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Precalculus JumpStart

Section 10.4 Basic Identities

Letโ€™s recap some familar relationships between the trigonometric functions.
The most famous identity satisfied by trigonometric functions is the so-called Pythagorean Identity. From Definitionย 10.1.1, suppose that \((x,y)\) is a point on the terminal ray of \(\theta\) and \(r = \sqrt{x^2+y^2}\)
\begin{align*} \left[\sin(\theta)\right]^2 + \left[\sin(\theta)\right]^2 &= \left[\frac{y}{r}\right]^2 + \left[\frac{x}{r}\right]^2 \\ &= \frac{y^2}{r^2} + \frac{x^2}{r^2}\\ &= \frac{y^2+x^2}{r^2}\\ &= \frac{r^2}{r^2}\\ &= 1. \end{align*}
Thus,
\begin{equation*} \left[\sin(\theta)\right]^2 + \left[\cos(\theta)\right]^2 = 1. \end{equation*}
Itโ€™s customary to write positive integer powers of trigonometric function values like \(\left[\sin(\theta)\right]^2\) as \(\sin^2(\theta)\text{.}\) So, the identity becomes
\begin{equation*} \sin^2(\theta) +\cos^2(\theta) =1. \end{equation*}

Proof.

We have already proved the first indentity. Starting from here and dividing every term by \(\cos^2(\theta)\text{,}\)
\begin{gather*} \sin^2(\theta) +\cos^2(\theta) =1\\ \frac{\sin^2(\theta)}{\cos^2(\theta)} +\frac{\cos^2(\theta)}{\cos^2(\theta)} = \frac{1}{\cos^2(\theta)}\\ \left[\frac{\sin(\theta)}{\cos(\theta)}\right]^2 +1 = \left[\frac{1}{\cos(\theta)}\right]^2 \end{gather*}
\begin{equation*} \tan^2(\theta) +1 = \sec^2(\theta). \end{equation*}
The third identity is left to you to prove.

Example 10.4.3. Finding Trigonometric Function Values Using Identities.

Suppose \(\cos(\theta) = -0.7\text{.}\) Letโ€™s find the remaining trigonometric function values of \(\theta\text{,}\) given that \(\theta\) is a Quadrant III angle.
By the Pythagorean Identity,
\begin{equation*} \sin^2(\theta) = 1-\cos^2(\theta) \end{equation*}
so that
\begin{equation*} \sin(\theta) = \pm \sqrt{1-\cos^2(\theta)}. \end{equation*}
But sine is negative in Quadrant II, so
\begin{equation*} \sin(\theta) = {\color{red}-}\sqrt{1-\cos^2(\theta)} = \sqrt{1-(0.7)^2} \approx -0.51. \end{equation*}
Everything falls into place now using ratio and reciprocal identities:
\begin{gather*} \tan(\theta) \approx \frac{-0.51}{-0.7} \approx 0.73\\ \csc(\theta) \approx \frac{1}{-0.51} \approx -1.96\\ \sec(\theta) = \frac{1}{-0.7} \approx -1.43\\ \cot(\theta) \approx \frac{-0.7}{-0.51} \approx 1.37 \end{gather*}