The tangent to graph of $y=\cos(x)$ is drawn at $x=\frac\pi6$. This tangent intersects graph of $y=\cos(x)$ again at $x=k$. Determine value of $k$

Question

The tangent to the graph of $y=\cos(x)$ is drawn at $x=\frac\pi6$. This tangent intersects the graph of $y=\cos(x)$ again at $x=k$.

a. Determine the equation of the tangent at $x=\frac \pi6$.

b. Determine the value of $k$, correct to three decimal places.

c. Write an expression for the area enclosed between the tangent at $x=\frac\pi6$ and the curve $y=\cos(x)$.

d. Hence evaluate the area correct to two decimal places.

I could solve part a. The equation of the tangent at $x=\frac\pi6$ is: $$y= -\frac{x}{2} + \frac {\sqrt 3}{2} + \frac{\pi}{12}$$

I am stuck at part b. We need to find where the tangent cuts the $y=\cos(x)$ curve at another point. I could find a graphical solution from Desmos but I need the analytical solution.

Answer 1

The intuition behind the Newton algorithm is that if You want to solve an equation $f(x)=0$ for a real valued differentiable function $f:D\rightarrow\mathbb{R}$ where in our case $D=\mathbb{R}$ and $f(x)=\cos(x)+\frac{x}{2}-\frac{\sqrt{3}}{2}-\frac{\pi}{12}$ and You guessed a zero $x_0\in\mathbb{R}$, then the zero of the tangent in $(x_0,f(x_0))$, that is where the tangent intersects the $x$-axis, is even closer to the real zero, since the tangent is a local approximation of the function $f$. This zero is easily computed to be: $$x_1=x_0-\frac{f(x_0)}{f'(x_0)}.$$ Now this can be iterated of course.

After trying to do this by hand and noticing I made a mistake I wrote a Python code:

import math
def f(x):   
    return math.cos(x)+0.5*x-0.5*math.sqrt(3)-math.pi/12    
def derivative_f(x):   
    return -math.sin(x)+0.5
x=math.pi+math.pi/6
n=1
while n<6:
    x=x-f(x)/derivative_f(x)  
    n=n+1
    print(x)

Sorry for the sloppy presentation, which isn't sloppy anymore thanks to Trebor. However the output is: $$3.826445909962073$$ $$3.816833801086834$$ $$3.816801916445538$$ $$3.816801916092871$$ $$3.816801916092871$$ There is no change for further iterations in the first fifteen decimal places. It is certainly possible to do these five iterations by hand, just after two You are already correct to three decimal places. I chose as a starting value $x_0=\pi+\frac{\pi}{6}$.