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 ≤ α_{1} ≤ α_{2} ≤ ...≤ α_{n} ≤ ½π.
Let A = 

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)
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