Question

1. (i) find the linear approximation to $f(x)=sin^{2}x$ at $a = pi/4$. (ii) use it to approximate $sin^{2}(\frac{pi}{4}+0.01)$.

Explanation:

Step1: Recall linear - approximation formula

The linear approximation of a function $y = f(x)$ at $x = a$ is given by $L(x)=f(a)+f^{\prime}(a)(x - a)$. First, find the derivative of $f(x)=\sin^{2}x$. Using the chain - rule, if $u = \sin x$, then $f(x)=u^{2}$ and $\frac{df}{dx}=2\sin x\cos x=\sin2x$.

Step2: Evaluate $f(a)$ and $f^{\prime}(a)$ at $a=\frac{\pi}{4}$

When $a = \frac{\pi}{4}$, $f(\frac{\pi}{4})=\sin^{2}\frac{\pi}{4}=(\frac{\sqrt{2}}{2})^{2}=\frac{1}{2}$, and $f^{\prime}(\frac{\pi}{4})=\sin(2\times\frac{\pi}{4})=\sin\frac{\pi}{2}=1$.

Step3: Write the linear approximation

The linear approximation $L(x)$ of $f(x)=\sin^{2}x$ at $a = \frac{\pi}{4}$ is $L(x)=f(\frac{\pi}{4})+f^{\prime}(\frac{\pi}{4})(x-\frac{\pi}{4})=\frac{1}{2}+1\times(x - \frac{\pi}{4})=\frac{1}{2}+x-\frac{\pi}{4}$.

Step4: Use the linear approximation to approximate $\sin^{2}(\frac{\pi}{4}+0.01)$

Let $x=\frac{\pi}{4}+0.01$. Then $L(\frac{\pi}{4}+0.01)=\frac{1}{2}+(\frac{\pi}{4}+0.01)-\frac{\pi}{4}=\frac{1}{2}+0.01 = 0.51$.

Answer:

The linear approximation of $f(x)=\sin^{2}x$ at $a=\frac{\pi}{4}$ is $L(x)=\frac{1}{2}+x - \frac{\pi}{4}$, and $\sin^{2}(\frac{\pi}{4}+0.01)\approx0.51$.