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 the 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 $f(a)$ and $f^{\prime}(x)$.
Given $f(x)=\sin^{2}x$, then $f(a)=f(\frac{\pi}{4})=\sin^{2}\frac{\pi}{4}$.
Since $\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}$, $f(\frac{\pi}{4})=(\frac{\sqrt{2}}{2})^{2}=\frac{1}{2}$.

Step2: Differentiate $f(x)$ using the chain - rule

Let $u = \sin x$, then $f(x)=u^{2}$. By the chain - rule $\frac{df}{dx}=\frac{df}{du}\cdot\frac{du}{dx}$.
$\frac{df}{du} = 2u$ and $\frac{du}{dx}=\cos x$, so $f^{\prime}(x)=2\sin x\cos x=\sin2x$.
Then $f^{\prime}(a)=f^{\prime}(\frac{\pi}{4})=\sin(2\times\frac{\pi}{4})=\sin\frac{\pi}{2}=1$.

Step3: Find the linear approximation $L(x)$

Substitute $f(\frac{\pi}{4})=\frac{1}{2}$ and $f^{\prime}(\frac{\pi}{4}) = 1$ into the linear - approximation formula $L(x)=f(a)+f^{\prime}(a)(x - a)$.
$L(x)=\frac{1}{2}+1\times(x-\frac{\pi}{4})=\frac{1}{2}+x-\frac{\pi}{4}$.

Step4: Approximate $\sin^{2}(\frac{\pi}{4}+0.01)$

Let $x=\frac{\pi}{4}+0.01$. Then, using the linear approximation $L(x)$:
$L(\frac{\pi}{4}+0.01)=\frac{1}{2}+(\frac{\pi}{4}+0.01)-\frac{\pi}{4}=\frac{1}{2}+0.01 = 0.51$.

Answer:

(i) The linear approximation is $L(x)=\frac{1}{2}+x - \frac{\pi}{4}$.
(ii) The approximation of $\sin^{2}(\frac{\pi}{4}+0.01)$ is $0.51$.