simplifying trigonometric expressions, especially in writing expressions as a function of either \(\sin\) or \(\cos\), as in statements of the double angle formulas.

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## Derivation of Fundamental Pythagorean Identity

By using the Pythagorean theorem, which states that \(a^2+b^2=c^2\), and the definitions of the basic trigonometric functions,

\[ a^2+b^2=c^2 \implies \frac{a^2 + b^2}{c^2} = \left( \frac{a}{c} \right)^2 + \left( \frac{b}{c} \right)^2 = 1 \implies \cos^2 \theta + \sin^2 \theta = 1.\]

This gives the following important identity of the basic trigonometric functions.

Pythagorean Identity:For any angle \(\theta,\) we have \(\cos^2 \theta + \sin^2 \theta = 1. \)

If \(\sin(30^\circ) = \frac{1}{2},\) what is \(\cos(30^\circ)?\)

By the Pythagorean identity, \(\sin^2(30^\circ) + \cos^2 (30^\circ) = 1,\) so \[\cos^2(30^\circ) = 1 - \frac{1}{4} = \frac{3}{4} \quad \implies \quad \cos^2(30^\circ) = \pm\frac{\sqrt{3}}{2}.\] However, since \(30^\circ\) is an acute angle, we must have \(\cos(30^\circ) = \frac{\sqrt{3}}{2}.\) \(_\square\)

## Other Forms of Pythagorean Identity

From the Pythagorean Identity, we have the following corollaries:

Corollaries:By dividing both sides by \(\cos^2\theta,\) we have the following identity:

\[ \frac{ \cos^2 \theta + \sin^2 \theta }{\cos^2 \theta} = \frac{1}{\cos^2 \theta} \quad \text{ or } \quad 1 + \tan^2 \theta = \sec^2 \theta. \]

Similarly, by dividing both sides by \(\sin^2\theta,\) we have the following identity:

\[ \frac{ \cos^2 \theta + \sin^2 \theta }{\sin^2 \theta} = \frac{1}{\sin^2 \theta} \quad \text{ or } \quad \cot^2 \theta + 1 = \csc^2 \theta. \]

## Applications and Problem-Solving

Evaluate \( \sin 225^\circ \) exactly.

Notice that the problem reduces to finding the value of \( y \) in the above picture.

Since \( 225^\circ-180^\circ = 45^\circ \), \( \theta \) makes an angle of \( 45^\circ \) with the negative \( x \)-axis. Further, since this is a right triangle, the angles must be \( 90^\circ + 45^\circ + 45^\circ = 180^\circ \), which means it is isosceles. Therefore, \( | x | = |y| \).

By the Pythagorean theorem,

\[ \begin{align} x^2 + y^2 &= 1 \\ 2y^2 &= 1 \\ y^2 &= \frac{1}{2} \\ y &=\pm \frac{\sqrt{2}}{2}. \end{align} \]

By inspection, \( y \) is negative, so \( \sin 225^\circ = -\frac{\sqrt{2}}{2} \). \(_\square\)

Note:most values for \( \theta \) will be difficult or impossible to evaluate exactly in this way, so we often use a calculator to evaluate the approximate value of the function.

In most problems, it is best to convert all trigonometric functions into sines and cosines, in order to take advantage of the Pythagorean identities.

If \( \tan \theta =\frac{1}{3},\) what is \(\frac{1}{\sin^2\theta} + \frac{1}{\cos^2\theta}?\)

We have \[ \begin{align} \tan^2 \theta + 1 &= \sec^2 \theta = \frac{1}{\cos^2 \theta} &\qquad (1) \\\\ \cot^2 \theta + 1 &= \csc^2 \theta = \frac{1}{\sin^2 \theta}. &\qquad (2) \end{align} \]

Since \(\tan \theta= \frac{1}{3}, \) implying \(\cot \theta=3,\) from \( (1)\) and \( (2)\) we have

\[ \begin{align} \frac{1}{\sin^2 \theta} + \frac{1}{\cos^2 \theta} &= \cot^2 \theta + 1 + \tan^2 \theta +1 \\ &= {3}^2 + 1 + {\left (\frac{1}{3}\right )}^2 + 1 \\ &= \frac{100}{9}. \ _\square \end{align} \]

The Pythagorean identities are so useful that it is often worth squaring expressions to take advantage of them:

Select the equivalent of

\[\left(\frac{1}{\sec^{2}\alpha-\cos^{2}\alpha}+\frac{1}{\csc^{2}\alpha-\sin^{2}\alpha}\right)\cdot\cos^{2}\alpha\sin^{2}\alpha.\]

This problem is part of the set Trigonometry.

If \( \sin \theta + \cos \theta = \frac{1}{2} ,\) what is \( \sin \theta \cdot \cos \theta ?\)

Squaring both sides of the above equation gives

\[ \begin{align} (\sin \theta + \cos \theta)^2 &= \left ( \frac{1}{2} \right )^2 \\ \sin^2 \theta + 2\cdot \sin \theta \cdot \cos \theta + \cos^2 \theta &= \frac{1}{4}. \end{align} \]

Since \( \sin^2 \theta + \cos^2 \theta = 1,\) it follows that

\[ \begin{align} \sin^2 \theta + 2\cdot \sin \theta \cdot \cos \theta + \cos^2 \theta &= \frac{1}{4} \\ 1 + 2\cdot \sin \theta \cdot \cos \theta &= \frac{1}{4} \\ 2 \cdot \sin \theta \cdot \cos \theta &= \frac{1}{4}-1 \\ \Rightarrow \sin\theta \cdot \cos \theta &= -\frac{3}{8}. \ _\square \end{align} \]

If \( \tan \theta + \frac{1}{\tan \theta} = 8 ,\) where \( 0 < \theta < \frac{\pi}{2} ,\) what is \( \sin \theta + \cos \theta ?\)

We have

\[ \begin{align} \tan \theta + \frac{1}{\tan \theta} &= 8 \\ \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} &= 8 \\ \frac{\sin^2\theta + \cos^2 \theta}{\sin\theta \cdot \cos \theta} &=8 \\ \frac{1}{\sin\theta \cdot \cos \theta} &= 8 \\ \sin\theta \cdot \cos \theta &= \frac{1}{8}. \end{align} \]

Since \( \sin \theta \cdot \cos \theta=\frac{1}{8}, \) thus we have \[ \begin{align} ( \sin \theta + \cos \theta)^2 &= 1 + 2 \cdot \sin \theta \cdot \cos \theta \\ &= 1+2\times \frac{1}{8} \\ &= \frac{5}{4}. \end{align} \]

Since \( \theta\) lies in the interval \( \left(0,\frac{\pi}{2}\right) ,\) the value of \( \sin \theta + \cos \theta \) is positive. Thus, \[ \sin \theta + \cos \theta = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}. \ _\square\]

If the two roots of the equation \( 2x^2 +px-1=0\) are \( \sin \theta\) and \(\cos \theta ,\) what is \(p?\)

From Vieta's formula, we have \[\begin{align} \sin \theta + \cos \theta &= -\frac{p}{2} &\qquad (1) \\ \sin \theta \cdot \cos \theta &= -\frac{1}{2}. &\qquad (2) \end{align} \]

Squaring both sides of \((1),\) we have

\[ \begin{align} \sin^2 \theta + 2\cdot \sin \theta \cdot \cos \theta + \cos^2 \theta &= \frac{p^2}{4} \\ 1+ 2 \cdot \sin \theta \cdot \cos \theta &= \frac{p^2}{4}. \qquad(3) \end{align} \]

Substituting \((2)\) to \((3),\) we have

\[ \begin{align} 1 + 2 \cdot \left ( -\frac{1}{2} \right ) &= \frac{p^2}{4} \\ \frac{p^2}{4} &=0 \\ \Rightarrow p &= 0. \ _\square \end{align} \]

## See Also

**Cite as:**Pythagorean Identities.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/pythagorean-identities/