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.

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 \(\cos(30^\circ)?\)

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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 \(\cos(30^\circ) = \frac{\sqrt{3}}{2}.\) \(_\square\)

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:

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

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

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Unit circle with angle greater than \(2 \pi\)

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 \(\frac{1}{\sin^2\theta} + \frac{1}{\cos^2\theta}?\)

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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:

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

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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 \( \sin \theta + \cos \theta ?\)

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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 \(p?\)

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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

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

• Trigonometry