1/05/2023

Arc Length, Radius, Radians, and Trigonometric

 1.  Find the length of an arc in a circle of radius 10 centimeters subtended by the central angle of 50°

An arc length S is the length of the curve along the arc. Just as the full circumference of a circle always has a constant ratio to the radius, the arc length produced by any given angle also has a constant relation to the radius, regardless of the length of the radius. The radian measure is also depends only on the angle. A full revolution (360°) equals 2π radians. A half revolution (180°) is equivalent to π radians.

Convert the 50° to radians 
  \frac{50^\circ }{360^\circ} = \frac{ \theta}{2 \pi }
  \theta= \frac{5 \pi }{18}
 S=r \theta=10( \frac{5 \pi }{18}) = \frac{25 \pi }{9} \approx8.72664



2.  Graph f(x)=x\ sin\ x on [-4π, 4π] and verbalize how the graph varies from the graphs of f(x)= \pm x




A function that has the same general shape as a sine or cosine function is known as a sinusoidal function. One of the general forms of sinusoidal functions is y = Asin(Bx − C) + D, where A represents its Amplitude, B represents its Periods, C and D represents its shifts. In this case,  f(x)=xsinx  has only inconstant Amplitude X. Therefore, this function is expected to be a function that has the same period of  y=sinx   but with larger and larger Amplitudes. And, since the function  y=sinx  is repeating its values between the interval [-1, 1], when  sinx=1   or  sinx=-1  , the function  y=xsinx  will repeatedly has the value of X and -X. In fact, they are the intersections of these functions,  y=xsinx  ,  y=x  , and  y=-x

Graph f(x)= \frac{sin\ x}{x}  on the window [−5π, 5π] and describe freely what the graph shows.



Similar to the previous sinusoidal functions. One of the general forms of 
sinusoidal functions is y = Asin(Bx − C) + D, where A represents its Amplitude, B represents its Periods, C and D represents its shifts. This time,   \frac{sinx}{x}  has only inconstant Amplitude   \frac{1}{x}  Therefore, this function is expected to be a function that has the same period of  y=sinx   but with smaller and smaller Amplitudes. And, since the function  y=sinx  is repeating its values between the interval [-1, 1], when  sinx=1  or  sinx=-1  , the function  \frac{sinx}{x}  will repeatedly has the value of 1/X and -1/X. In fact, they are the intersections of these functions,  \frac{sinx}{x}  ,  \frac{1}{x}  , and
 \frac{-1}{x}




3. A 23-ft ladder leans against a building so that the angle between the ground and the ladder is 80°. How high does the ladder reach up the side of the building? 

If the ladder is lean against the TOP of the building and the building is vertically standing above the ground, we can use the sine function to solve this question. That is the heigh of the building is  23sin80^\circ \approx22.6505

If the ladder is NOT lean against the TOP of the building and the building is vertically standing above the ground, then the height we just calculated is just a portion of the building and it represents where the ladder reach up the side of the building.


1/04/2023

How can De Moivre's theorem be described? What is the scope of this theorem? Examples for roots and powers.

Finding powers of complex numbers is greatly simplified using De Moivre’s Theorem. 

According to the De Moivre’s Theorem 

If Z = r(cos θ + isin θ) is a complex number, then Z= rn[cos(nθ) + isin(nθ)] 

where n is a positive integer. Zn = rn cis(nθ) 

How does it come from? To understand this theorem, we must know the products of complex numbers in polar form and the quotients of complex numbers in polar form first.

Recall that : 

    sin(α + β) = sin(α)cos(β) + cos(α)sin(β)

    sin(α – β) = sin(α)cos(β) – cos(α)sin(β)

    cos(α + β) = cos(α)cos(β) – sin(α)sin(β)

    cos(α – β) = cos(α)cos(β) + sin(α)sin(β)


 


 (1+ \sqrt[]{3}i)^3

 r= \sqrt[]{ 1^{2}+( \sqrt[]{3})^2 } =2

 tan \theta= \sqrt[]{3}   ,   3(\frac{ \pi }{3})= \pi

 (1+ \sqrt[]{3}i)^3=2^3(cos\pi +sin \pi)=-8


 (\sqrt[]{2}+ \sqrt[]{2}i)^3

 r= \sqrt[]{ (\sqrt[]{2})^{2}+( \sqrt[]{2})^2 } =2

 tan \theta= 1  ,   \theta= \pi/4

 (\sqrt[]{2}+ \sqrt[]{2}i)^3=2^3(cos3\pi/4 +sin3\pi/4)=\sqrt[]{2}(-4+4i)


 (1+ \sqrt[]{3}i)^{ \frac{1}{2}}

 r= \sqrt[]{(1)^2+ (\sqrt[]{3})^2 } =2

 tan \theta= \sqrt[]{3}  ,    \theta= \frac{ \pi }{3}

 (1+ \sqrt[]{3}i)^{ \frac{1}{2}}   =(2^{ \frac{1}{2} })(cos(\frac{1}{2})(\frac{ \pi }{3})+isin(\frac{1}{2})(\frac{ \pi }{3}))  =   \sqrt[]{2} ( \frac{\sqrt[]{3}}{2} + \frac{1}{2} i)


  (\sqrt[]{2}+ \sqrt[]{2}i)^{ \frac{1}{2}}

 r= \sqrt[]{( \sqrt[]{2})^2+(\sqrt[]{2})^2} =2

 tan \theta=1   ,   \theta= \frac{ \pi }{4}

 (\sqrt[]{2}+ \sqrt[]{2}i)^{ \frac{1}{2}}   = \sqrt[]{2}(cos \frac{ \pi }{8} +isin \frac{ \pi }{8})= \sqrt[]{2}( \sqrt[]{\frac{1+ \frac{\sqrt[]{2}}{2} }{2} } +\sqrt[]{\frac{2-\sqrt[]{2} }{2} })

The similarity of triangles gives rise to trigonometry.

One of the largest issues in ancient mathematics was accuracy—nobody had calculators that went out ten decimal places, and accuracy generally got worse as the numbers got larger. The famous Eratosthenes experiment, that can be found at https://www.famousscientists.org/eratosthenes/, relied on the fact known to Thales and others that a beam of parallels cut by a transverse straight line determines equal measure for the corresponding angles.  Given two similar triangles, one with small measurements that can be accurately determined, and the other with large measurements, but at least one is known with accuracy, can the other two measurements be deduced? Explain and give an example.

The similarity of triangles gives rise to trigonometry. 

How could we understand that the right triangles of trigonometry with a hypotenuse of measure 1 represent all possible right triangles? Ultimately, the similarity of triangles is the basis for proportions between sides of two triangles, and these proportions allow for the calculations of which we are speaking here. The similarity of triangles is the foundation of trigonometry.



Similar triangles are triangles that have the same shape, but their sizes may vary. If two triangles are similar, then their corresponding angles are congruent and corresponding sides are in equal proportion. Two triangles are similar if they have the same shape but are of different sizes. Thus mathematically, if two triangles are similar, then their corresponding sides are proportional and their corresponding angles are congruent.



Since these two triangles are similar triangles, \frac{h}{y} = \frac{1}{x} ,  h= \frac{y}{x}

For the smaller triangle, tan \theta = \frac{y}{x}

For the larger triangle, tan \theta = \frac{h}{1}











Reference

Abramson, J. (2017). Algebra and trigonometry. OpenStax, TX: Rice University. Retrieved from https://openstax.org/details/books/algebra-and-trigonometry


Admin. (2021, March 10). Similar triangles- Formula, Theorem & Proof of SSS, SAS AAA Similarity. BYJUS. Retrieved December 26, 2022, from https://byjus.com/maths/similar-triangles/ 


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