Trigonometry

\begin{array}% \cos 3x+\cos x=4 \cos ^{3} x-3 \cos x + \cos x \\ \therefore 4 \cos ^{3} x-2 \cos x=0 \\ \cos x\left(\cos ^{2} x-\frac{1}{2}\right)=0 \end{array}

Either \(\cos x=0\) or \(\cos x = \pm\frac{1}{\sqrt{2}}\).

\begin{align} \cos x&=0 \text { implies } x \in\left(\frac{\pi}{2},-\frac{\pi}{2}\right) \\ \cos x&=\frac{1}{\sqrt{2}} \text { implies } x \in\left(\frac{\pi}{4},-\frac{\pi}{4}\right) \\ \cos x&=-\frac{1}{\sqrt{2}} \text { imples } x \in\left(\frac{3 \pi}{4},-\frac{3 \pi}{4}\right) \end{align}

Therefore, there are six solutions to the given equation.

(a) True.
(b) True. See below.
(c) False. \(g(-v) = - \int_{-v}^0 f(x) dx = g(v)\). So \(g\) is convex.
(d) False. \(g\) is an even function.
Explanation for (b). Let \( \arctan(1)=x ; \arctan (2)=y ;\) and \(\arctan (3)=z\).
\[\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \cdot \tan y}=\frac{1+2}{1-2}=-3\] Put \(u = x+y\) and evaluate: \[\tan (z+u)=\frac{\tan z+\tan u}{1-\tan z \cdot \tan u}=\frac{-3+3}{1-9}=0\] So \( \tan( x+y+z ) = 0\). Hence, \(\arctan (1)+\arctan (2)+\arctan (3)=x+y+z=\pi\).

Since the equation is symmetric, if \(r\) is a solution then \(-r\) is also a solution. Hence, we restrict to finding the positive roots. Note that \(x=0\) is also a root.

The function \(99 \sin \pi x\) is positive when \(2k < x < 2k+1 \). It equals \(x\) when \(x\leq 99\). The positive cycle of \(\sin\pi x\) meets \(y=x\) twice, so there are 198 positive solutions. Counting the zero gives 199 solutions.

We want the number of intersection points between the line \(y=1\) and the periodic function \(f(x) = 4\sin (3x+2) \). The function \(f\) has period \(2\pi/3\). Between every maxima and a minima, there is exactly one solution.

At \(x=0\) and \(x=2\pi\), \(f\) is close to maxima. So all we need to do is find the number of minimas of \(f\) in the given domain and multiply it by 2.

The minimas occur at \(\displaystyle x=\left( \frac{\pi}{2}-\frac{2}{3} + \frac{2\pi k}{3} \right)\) for \( k\in\{0,1,2\} \). Hence, there are six solutions to the given equation.

The graph of \(f\) is periodic with \(f(x)=f(x+2)\). It is defined as follows in the interval \((0,2)\).

\[ f(x)= \begin{cases} \pi x & 0\leq x \leq 0.5 \\ \pi-\pi x & 0.5 \leq x \leq 1.5 \\ \pi x & 1.5 \leq x \leq 2 \end{cases} \]

We can answer the questions by consulting the graph:

1. \(f(2.7)=f(0.7)=\pi- 0.7\pi=0.3\pi\)
2. \(f^{\prime}(2.7)=-\pi\)
3. \(\int_{0}^{2.5} f(x) d x=\int_{0}^{0.5} f(x) dx=\pi / 8\)
4. the smallest positive \(x\) at which \(f^{\prime}(x)\) does not exist is \(x=1/2\)

From the graph, we see that only the last statement is false.

A solution without using the hint.

Let \( x = \frac{\pi}{101} \).

Since \( \cos^2x = (\cos 2x + 1 )/2 \), the given expression is:

\[ S = \frac{n}{2} + \frac{1}{2}\cdot \left( \cos 2x + \cos 4x + \cdots + \cos 2nx \right)\quad \text{where } n=100 \]

The following identity is useful:

\[ c=\cos 2 x+\cos 4 x+\cdots+\cos 2 n x=\frac{\sin n x \cos (n+1) x}{\sin x} \]

We can prove this as follows:

\(2 \sin x(\cos 2 x+\cos 4 x+\cdots+\cos 2 n x) \) \(=[\sin 3 x-\sin x]+[\sin 5 x-\sin 3 x]+\cdots+[\sin (2 n+1) x-\sin (2 n-1) x]\) \(=\sin (2 n+1) x-\sin x=2 \sin n x \cos (n+1) x \)

With this simplification the original expression becomes:

Squaring both the equations and adding gives:

The maximum value of \(\cos \theta\) is 1. So, the above equation can be satisfied with equality only when each of the terms is 1. This implies that the only solutions are those that are given in the problem.

Altenatively, we can use complex numbers and triangle inequality to prove the statement

Equalit holds only when \(\theta_i\)s are multiples of \( 2\pi \), a contradiction.

We look at \(\cos n A\) as the real part of \(e^{i n A}\).

\begin{align} S &=\sum_{n=0}^{8} \cos n A-\sum_{n=1}^{2} \cos (3 n A) \\ &=\operatorname{Re}\left(\sum_{n=0}^{8} e^{i n A}-\sum_{n=1}^{2} e^{i n \frac{2 \pi}{3}}\right) \\ &=\operatorname{Re}\left(0-\omega-\omega^{2}\right) \quad \text{where } \omega \text{ is a complex cube root of unity} \\ &=1 \end{align}

The RHS of equations hints that a right angle triangle is involved. The LHS expressions resemble the cosine formulas. All we have to do is find an interpretation for \(x,y\) and \(z\).

Consider the right angled triangle \(ABC\) with sides 3,4,5 and an interior point \(O\) such that \(A O=x, \angle A O B=90\) and \(C O=z, \angle C O A=150\) and \(B O=\) \(y, \angle B O C=120\). Then the three given equations are in fact cosine rule for each of the triangle prescribed above.

For example, in \(\Delta BOC\) we have:

If we divide the expression \(R\) by 4, we get: \[ \frac{1}{2}xy\sin 90+\frac{1}{2}xz\sin 60 + \frac{1}{2}yz\sin 60 \] The three terms are the areas of the interior triangles which should be equal to the area of triangle \(ABC\). \[ 6=\frac{1}{2} x y+\frac{1}{2} y z \sin 60+\frac{1}{2} \sin 30 \] So the answer is 24.