*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+ 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

*is any integer number.*

**N***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)

cos(x)·cos(y) − sin(x)·sin(y)

and

*cos(x−y) =*

cos(x)·cos(y) + sin(x)·sin(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

*has two solutions for*

**cos(x+y)=−1/2***within an interval [*

**(x+y)***0,2π*]:

**x + y = ±2π/3**The equation

*has one solution for*

**cos(x−y)=1***within an interval [*

**(x−y)***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·**Nwhere

*M*and

*N*are any integers, and

**x + y = −2π/3 + 2π·**M

**x − y = 2π·**NEach one of these systems can be easily solved by adding and subtracting the equations, which leads to the first series of solutions

**x**(M+N)_{1}= π/3 + π·

**y**(M−N)_{1}= π/3 + π·and the second series of solutions

**x**(M+N)_{2}= −π/3 + π·

**y**(M−N)_{2}= −π/3 + π·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.

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