Find the Linearization at a=p/3 f(x)=sin(x) , a=pi/3

$f (x) = \sin (x)$ , $a = \frac{π}{3}$

Consider the function used to find the linearization at $a$ .

$L (x) = f (a) + f' (a) (x - a)$

Substitute the value of $a = \frac{π}{3}$ into the linearization function.

$L (x) = f (\frac{π}{3}) + f' (\frac{π}{3}) (x - \frac{π}{3})$

Evaluate $f (\frac{π}{3})$ .

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Replace the variable $x$ with $\frac{π}{3}$ in the expression.

$f (\frac{π}{3}) = \sin (\frac{π}{3})$

Simplify $\sin (\frac{π}{3})$ .

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Remove parentheses.

$\sin (\frac{π}{3})$

The exact value of $\sin (\frac{π}{3})$ is $\frac{\sqrt{3}}{2}$ .

$\frac{\sqrt{3}}{2}$

Find the derivative and evaluate it at $a = \frac{π}{3}$ .

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The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$ .

$\cos (x)$

Replace the variable $x$ with $\frac{π}{3}$ in the expression.

$\cos (\frac{π}{3})$

The exact value of $\cos (\frac{π}{3})$ is $\frac{1}{2}$ .

$\frac{1}{2}$

Substitute the components into the linearization function in order to find the linearization at $a$ .

$L (x) = \frac{\sqrt{3}}{2} + \frac{1}{2} (x - \frac{π}{3})$

Simplify each term.

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Apply the distributive property.

$L (x) = \frac{\sqrt{3}}{2} + \frac{1}{2} x + \frac{1}{2} \cdot (- \frac{π}{3})$

Combine $\frac{1}{2}$ and $x$ .

$L (x) = \frac{\sqrt{3}}{2} + \frac{x}{2} + \frac{1}{2} \cdot (- \frac{π}{3})$

Multiply $\frac{1}{2} (- \frac{π}{3})$ .

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Multiply $\frac{1}{2}$ and $\frac{π}{3}$ .

$L (x) = \frac{\sqrt{3}}{2} + \frac{x}{2} - \frac{π}{2 \cdot 3}$

Multiply $2$ by $3$ .

$L (x) = \frac{\sqrt{3}}{2} + \frac{x}{2} - \frac{π}{6}$

Write the equation in $y = m x + b$ form.

$L (x) = \frac{1}{2} x + \frac{3 \sqrt{3} - π}{6}$

Find the Linearization at a=p/3 f(x)=sin(x) , a=pi/3

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