Chapter 1, Problem 1.17P

To determine

Two linear approximations of the function $f(\theta )=sin\theta$.

Answer to Problem 1.17P

The linear approximation, $f(\theta )$ at $\theta \text{=}\frac{\pi }{4}\text{rad }$ is $0.7071\left(1+x-\frac{\pi }{4}\right)$.

The linear approximation, $f(\theta )$ at $\theta \text{=}\frac{3\pi }{4}\text{rad }$ is $0.7071\left(1-x+\frac{3\pi }{4}\right)$.

Explanation of Solution

Given Information:

Function $f(\theta )=sin\theta$

Consider the following data.

Function $f(\theta )=sin\theta$

Angles,

$\theta =\frac{\pi }{4}$

And

$\theta =\frac{3\pi }{4}$

Differentiate given equation with respect to $\theta$

$\begin{array}{l}\frac{d}{d\theta }f(\theta )=\frac{d}{d\theta }\sin \theta \\ f^1(\theta )=cos\theta \left[\frac{d}{d\theta }f(\theta )isrespresentedas{f}^1(\theta )\right]\end{array}$

The linear approximation near $\theta =\frac{\pi }{4}\text{rad}$ is,

$\begin{array}{c}f(\theta )=f(\theta )+f^1(\theta )(x-\theta )\\ =f\left(\frac{\pi }{4}\right)+f^1\left(\frac{\pi }{4}\right)\left(x-\frac{\pi }{4}\right)\\ =\sin \left(\frac{\pi }{4}\right)+\cos \left(\frac{\pi }{4}\right)\left(x-\frac{\pi }{4}\right)\\ =0.7071+\left(0.7071\right)\left(x-\frac{\pi }{4}\right)\\ =0.7071\left(1+x-\frac{\pi }{4}\right)\\ \end{array}$

Calculate the linear approximation, near $\theta =\frac{3\pi }{4}\text{rad}$.

$\begin{array}{c}f(\theta )=f(\theta )+f^1(\theta )(x-\theta )\\ =f\left(\frac{3\pi }{4}\right)+f^1\left(\frac{3\pi }{4}\right)\left(x-\frac{3\pi }{4}\right)\\ =\sin \left(\frac{3\pi }{4}\right)+\cos \left(\frac{3\pi }{4}\right)\left(x-\frac{3\pi }{4}\right)\\ =0.7071+\left(-0.7071\right)\left(x-\frac{3\pi }{4}\right)\\ =0.7071\left(1-x+\frac{3\pi }{4}\right)\end{array}$.

Conclusion:

The linear approximation, $f(\theta )$ at $\theta \text{=}\frac{\pi }{4}\text{rad }$ is $0.7071\left(1+x-\frac{\pi }{4}\right)$.

The linear approximation, $f(\theta )$ at $\theta \text{=}\frac{3\pi }{4}\text{rad }$ is $0.7071\left(1-x+\frac{3\pi }{4}\right)$.