Trigonometry+ 08: UNIZOR.COM - Math+ & Problems - Trigonometry

Notes to a video lecture on http://www.unizor.com

Trigonometry+ 08


Problem A

Prove the following inequality
cos(36°) ≥ tan(36°)

Hint A
Find the point where left side equals to the right side and compare it with 36°.
Use the following values:
π/5≅0.628 and
arcsin(½(√5−1))≅0.666.


Problem B

Solve the equation
a·sin²(x) + b·sin(x)·cos(x) +
+ c·cos²(x) = d
where a ≠ d.

Hint B
For the right side of this equation use the identity
sin²(x) + cos²(x) = 1

Answer B
x = arctan{R/[2(a−d)]}+π·N
where
R=[−b±√b²−4·(a−d)·(c−d)]
and N is any integer number.


Problem C

Solve the following system of equations
tan(x)·tan(y) = 3
sin(x)·sin(y) = 3/4

Hint C
Convert this system into form
cos(x+y)=...
cos(x−y)=...

Solution C
Since tan()=sin()/cos(), substitute the second equation's value 3/4 into numerator of the first and invert the fraction
(3/4)/[cos(x)·cos(y)]=3
cos(x)·cos(y) = 1/4
As we know, cos(x+y) =
cos(x)·cos(y) − sin(x)·sin(y)
and cos(x−y) =
cos(x)·cos(y) + sin(x)·sin(y)
Since sin(x)·sin(y) = 3/4
and cos(x)·cos(y) = 1/4
we can find
cos(x+y) = 1/4 − 3/4 = −1/2
cos(x−y) = 1/4 + 3/4 = 1
Function cos() is periodical with a period of 2π.
The equation cos(x+y)=−1/2 has two solutions for (x+y) within an interval [0,2π]:
x + y = ±2π/3
The equation cos(x−y)=1 has one solution for (x−y) within an interval [0,2π]:
x − y = 0
Adding periodicity, we come up with two systems of equations, each depending on some integer parameters
x + y = 2π/3 + 2π·M
x − y = 2π·N
where M and N are any integers, and
x + y = −2π/3 + 2π·M
x − y = 2π·N
Each one of these systems can be easily solved by adding and subtracting the equations, which leads to the first series of solutions
x₁ = π/3 + π·(M+N)
y₁ = π/3 + π·(M−N)
and the second series of solutions
x₂ = −π/3 + π·(M+N)
y₂ = −π/3 + π·(M−N)
In both series M and N can independently take any integer value.

Note C
Since original system of equation contained tan(x) and tan(y), we have to make sure that by getting rid of cos() in the denominator we have not added extraneous solutions.
Function cos() is zero at π/2+π·K, where K can be any integer number. If any of our solutions falls in this set, it must be excluded. Fortunately, none of our solutions coincides with this set.

Trigonometry+ 07: UNIZOR.COM - Math+ & Problems - Trigonometry

Notes to a video lecture on http://www.unizor.com

Trigonometry+ 07


Problem A

Prove the following identity
2·arccos[√(1+x)/2] = arccos(x)

Proof A

By definition of function arccos(x), it's an angle in interval [0,π], whose cosine is x.
That is, cos(arccos(x))=x.
Therefore, we have to prove that cosine of the left side of an equality above equals to x.
Let's use a known identity
cos(2α)=2cos²(α)−1
Now
cos{2·arccos[√(1+x)/2]} =
= 2cos²{arccos[√(1+x)/2]}−1 =
= 2·[√(1+x)/2]² = x


Problem B
Simplify the expression
tan[½arctan(x)]

Hint B
Express tan(φ/2) in terms of tan(φ).

Solution B
Let angle φ = arctan(x).
We know that, by definition of function arctan(), its domain is all real values of x, its value, angle φ, is in the interval from −π/2 to π/2 and tan(φ)=x.
Hence, this problem can be formulated as
If a tangent of an angle φ is x, what is the tangent of the half of this angle?
Our first task is to express a tangent of the half of an angle in terms of a tangent of the whole angle.
As we know from UNIZOR.COM - Math 4 Teens - Trigonometry - Sum of Angles Problem 1 : tan(φ/2),
tan(φ)=2tan(φ/2)/[1−tan²(φ/2)]
Let's resolve this equation for tan(φ/2) in terms of tan(φ)
If A = 2B/(1−B²) then
A·B² + 2·B −A = 0
Solving this for B, we obtain
B_1,2 = (−2±√4+4A²)/2A =
= (−1±√1+A²)/A
Using this for A=tan(φ) and B=tan(φ/2), we get the expression for tan(φ/2) in terms of tan(φ).
An important detail is that on interval (−π/2,π/2) the sign of tan(φ) and tan(φ/2) are the same (positive for positive angle and negative for negative angle).
Therefore, sign ± in the formula above should be replaced with + and the final formula expressing tan(φ/2) in terms of function tan(φ) is
tan(φ/2) =
= [−1+√1+tan²(φ)] /tan(φ)
Since φ=arctan(x) and tan(φ)=x, we can state the following:
tan[½arctan(x)] =
= [−1+√1+x²] /x
The value x=0 should be excluded from this formula, we can, obviously, say that in this case
tan[½arctan(x)] = 0


Problem C
Prove geometrically that for an acute angle θ, measured in radians, the following inequalities are true:
sin(θ) ≤ θ ≤ tan(θ)

Hint
Use the unit circle and a geometric interpretation of components of the inequality to be proven.

Proof

Since the radius of a circle is 1, the length of an arc AB is θ - a measure in radians of an angle ∠AOB.
By definitions of sin() and cos(), abscissa of point A (segment OD) equals to cos(θ) and the ordinate of point A (segment AD) equals to sin(θ).
Comparing the length of segment AD (that is, sin(θ)) and arc AB (that is, θ), taking into consideration that AD is a perpendicular to radius OB, while arc AB is a curve from the same original point A to OB, we conclude that the length of AD is less or equal to the length of arc AB with equality held only if point A coincides with point B, that is when angle θ is zero.
Therefore, we have proven that sin(θ) is less or equal to θ.
By definition of function tan(), it's a ratio of sin() to cos().
In our case it's a ratio of AD to OD.
Draw a perpendicular to OB at point B, and let C be an intersection of this perpendicular with a continuation of OA.
Since ΔOAD is similar to ΔOCB, the ratio of AD to OD is the same as the ratio of CB to OB. Since OB=1, tan(θ)=CB.
Area of circular sector AOB is less than area of right triangle COB.
Area of circular sector AOB of radius 1 equals to the area of a circle (that is, π) times θ/(2π), that is ½θ.
Area of a right triangle COB equals to ½CB·OB=½tan(θ).
Comparing these two areas, we conclude that θ is less or equal to tan(θ) with equality held only if point C coincides with point B, that is when angle θ is zero.

Trigonometry+ 06: UNIZOR.COM - Math+ & Problems - Trigonometry

Notes to a video lecture on http://www.unizor.com

Trigonometry+ 06


Problem A

Given ∠α, ∠β and ∠γ are acute angles of a triangle.
Prove that
cos(α)+cos(β)+cos(γ) ≤ 3/2

Hint A
Using α + β + γ = π
reduce the left side of the inequality to a function of sin(½γ).

Solution A
Recall the transformation of a sum of two cosines into a product of other cosines
cos(α)= cos[½(α+β)+½(α−β)]=
= cos(½(α+β))·cos(½(α−β)) −
− sin(½(α+β))·sin(½(α−β))
cos(β)= cos[½(α+β)−½(α−β)]=
= cos(½(α+β))·cos(½(α−β)) +
+ sin(½(α+β))·sin(½(α−β))
Therefore,
cos(α) + cos(β) =
= 2·cos(½(α+β))·cos(½(α−β)) =
= 2·cos(½(π−γ))·cos(½(α−β)) =
= 2·cos(½π−½γ))·cos(½(α−β))
But cos(½π−½γ) = sin(½γ).
Also, cos(½(α−β)) ≤ 1.
Therefore,
cos(α) + cos(β) ≤
≤ 2·sin(½γ))
Hence,
cos(α)+cos(β)+cos(γ) ≤
≤ 2·sin(½γ)) + cos(γ) =
= 2·sin(½γ)) + cos(2·½γ) =
= 2·sin(½γ)) + 1 − 2·sin²(½γ) =
= −2X² + 2X + 1
where X=sin(½γ)
The quadratic function −2X²+2X+1 has a maximum of 3/2 at X=½.


Problem B

Given
0 ≤ α₁ ≤ α₂ ≤ ...≤ α_n ≤ ½π.

Let A =

Σi∈[1,n]sin(αi) Σi∈[1,n]cos(αi)

Prove: tan(α₁) ≤ A ≤ tan(α_n)


Problem C

Solve the equation
sin(A·x) + sin(B·x) = 0
where A and B are some real numbers.

Hint C
Convert the left side of an equation into a product.

Solution
Recall the transformation of a sum of two sines into a product of other trigonometric functions
sin(α)= sin[½(α+β)+½(α−β)]=
= sin(½(α+β))·cos(½(α−β)) +
+ cos(½(α+β))·sin(½(α−β))
sin(β)= sin[½(α+β)−½(α−β)]=
= sin(½(α+β))·cos(½(α−β)) −
− cos(½(α+β))·sin(½(α−β))
Therefore,
sin(α) + sin(β) =
= 2·sin(½(α+β))·cos(½(α−β))
Using this for α=A·x and β=B·x, the left side of our equation can be invariantly transformed into
2·sin(½(A+B)x)·cos(½(A−B)x)
It can be equal to zero if
sin(½(A+B)x) = 0
from which follows
½(A+B)x = π·N
x = 2πN/(A+B)
(for A≠−B and where N is any integer)
or if
cos(½(A−B)x) = 0
from which follows
½(A−B)x = ½π+π·N
x = π·(2πN+1)/(A−B)
(for A≠B and where N is any integer)

Algebra+ 06: UNIZOR.COM - Math+ &Problems - Algebra

Notes to a video lecture on http://www.unizor.com

Algebra+ 06

Problem A

Prove that sum of square roots of 2, 3 and 5 is an irrational number.

Hint A
Assume, this sum is rational, that is
√2 + √3 + √5 = p/q
where p and q are integer numbers without common divisors (if they do, we can reduce the fraction by dividing a numerator p and denominator q by a common divisor without changing the value of a fraction).
Then simplify the above expression by getting rid of square roots and prove that p must be an even number and, therefore, can be represented as p=2r.
Then prove that q must be even as well, and, therefore, p and q have a common divisor 2, which we assumed they don't.


Problem B

Given a polynomial represented as a product
P = (5−4x)¹⁰⁰⁰·(3x−4)¹⁰⁰¹
Assume, we open all the parenthesis and combine all similar terms to express this polynomial in a canonical form
P = Σ_n∈[0,2001]A_n·x²⁰⁰¹⁻ⁿ
What would be a sum of all the coefficients A_n?

Hint B:
Do not attempt to use Newton's binomial and find the answer by explicitly performing all the operations to convert the given expression into canonical polynomial form.
There is a better and very quick way.

Answer B:
Sum of all the coefficients will be equal to −1.


Problem C

Find all prime integer x and y, for which the following is true
13·(x + y) = 3·(x² − x·y + y²)

Hint C
Express the given equation as a quadratic equation for x with coefficients as functions of y.
Then, to have solutions for x, a discriminant must be non-negative, which reduces the possible values for y to be in an interval (0,10).

Answer C
Only a pair of numbers 2 and 7 satisfies the condition of this problem.
So, the solutions are (x=2,y=7) or (x=7,y=2).

Algebra+ 05: UNIZOR.COM - Math+ & Problems- Algebra

Notes to a video lecture on http://www.unizor.com

Algebra+ 05

Problem A

Given a system of two equations with three unknown variables x, y and z:
x + y + z = A
x⁻¹ + y⁻¹ + z⁻¹ = A⁻¹
Prove that one of the unknown variables equals to A.

Hint A

System of equations
x + y = p
x · y = q
fully defines a pair of numbers (generally speaking, complex numbers) as solutions to a quadratic equation
X² − p·X + q.
Indeed, if X₁ and X₂ are the solution of the equation, then, according to the Vieta's Theorem,
X₁ + X₂ = −(−p) = p and
X₁ · X₂ = q
(See a lecture Math 4 Teens - Algebra - Quadratic Equations - Lecture on UNIZOR.COM)

The only unresolved issue is:
which unknown variable takes which value from a pair
is it (x=X₁,y=X₂) or (x=X₂,y=X₁).

From this follows that, if the following system of equations is given
x + y = a + b
x · y = a · b
then either (x=a,y=b)
or (x=b,y=a).

Proof A

x + y = A − z
x⁻¹ + y⁻¹ = A⁻¹ − z⁻¹
None of the unknown variables can be equal to zero, since each is represented in the second equation in the denominator.
Therefore, we can multiply the second equation by x·y getting
y + x = x·y/A − x·y/z

Using the first equation, substitute x+y into the second getting a system of equations
x + y = A − z
A − z = x·y·(1/A − 1/z)
or
x + y = A − z
x·y = (A−z)/(1/A − 1/z)
or
x + y = A − z
x·y = −A·z
or
x + y = A + (−z)
x·y = A·(−z)
Therefore, either x=A, y=−z
or y=A, x=−z.


Problem B

Prove that
(x + y)⁴ ≤ 8·x⁴ + 8·y⁴

Proof B

Let's start with analysis of this problem.
Assume, this inequality (call it "statement A") is true and make invariant (reversible and equivalent) transformations to it, trying to get to an obviously true statement B.
Then, using the fact that our transformations were invariant, we can say that we can start with obviously true statement B and, using the reverse transformations, derive statement A, that is we will prove that A is true.

Notice that, if we divide both sides of this inequality by y⁴ and assign t=x/y, we will reduce the number of variables from two to one, which seems to simplify the task.
Dividing by positive y⁴ is an invariant transformation of an inequality, except a case of y=0. The case of y=0 can be considered separately, and in this case the inequality is obviously true since x⁴ ≤ 8·x⁴.

After dividing by y⁴ and substituting t=x/y the new inequality looks like
(t + 1)⁴ ≤ 8t⁴ + 8
which seems to be simpler to prove.

Let's open the parenthesis and bring all items to one side of an inequality - obviously invariant transformation
7t⁴ − 4t³ − 6t² − 4t + 7 ≥ 0
Notice that the sum of coefficients of a polynomial on the left is zero. That means that t=1 is a root of this polynomial, that is it's equal to zero for t=1.

Recall the Fundamental Theorem of Algebra (see Math 4 Teens course on UNIZOR.COM, menu items Algebra - Fundamental Theorem of Algebra and its Corollary 1) that states that if x=a is a root of a polynomial P⁽ⁿ⁾(x) of n^th degree, then this polynomial is divisible by x−a, that is
P⁽ⁿ⁾(x) = (x−a)·Q⁽ⁿ⁻¹⁾(x)
where Q⁽ⁿ⁻¹⁾(x) is a polynomial of a degree lower by 1 than P⁽ⁿ⁾(x).

Therefore, since t=1 is a root of the polynomial of the 4th degree above, we can represent that polynomial as (t−1) multiplied by another polynomial of the 3rd degree.
7t⁴ − 4t³ − 6t² − 4t + 7 =
= (t − 1)·(7t³ + 3t² − 3t − 7)

Consider the polynomial
7t³ + 3t² − 3t − 7
The sum of its coefficient is zero too. Therefore, we can represent it as a product of (t−1) and a polynomial of the second degree
7t³ + 3t² − 3t − 7 =
= (t − 1)·(7t² + 10t + 7)

So, the inequality we have to prove was transformed into this one:
(t − 1)²·(7t² + 10t + 7) ≥ 0
In this inequality the member (t−1)² is always greater or equal to zero.
Quadratic polynomial 7t²+10t+7 has discriminant Δ=10²−4·7·7=−96, which is negative and, consequently, it has no roots, it's always not equal to zero.
It can only be greater than zero since the coefficient at t² is positive.
Therefore, this polynomial is always greater than zero.
That concludes the analysis of our problem.

The proof proper is to start from an obviously truthful statement
(t − 1)²·(7t² + 10t + 7) ≥ 0
and transform it into
(t + 1)⁴ ≤ 8t⁴ + 8
Replacing t with x/y (recall, a trivial case y=0 was already checked, so now we assume that y≠0) and multiplying by y⁴ finishes the proof.


Problem C

Prove the following inequality
x¹² − x⁹ + x⁴ − x + 1 > 0

Proof C

Consider a polynomial
x¹² − x⁹ + x⁴ − x
It can be invariantly transformed into
x⁹·(x³ − 1) + x·(x³ − 1) or
(x³ − 1)·(x⁹ + x) or
(x³ − 1)·x·(x⁸ + 1)

This polynomial has only two roots: x=0 and x=1
As easily checked, values outside interval (0,1) are non-negative and inside this interval the values of a polynomial are negative.
Since we are interested in the values of this polynomial +1, the only interval where it's not obvious whether after adding 1 it is positive or not is inside the interval (0,1).

Inside interval (0,1)
x¹² − x⁹ + x⁴ − x + 1 =
= x¹² + x⁴·(1−x⁵) + (1−x)
with every item in parenthesis and every other participant in the above expression is positive, which results in a positive value of an entire expression.

End of proof.