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

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

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

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

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

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

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

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

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

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