Trigonometry

Impossible polynomial

B4, 2021

Show that there is no polynomial such that for every .

Solution

Put

and

. Then

and

which cannot be the case.

An easy problem

A1, 2010

Find all such that .

Solution

Either or .

Therefore, there are six solutions to the given equation.

Approximations help

A5, 2022

Which of these are true?

Solution

True.

True. Can solve this problem by using the on-screen calculator. Alternatively, you may use Taylor expansion.

False.

True.

Three arctans

A10, 2021

Let and .
(a)
(b)
(c) is an increasing function in .
(d) is an odd function.

Solution

(a) True.
(b) True. See below.
(c) False. . So is convex.
(d) False. is an even function.
Explanation for (b). Let and .
Put and evaluate: So . Hence, .

Intersection of a line and periodic function I

A1, 2012

Find the number of real solutions to the equation .

Solution

Since the equation is symmetric, if is a solution then is also a solution. Hence, we restrict to finding the positive roots. Note that is also a root.

The function is positive when . It equals when . The positive cycle of meets twice, so there are 198 positive solutions. Counting the zero gives 199 solutions.

Intersection of a line and periodic function II

A3, 2011

Find the number of solutions to the equation in the domain .

Solution

We want the number of intersection points between the line and the periodic function . The function has period . Between every maxima and a minima, there is exactly one solution.

At and , is close to maxima. So all we need to do is find the number of minimas of in the given domain and multiply it by 2.

The minimas occur at for . Hence, there are six solutions to the given equation.

A Saw-tooth function

A9, 2015

Let . Write the values of the following.

The smallest positive at which does not exist.

Solution

The graph of is periodic with . It is defined as follows in the interval .

We can answer the questions by consulting the graph:

the smallest positive at which does not exist is

Repeated saw-tooth

A10, 2018

Recall that arcsin is a function with domain [-1,1] and range Consider the function and answer the following questions as a series of four letters ( for True and for False) in order.

The function is well defined for all real numbers .
The function is continuous wherever it is defined.
The function is differentiable wherever it is continuous.

Solution

The problem is nearly the same as the previous problem from 2015 paper.

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

Use of telescoping

A5, 2016

Find the value of the following sum of 100 terms.

You may use this hint: also consider the same sum with instead of .)

Solution

A solution without using the hint.

Let .

Since , the given expression is:

The following identity is useful:

We can prove this as follows:

With this simplification the original expression becomes:

Trignometric triangle inequality

A10, 2010

Consider the following equations:

Show that for some .

Solution I

Solution due to Piyush Jha.

For any real numbers

and

, we know from Cauchy-Swarchz inequality that

and equality holds if and only if there exists a nonzero constant

such that for all

.
Applying Cauchy-Swarchz inequality on

and on

we get

Now applying same on

and

gives:

which implies that

, that means Cauchy-Schwarz inequality is tight and there must exist a

such that

. From the problem statement

and the solution follows.

Solution II

Squaring both the equations and adding gives:

The maximum value of 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

Solution III

Solution due to Aryan Komarla.

We will prove by contradiction. Let

and

, where

, and

are not integer multiples of

. Then, from the given equations, we get:

Multiplying the first equation by

and adding, we get,

Equalit holds only when s are multiples of , a contradiction.

Roots of unity to rescue

B1, 2017

Let i.e., degrees. Calculate the following

Solution

We look at as the real part of .

Geometric interpretation of algebraic expressions

B5, 2019

Three positive real numbers satisfy

Find the value of

Solution

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 and .

Consider the right angled triangle with sides 3,4,5 and an interior point such that and and . Then the three given equations are in fact cosine rule for each of the triangle prescribed above.

For example, in we have:

If we divide the expression by 4, we get: The three terms are the areas of the interior triangles which should be equal to the area of triangle . So the answer is 24.

This is an artificial problem. Students are expected to play guess-the-pattern game with the examiners.