*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= 2cos²

*[*

**arccos***]}*

**√(1+x)/2***[*

**−1 =**

= 2·= 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 B*

Let angle

*.*

**φ = arctan(x)**We know that, by definition of function

*arctan()*, its domain is all real values of

*x*, its value, angle

*φ*, is in the interval from

*−π/2*to

*π/2*and

*tan(φ)=x*.

Hence, this problem can be formulated as

*If a tangent of an angle*

**φ**is**x**, what is the tangent of the half of this angle?Our first task is to express a tangent of the half of an angle in terms of a tangent of the whole angle.

As we know from

*UNIZOR.COM - Math 4 Teens - Trigonometry - Sum of Angles Problem 1 : tan(φ/2)*,

*[*

**tan(φ)=2tan(φ/2)/***]*

**1−tan²(φ/2)**Let's resolve this equation for

*tan(φ/2)*in terms of

*tan(φ)*

If

*A = 2B/(1−B²)*then

*A·B² + 2·B −A = 0*

Solving this for

*B*, we obtain

**B**_{1,2}**= (−2±√4+4A²)/2A =**

= (−1±√1+A²)/A= (−1±√1+A²)/A

Using this for

*and*

**A=tan(φ)***, we get the expression for*

**B=tan(φ/2)***in terms of*

**tan(φ/2)***.*

**tan(φ)**An important detail is that on interval (

*−π/2,π/2*) the sign of

*tan(φ)*and

*tan(φ/2)*are the same (positive for positive angle and negative for negative angle).

Therefore, sign ± in the formula above should be replaced with + and the final formula expressing

*in terms of function*

**tan(φ/2)***is*

**tan(φ)***[*

**tan(φ/2) =**

==

*]*

**−1+√1+tan²(φ)**

**/tan(φ)**Since

*and*

**φ=arctan(x)***, we can state the following:*

**tan(φ)=x***[*

**tan***]*

**½arctan(x)***[*

**=**

==

*]*

**−1+√1+x²**

**/x**The value

*should be excluded from this formula, we can, obviously, say that in this case*

**x=0***[*

**tan***]*

**½arctan(x)**

**= 0***Problem C*

Prove geometrically that for an acute angle

*, measured in radians, the following inequalities are true:*

**θ**

**sin(θ) ≤ θ ≤ tan(θ)***Hint*

Use the unit circle and a geometric interpretation of components of the inequality to be proven.

*Proof*

Since the radius of a circle is

*, the length of an arc*

**1***is*

**AB***- a measure in radians of an angle ∠*

**θ***.*

**AOB**By definitions of

*sin()*and

*cos()*, abscissa of point

*(segment*

**A***) equals to*

**OD***and the ordinate of point*

**cos(θ)***(segment*

**A***) equals to*

**AD***.*

**sin(θ)**Comparing the length of segment

*(that is,*

**AD***) and arc*

**sin(θ)***(that is,*

**AB***), taking into consideration that*

**θ***is a perpendicular to radius*

**AD***, while arc*

**OB***is a curve from the same original point*

**AB***to*

**A***, we conclude that the length of*

**OB***is less or equal to the length of arc*

**AD***with equality held only if point*

**AB***coincides with point*

**A***, that is when angle*

**B***is zero.*

**θ**Therefore, we have proven that

*is less or equal to*

**sin(θ)***.*

**θ**By definition of function

*tan()*, it's a ratio of

*sin()*to

*cos()*.

In our case it's a ratio of

*to*

**AD***.*

**OD**Draw a perpendicular to

*at point*

**OB***, and let*

**B***be an intersection of this perpendicular with a continuation of*

**C***.*

**OA**Since Δ

*is similar to Δ*

**OAD***, the ratio of*

**OCB***to*

**AD***is the same as the ratio of*

**OD***to*

**CB***. Since*

**OB***,*

**OB=1***.*

**tan(θ)=CB**Area of circular sector

*is less than area of right triangle*

**AOB***.*

**COB**Area of circular sector

*of radius*

**AOB***equals to the area of a circle (that is,*

**1***) times*

**π***, that is*

**θ/(2π)***.*

**½θ**Area of a right triangle

*equals to*

**COB***.*

**½CB·OB=½tan(θ)**Comparing these two areas, we conclude that

*is less or equal to*

**θ***with equality held only if point*

**tan(θ)***coincides with point*

**C***, that is when angle*

**B***is zero.*

**θ**