Question

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

Explanation:

Step1: Recall linear - approximation formula

The linear - approximation formula is $L(x)=f(a)+f^{\prime}(a)(x - a)$, where $f^{\prime}(x)$ is the derivative of $f(x)$.
First, find the derivative of $y = f(x)=\sin^{2}x$. Using the chain - rule, if $u=\sin x$, then $y = u^{2}$. So, $y^{\prime}=2u\cdot u^{\prime}=2\sin x\cos x=\sin2x$.

Step2: Choose $a$ and find $f(a)$ and $f^{\prime}(a)$

We want to approximate $f(x)$ near a known value. Let $a = \frac{\pi}{4}$. Then $f(\frac{\pi}{4})=\sin^{2}\frac{\pi}{4}=(\frac{\sqrt{2}}{2})^{2}=\frac{1}{2}$, and $f^{\prime}(\frac{\pi}{4})=\sin(2\times\frac{\pi}{4})=\sin\frac{\pi}{2}=1$.

Step3: Write the linear - approximation formula

The linear approximation $L(x)$ of $y = f(x)=\sin^{2}x$ at $x = \frac{\pi}{4}$ 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: Approximate $\sin^{2}(\frac{\pi}{4}+0.01)$

Let $x=\frac{\pi}{4}+0.01$. Then $L(\frac{\pi}{4}+0.01)=\frac{1}{2}+(\frac{\pi}{4}+0.01)-\frac{\pi}{4}$.
Simplify the expression: $L(\frac{\pi}{4}+0.01)=\frac{1}{2}+0.01 = 0.5 + 0.01=0.51$.

Answer:

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