Question 12 (1 point) If f(x)=sinx use the linear approximation of f(x) to evaluate sin((pi )/(6)+(1)/(10)) (10-sqrt (3))/(20) (2)/(5) (9)/(20) (5-sqrt (3))/(10) (10+sqrt (3))/(20)

## Step 1: Find the derivative of $f(x)$<br />### The derivative of $f(x) = \sin(x)$ is $f'(x) = \cos(x)$.<br /><br />## Step 2: Calculate $f(a)$ and $f'(a)$ at $a = \frac{\pi}{6}$<br />### $f(a) = f(\frac{\pi}{6}) = \sin(\frac{\pi}{6}) = \frac{1}{2}$<br />### $f'(a) = f'(\frac{\pi}{6}) = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$<br /><br />## Step 3: Write the linear approximation formula<br />### The linear approximation of $f(x)$ at $x=a$ is given by $L(x) = f(a) + f'(a)(x-a)$. In our case, $a = \frac{\pi}{6}$ and $x = \frac{\pi}{6} + \frac{1}{10}$.<br /><br />## Step 4: Substitute the values and simplify<br />### $L(\frac{\pi}{6} + \frac{1}{10}) = f(\frac{\pi}{6}) + f'(\frac{\pi}{6})(\frac{\pi}{6} + \frac{1}{10} - \frac{\pi}{6})$<br />### $L(\frac{\pi}{6} + \frac{1}{10}) = \frac{1}{2} + \frac{\sqrt{3}}{2}(\frac{1}{10})$<br />### $L(\frac{\pi}{6} + \frac{1}{10}) = \frac{1}{2} + \frac{\sqrt{3}}{20}$<br />### $L(\frac{\pi}{6} + \frac{1}{10}) = \frac{10 + \sqrt{3}}{20}$