## Tuesday, April 23, 2024

### 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

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 .
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.