Unizor | Creative Mind through Art of Mathematics
376 FOLLOWERS
Unizor is a site where students can learn the high school math (and in the future some other subjects) in a thorough and rigorous way. It allows parents to enroll their children in educational programs and to control the learning process.
Unizor | Creative Mind through Art of Mathematics
2d ago
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 sy ..read more
Unizor | Creative Mind through Art of Mathematics
4d ago
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 ..read more
Unizor | Creative Mind through Art of Mathematics
6d ago
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)) +
+ si ..read more
Unizor | Creative Mind through Art of Mathematics
1w ago
Notes to a video lecture on http://www.unizor.com
Trigonometry+ 05
Problem A
Given ∠α, ∠β and ∠γ are acute angles of a triangle.
Evaluate the expression
tan(α)·tan(β) + tan(β)·tan(γ) +
+ tan(γ)·tan(α)
Hint A
α + β + γ = π/2
Solution A
γ = π/2 − (α+β)
tan(γ) = sin(γ)/cos(γ) =
= cos(π/2−γ)/sin(π/2−γ) =
= cos(α+β)/sin(α+β) =
= [cos(α)·cos(β)−sin(α)·sin(β)]/
/[sin(α)·cos(β)+cos(α)·sin(β)] =
divide both numerator and denominator by cos(α)·cos(β)
= [1−tan(α)·tan(β)]/
/[tan(α)+tan(β)]
Therefore,
tan(α)·tan(β) + tan(β)·tan(γ) +
+ tan(γ)·tan(α) =
= tan(α)·tan(β) +
+ [tan(α)+tan(β)]·tan(γ) Substi ..read more
Unizor | Creative Mind through Art of Mathematics
2w ago
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 ..read more
Unizor | Creative Mind through Art of Mathematics
3w ago
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 ..read more
Unizor | Creative Mind through Art of Mathematics
1M ago
Notes to a video lecture on http://www.unizor.com
Geometry+ 07
Problem A
Given any circle with a center at point O, its diameter MN and any point P on this circle not coinciding with the ends M, N of a given diameter.
Let point Q be a projection of point P on a diameter MN.
This point Q divides diameter MN into two parts:
MQ = a and QN= b
Prove that
(1) Radius of a OP circle is an arithmetic average of a and b.
(2) Projection segment PQ is a geometric average of a and b.
(3) Based on these proofs, conclude that geometric average of two non-negative real numbers is less or equal to their a ..read more
Unizor | Creative Mind through Art of Mathematics
1M ago
Notes to a video lecture on http://www.unizor.com
Logic+ 06
Problem A
There are 5 towns.
Some of them are connected by direct roads, that is by roads not going through other towns.
It's known that among any group of 4 towns out of these 5 there is always one town connected by direct roads with each of the other 3 towns of this group.
Prove that there is at least one town connected with all 4 others by direct roads.
Proof A
Choose any 4 towns from given 5 as the first group towns.
One of these towns is connected to 3 others, as the problem states. Let's call this town A and the others wi ..read more
Unizor | Creative Mind through Art of Mathematics
1M ago
Notes to a video lecture on http://www.unizor.com
Geometry+ 06
Problem A
Given an isosceles triangle ΔABC with AB=BC and ∠ABC=20°.
Point D on side BC is chosen such that ∠CAD=60°.
Point E on side AB is chosen such that ∠ACE=50°.
Find angle ∠ADE.
Solution A
Problem B
Given two triangles ΔA1B1C1 and ΔA2B2C2 with the following properties:
(a) side A1B1 of the first triangle equals to side A2B2 of the second;
(b) angles opposite to these sides, ∠A1C1B1 and ∠A2C2B2, are equal to each other;
(c) bisectors of these angles, C1X1 and C2X2, are also equal to each other.
Prove that these triang ..read more
Unizor | Creative Mind through Art of Mathematics
1M ago
Notes to a video lecture on http://www.unizor.com
Trigonometry+ 04
Problem A
Find the sums
Σk∈[0,n−1]sin²(x+k·π/n)
Solution A
First, convert sin²(...) into cos(...) by using the identity
cos(2φ) = cos²(φ) − sin²(φ) =
= 1 − 2sin²(φ)
from which follows
sin²(φ) = ½(1 − cos(2φ))
Now our sum looks like this
Σk∈[0,n−1]
½(1−cos(2x+2k·π/n)) =
= n/2 −
− ½Σk∈[0,n−1]cos(2x+2k·π/n)
To calculate Σk∈[0,n−1] above, let's use the result of the previous lecture Trigonometry 03 that proved the following
Σk∈[0,n]cos(x+k·y) =
= [sin(x+(2n+1)·y/2) −
− sin(x−y/2)] /
/ [2·sin(y/2 ..read more