Chapter 4 - Applications of the Derivative - 4.1 Linear Approximation and Applications - Exercises - Page 174: 47

$$L(\theta)\approx \theta-\frac{\pi}{4}+\frac{1}{2} $$

Given $$ f(\theta)=\sin ^{2} \theta, \quad a=\frac{\pi}{4}$$ Since $f(\pi/4)= \dfrac{1}{2}$ and \begin{align*} f'(\theta)&=2\sin \theta\cos \theta\\ f'(\pi/4)&= 2 \sin (\pi/4)\cos (\pi/4)\\ &=1 \end{align*} Then the linearization at $a=\pi/4$ given by \begin{align*} L(\theta)&\approx f^{\prime}(a)(\theta-a)+f(a)\\ &=1\left(\theta-\frac{\pi}{4}\right)+\frac{1}{2}\\ &=\theta-\frac{\pi}{4}+\frac{1}{2} \end{align*}