Sunday, January 16, 2011

Math : Trigonometry

INTRODUCTION

Trigonometry (from Greek trigōnon "triangle" + metron "measure"[1] or from Sanskrit trikon "triangle" + miti "measurement" = trikonmiti[2]) is a branch of mathematicsthat studies triangles and the relationships between their sides and the angles between the sides. Trigonometry defines the trigonometric functions, which describe those relationships and have applicability to cyclical phenomena, such as waves. The field evolved during the third century BC as a branch of geometry used extensively for astronomical studies.[3]
Trigonometry is usually taught in middle and secondary schools either as a separate course or as part of a precalculus curriculum. It has applications in both pure mathematics and applied mathematics, where it is essential in many branches of science and technology. A branch of trigonometry, called spherical trigonometry, studies triangles on spheres, and is important in astronomy and navigation.

HISTORY

Ancient Egyptian and Babylonian mathematicians lacked the concept of an angle measure, but they studied the ratios of the sides of similar triangles and discovered some properties of these ratios. The ancient Greeks transformed trigonometry into an ordered science.[5]
Ancient Greek mathematicians such as Euclid and Archimedes studied the properties of the chord of an angle and proved theorems that are equivalent to modern trigonometric formulae, although they presented them geometrically rather than algebraically. Claudius Ptolemy expanded upon Hipparchus' Chords in a Circle in hisAlmagest.[6] The modern sine function was first defined in the Surya Siddhanta, and its properties were further documented by the 5th century Indian mathematicianand astronomer Aryabhata.[7] These Greek and Indian works were translated and expanded by medieval Islamic mathematicians. By the 10th century, Islamic mathematicians were using all six trigonometric functions, had tabulated their values, and were applying them to problems in spherical geometry. At about the same time, Chinese mathematicians developed trigonometry independently, although it was not a major field of study for them. Knowledge of trigonometric functions and methods reached Europe via Latin translations of the works of Persian and Arabic astronomers such as Al Battani and Nasir al-Din al-Tusi.[8] One of the earliest works on trigonometry by a European mathematician is De Triangulis by the 15th century German mathematician Regiomontanus. Trigonometry was still so little known in 16th century Europe that Nicolaus Copernicus devoted two chapters of De revolutionibus orbium coelestium to explaining its basic concepts.
Driven by the demands of navigation and the growing need for accurate maps of large areas, trigonometry grew to be a major branch of mathematics.[9]Bartholomaeus Pitiscus was the first to use the word, publishing his Trigonometria in 1595.[10] Gemma Frisius described for the first time the method of triangulationstill used today in surveying. It was Leonhard Euler who fully incorporated complex numbers into trigonometry. The works of James Gregory in the 17th century andColin Maclaurin in the 18th century were influential in the development of trigonometric series.[11] Also in the 18th century, Brook Taylor defined the general Taylor series.[12]

Applications of trigonometry

Sextants are used to measure the angle of the sun or stars with respect to the horizon. Using trigonometry and amarine chronometer, the position of the ship can be determined from such measurements.

There are an enormous number of uses of trigonometry and trigonometric functions. For instance, the technique of triangulation is used in astronomy to measure the distance to nearby stars, in geography to measure distances between landmarks, and in satellite navigation systems. The sine and cosine functions are fundamental to the theory of periodic functions such as those that describe sound and light waves.
Fields that use trigonometry or trigonometric functions include astronomy (especially for locating apparent positions of celestial objects, in which spherical trigonometry is essential) and hence navigation (on the oceans, in aircraft, and in space), music theory, acoustics, optics, analysis of financial markets,electronics, probability theory, statistics, biology, medical imaging (CAT scans and ultrasound), pharmacy, chemistry, number theory (and hence cryptology),seismology, meteorology, oceanography, many physical sciences, land surveying and geodesy, architecture, phonetics, economics, electrical engineering,mechanical engineering, civil engineering, computer graphics, cartography, crystallography and game development.

Trigonometric Functions

  • Sin, Cos, Tan Functions

  • Cosec, Sec and Cot Functions

Sin, Cos and Tan Functions

The trigonometric functions of angles are the ratios of the various sides of a triangle. Consider a right-angled triangle ABC as shown in the figure below.
Trigonometry: Triangle Image 1

The following terminology is useful.
  • Hypotenuse: The side opposite to the right angle in a triangle is called the hypotenuse. Here the side AC is the hypotenuse.

  • Opposite Side: The side opposite to the angle in consideration is called the opposite side. So, if we are considering angle A, then the opposite side is CB.

  • Base: The third side of the triangle, which is one of the arms of the angle under consideration, is called the base. If A is the angle under consideration, then the side AB is the base.

For angle A (sometimes referred to as angle CAB), the following fundamental trigonometric functions can be defined.



Sine of A = sin A = Opposite Side / Hypotenuse = CB/CA = a/b
Cosine of A = cos A = Base / Hypotenuse = AB/CA = c/b
Tangent of A = tan A = Opposite Side / Base = CB/AB = a/c

From the definitions, it can be seen that tan A = sin A / cos A.

Cosec, Sec and Cot Functions

A few more useful functions can be defined.

Cosecant of A = cosec A = 1 / sin A = b/a
Secant of A = sec A = 1 / cos A = b/c
Cotangent of A = cot A = 1 / tan A = c/a

From the definitions, it can be seen that cot A = cosec A / sec A

Values of Trigonometric Functions for Common Angles

It is useful to know the values of the trigonometric functions for certain common angles. The values of the trigonometric functions for angles 0, 30o, 45o, 60o and 90oare given in the table below.

Angle A
0
30o
45o
60o
90o
sin A
0
½
1/√2
√3/2
1
cos A
1
√3/2
1/√2
½
0
tan A
0
 1/√3
1
√3
 ∞

It is possible to obtain the values in the above table for 0, 45o and 90o by using the definitions from Trigonometry Module 1.

The values in the above table can be easily remembered using the following mnemonic. The values of sin 0, sin 30o, sin 45o, sin 60o and sin 90o are simply given by the square roots of 0/4, 1/4, 2/4, 3/4 and 4/4. The values of cos 0, cos 30o, cos 45o, cos 60o and cos 90o are given by the square roots of 4/4, 3/4, 2/4, 1/4 and 0/4. Obviously, the values of tan for any angle are obtained by dividing the sine value by the cosine value.

Important Trigonometric Identities

Some important trigonometric identities relating functions of a single angle (say, A) are given below.

sin2 A + cos2 A = 1
... (1)
 1 + tan2 A = sec2 A... (2)
   1+ cot2 A = cosec2 A ... (3)


These identities are useful in simplification and solution of problems. Their proofs are given below.
  • Based on the right-angled triangle in the figure alongside, sin A = a / b and cos A = c / b.
    Therefore sin2 A + cos2 A = (a2 + c2)/b2.
    By Pythagoras Theorem, a2 + c2 = b2 in a right-angled triangle.
    Thus sin2 A + cos2 A = 1 and equation (1) is proved.

  • On dividing both sides of equation (1) by cos2 A,
    we obtain tan2 A + 1 = 1 / cos2 A = sec2 A. Equation (2) is proved.

  • On dividing both sides of equation (1) by sin2 A, 
    we obtain 1 + cot2 A = 1/sin2 A = cosec2 A. Equation (3) is proved.

    Cofunction Property

    The cofunction property is very important and is given below.

    sin (90° − A) = cos A... (4)
    cos (90° − A) = sin A... (5)
    tan (90° − A) = cot A... (6)


    The cofunction property is easy to prove by starting with the fact that the sum of the three angles of a triangle is always 180o, i.e., A + B + C = 180o. For a right-angled triangle, angle B is 90o and A + C = 90o. On using 90o - C = A, the following are readily obtained:
    • sin (90o - A) = sin C = c / b = cos A
    • cos (90o - A) = cos C = a / b = sin A
    • tan (90o - A)

    • COMMON FORMULAS


      Triangle with sides a,b,c and respectively opposite angles A,B,C
      Certain equations involving trigonometric functions are true for all angles and are known as trigonometric identities. Some identities equate an expression to a different expression involving the same angles. These are listed in List of trigonometric identities. Triangle identities that relate the sides and angles of a given triangle are listed below.
      In the following identities, A, B and C are the angles of a triangle and a, b and c are the lengths of sides of the triangle opposite the respective angles.



      Law of sines

      The law of sines (also known as the "sine rule") for an arbitrary triangle states:
      \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R,
      where R is the radius of the circumscribed circle of the triangle:
      R = \frac{abc}{\sqrt{(a+b+c)(a-b+c)(a+b-c)(b+c-a)}}.
      Another law involving sines can be used to calculate the area of a triangle. Given two sides and the angle between the sides, the area of the triangle is:
      \mbox{Area} = \frac{1}{2}a b\sin C.
      All of the trigonometric functions of an angle θ can be constructed geometrically in terms of a unit circle centered at O.

      Law of cosines

      The law of cosines (known as the cosine formula, or the "cos rule") is an extension of the Pythagorean theorem to arbitrary triangles:
      c^2=a^2+b^2-2ab\cos C ,\,
      or equivalently:
      \cos C=\frac{a^2+b^2-c^2}{2ab}.\,

      Law of tangents

      The law of tangents:
      \frac{a-b}{a+b}=\frac{\tan\left[\tfrac{1}{2}(A-B)\right]}{\tan\left[\tfrac{1}{2}(A+B)\right]}


      Euler's formula

      Euler's formula, which states that eix = cosx + isinx, produces the following analytical identities for sine, cosine, and tangent in terms of 'e' and the imaginary unit 'i' :
      \sin x = \frac{e^{ix} - e^{-ix}}{2i}, \qquad \cos x = \frac{e^{ix} + e^{-ix}}{2}, \qquad \tan x = \frac{i(e^{-ix} - e^{ix})}{e^{ix} + e^{-ix}}.
       = tan C = c / a = cot A 


      TRIGONOMETRY - MEASURE OF AN ANGLE


      Any real number tex2html_wrap_inline183 may be interpreted as the radian measure of an angle as follows: If tex2html_wrap_inline185 , think of wrapping a length tex2html_wrap_inline183 of string around the standard unit circle C in the plane, with initial point P(1,0), and proceeding counterclockwise around the circle; do the same if tex2html_wrap_inline193 , but wrap the string clockwise around the circle. This process is described in Figure 1 below.
      Figure 1
      If Q(x,y) is the point on the circle where the string ends, we may think of tex2html_wrap_inline183 as being an angle by associating to it the central angle with vertex O(0,0) and sides passing through the points P and Q. If instead of wrapping a length s of string around the unit circle, we decide to wrap it around a circle of radius R, the angle tex2html_wrap_inline183 (in radians) generated in the process will satisfy the following relation:
        equation34
      Observe that the length s of string gives the measure of the angle tex2html_wrap_inline183 only when R=1.
      As a matter of common practice and convenience, it is useful to measure angles in degrees, which are defined by partitioning one whole revolution into 360 equal parts, each of which is then called one degree. In this way, one whole revolution around the unit circle measures tex2html_wrap_inline219 radians and also 360 degrees (or tex2html_wrap_inline223 ), that is:
        equation39
      Each degree may be further subdivided into 60 parts, called minutes, and in turn each minute may be subdivided into another 60 parts, called seconds:
        align48


      EXAMPLE 1 Express the angle tex2html_wrap_inline225 in Degree-Minute-Second (DMS) notation.
      Solution: We use Equation 3 to convert a fraction of a degree into minutes and a fraction of a minute into seconds:
      align59

      Therefore, tex2html_wrap_inline227 .

      The magic identity

      Trigonometry is the art of doing algebra over the circle. So it is a mixture of algebra and geometry. The sine and cosine functions are just the coordinates of a point on the unit circle. This implies the most fundamental formula in trigonometry (which we will call here the magic identity)

      displaymath91
      where tex2html_wrap_inline93 is any real number (of course tex2html_wrap_inline93 measures an angle).
      Example. Show that
      displaymath97
      Answer. By definitions of the trigonometric functions we have
      displaymath99
      Hence we have
      displaymath101
      Using the magic identity we get
      displaymath103
      This completes our proof.
      Remark. the above formula is fundamental in many ways. For example, it is very useful in techniques of integration.
      Example. Simplify the expression
      displaymath105
      Answer. We have by definition of the trigonometric functions
      displaymath107
      Hence
      displaymath109
      Using the magic identity we get
      displaymath111
      Putting stuff together we get
      displaymath113
      This gives
      displaymath115
      Using the magic identity we get
      displaymath117
      Therefore we have
      displaymath119


      The Addition Formulas

      The fundamental identities are very important for the analysis of trigonometric expressions and functions but they are a direct result of the intimate relation between trigonometry and geometry. The power behind the algebraic nature of trigonometry is hidden and can be measured only with the addition formulas

      displaymath133
      and
      displaymath135
      Of course, we used the fact that
      displaymath137


      Example. verify the identity
      displaymath139

      Answer. We have
      displaymath141
      which gives
      displaymath143
      But
      displaymath145
      and since
      displaymath147
      and tex2html_wrap_inline149 , we get finally
      displaymath151


      Remark. In general it is good to check whether the given formula is correct. One way to do that is to substitute some numbers for the variables. For example, if we take a=b = 0, we get
      displaymath155
      or we may take tex2html_wrap_inline157 . In this case we have
      displaymath159


      Example. Find the exact value of
      displaymath161

      Answer. We have
      displaymath163
      Hence, using the additions formulas for the cosine function we get
      displaymath165
      Since
      displaymath167
      we get
      displaymath169


      Example. Find the exact value for
      displaymath171

      Answer. We have
      displaymath173
      Since
      displaymath175
      we get
      displaymath177
      Finally we have
      displaymath179


      Remark. Using the addition formulas, we generate the following identities
      displaymath181

      More identities may be proved similar to the above ones. The bottom line is to remember the addition formulas and use them whenever needed.

      Double-Angle and Half-Angle Formulas

      Double-Angle and Half-Angle formulas are very useful. For example, rational functions of sine and cosine will be very hard to integrate without these formulas. They are as follow

      displaymath196


      Example. Check the identities
      displaymath198
      Answer. We will check the first one. the second one is left to the reader as an exercise. We have
      displaymath200
      Hence
      displaymath202
      which implies
      displaymath204
      Many functions involving powers of sine and cosine are hard to integrate. The use of Double-Angle formulas help reduce the degree of difficulty.


      Example. Write tex2html_wrap_inline206 as an expression involving the trigonometric functions with their first power.
      Answer. We have
      displaymath208
      Hence
      displaymath210
      Since tex2html_wrap_inline212 , we get
      displaymath214
      or
      displaymath216


      Example. Verify the identity
      displaymath218
      Answer.We have
      displaymath220
      Using the Double-Angle formulas we get
      displaymath222
      Putting stuff together we get
      displaymath224
      From the Double-Angle formulas, one may generate easily the Half-Angle formulas
      displaymath226
      In particular, we have
      displaymath228


      Example. Use the Half-Angle formulas to find
      displaymath230
      Answer. Set tex2html_wrap_inline232 . Then
      displaymath234
      Using the above formulas, we get
      displaymath236
      Since tex2html_wrap_inline238 , then tex2html_wrap_inline240 is a positive number. Therefore, we have
      displaymath242
      Same arguments lead to
      displaymath244


      Example. Check the identities
      displaymath246
      Answer. First note that
      displaymath248
      which falls from the identity tex2html_wrap_inline250 . So we need to verify only one identity. For example, let us verify that
      displaymath252
      using the Half-Angle formulas, we get
      displaymath254
      which reduces to
      displaymath256

      Product and Sum Formulas

      From the Addition Formulas, we derive the following trigonometric formulas (or identities)

      displaymath113


      Remark. It is clear that the third formula and the fourth are identical (use the property tex2html_wrap_inline115 to see it).
      The above formulas are important whenever need rises to transform the product of sine and cosine into a sum. This is a very useful idea in techniques of integration.


      Example. Express the product tex2html_wrap_inline117 as a sum of trigonometric functions.
      Answer. We have
      displaymath119
      which gives
      displaymath121
      Note that the above formulas may be used to transform a sum into a product via the identities
      displaymath123


      Example. Express tex2html_wrap_inline125 as a product.
      Answer. We have
      displaymath127
      Note that we used tex2html_wrap_inline129 .


      Example. Verify the formula
      displaymath131
      Answer. We have
      displaymath133
      and
      displaymath135
      Hence
      displaymath137
      which clearly implies
      displaymath131


      Example. Find the real number x such that tex2html_wrap_inline143 and
      displaymath145
      Answer. Many ways may be used to tackle this problem. Let us use the above formulas. We have
      displaymath147
      Hence
      displaymath149
      Since tex2html_wrap_inline143 , the equation tex2html_wrap_inline153 gives tex2html_wrap_inline155 and the equation tex2html_wrap_inline157 gives tex2html_wrap_inline159 . Therefore, the solutions to the equation
      displaymath145
      are
      displaymath163


      Example. Verify the identity
      displaymath165
      Answer. We have
      displaymath167
      Using the above formulas we get
      displaymath169
      Hence
      displaymath171
      which implies
      displaymath173
      Since tex2html_wrap_inline175 , we get
      displaymath177


      TRIGONOMETRIC EQUATIONS

      Some equations which involve trigonometric functions of the unknown may be readily solved by using simple algebraic ideas (as Equation 1 below), while others may be impossible to solve exactly, only approximately (e.g., Equation 2 below):

      gather44


      EXAMPLE 1: Find all solutions of the equation tex2html_wrap_inline77 .
      Solution: We can graphically visualize all the angles u which satisfy the equation by noticing that tex2html_wrap_inline81 is the y-coordinate of the point where the terminal side of the angle u (in standard position) intersects the unit circle (see Figure 1):
      We can see that there are two angles in tex2html_wrap_inline87 that satisfy the equation: tex2html_wrap_inline89 and tex2html_wrap_inline91 . Since the period of the sine function is tex2html_wrap_inline93 , it follows that all solutions of the original equation are:
      displaymath75


      Table of Trigonometric Identities

      Reciprocal identities

      displaymath161

      Pythagorean Identities
      displaymath162

      Quotient Identities
      displaymath163

      Co-Function Identities
      displaymath164

      Even-Odd Identities
      displaymath165

      Sum-Difference Formulasdisplaymath166

      Double Angle Formulas
      align99

      Power-Reducing/Half Angle Formulas
      displaymath167

      Sum-to-Product Formulas
      displaymath168

      Product-to-Sum Formulas
      displaymath169


      SOLVING TRIGONOMETRIC EQUATIONS

      This sections illustrates the process of solving trigonometric equations of various forms. It also shows you how to check your answer three different ways: algebraically, graphically, and using the concept of equivalence.The following table is a partial lists of typical equations.



      Solve for x in the following equations. 




      Example 1:         
      $2sin\left( x\right) -1=0$

      There are an infinite number of solutions to this problem. To solve for x, you must first isolate the sine term.

      \begin{eqnarray*}&& \\ 2\sin \left( x\right) -1 &=&0 \\ && \\ \sin \left( x\right) &=&\displaystyle \frac{1}{2} \\ && \\ && \end{eqnarray*}

      We know that the $\sin \left( \displaystyle \frac{\pi }{6}\right) =\displaystyle \frac{1}{2},$therefore $x=\displaystyle \frac{\pi }{6}.$ The sine function is positive in quadrants I and II. The $\sin \left( \pi -\displaystyle \frac{\pi }{6}\right) =\sin \displaystyle \frac{5\pi }{6}$is also equal to $\displaystyle \frac{1}{2}.$ Therefore, two of the solutions to the problem are $x=\displaystyle \frac{\pi }{6}$and $x=\displaystyle \frac{5\pi }{6}. $


      The period of the sin $\left( x\right) $ function is $2\pi .$ This means that the values will repeat every $2\pi $ radians in both directions. Therefore, the exact solutions are $x=\displaystyle \frac{\pi }{6}\pm n\left( 2\pi \right) $ and $x= \displaystyle \frac{5\pi }{6}\pm n\left( 2\pi \right) $ where n is an integer. The approximate solutions are $x=0.523598775598\pm n\left( 2\pi \right) $ and $ x=2.61799387799\pm n\left( 2\pi \right) $ where n is an integer.


      These solutions may or may not be the answers to the original problem. You much check them, either numerically or graphically, with the original equation.
      Numerical Check:
      Check answer . $x=\displaystyle \frac{\pi }{6}$
      • Left Side: $\qquad 2\sin \left( x\right) -1=2\sin \left( \displaystyle \frac{\pi }{6} \right) -1=2\left( \displaystyle \frac{1}{2}\right) -1=0 $
      • Right Side:        0
      Since the left side equals the right side when you substitute $\displaystyle \frac{\pi }{6 }$ for x, then $\displaystyle \frac{\pi }{6 }$ is a solution.


      Check answer . $x=\displaystyle \frac{5\pi }{6}$
      • Left Side: $\qquad 2\sin \left( x\right) -1=2\sin \left( \displaystyle \frac{5\pi }{6} \right) -1=2\left( \displaystyle \frac{1}{2}\right) -1=0 $
      • Right Side:        0
      Since the left side equals the right side when you substitute $\displaystyle \frac{5\pi }{ 6}$ for x, then $\displaystyle \frac{5\pi }{ 6}$ is a solution.


      Graphical Check:
      Graph the equation
      f (x) = 2 sin(x) - 1
      Note that the graph crosses the x-axis many times indicating many solutions. Note that it crosses at $\displaystyle \frac{ \pi }{6}\approx 0.5236$. Since the period is $2\pi \approx 6.2831$, it crosses again at 0.5236+6.283=6.81 and at0.5236+2(6.283)=13.09, etc. The graph crosses at $\displaystyle \frac{5\pi }{6}\approx 2.618$. Since the period is $ 2\pi \approx 6.283$, it will cross again at 2.618+6.283=8.9011 and at 2.618+2(6.283)=15.18, etc.



    •   



    • Example 2:        
      $2\sin\left( 3x\right) -1=0$



      There are an infinite number of solutions to this problem. To solve for x, you must first isolate the sine term.
      \begin{eqnarray*}&& \\ 2\sin \left( 3x\right) -1 &=&0 \\ && \\ \sin \left( 3x\right) &=&\displaystyle \frac{1}{2} \\ && \\ && \end{eqnarray*}

      If we restriction the domain of the sine function to $\left[ -\displaystyle \frac{\pi }{2} \leq 3x\leq ,\displaystyle \frac{\pi }{2}\... ...[ -\displaystyle \frac{\pi }{6}\leq x\leq ,\displaystyle \frac{\pi }{6}\right] $, we can use the inverse sine function to solve for reference angle 3x and then x. 
      \begin{eqnarray*}&& \\ \sin \left( 3x\right) &=&\displaystyle \frac{1}{2} \\ &... ...&=&\sin ^{-1}\left( \displaystyle \frac{1}{2 }\right) \\ && \\ \end{eqnarray*}
      \begin{eqnarray*}3x &=&\sin ^{-1}\left( \displaystyle \frac{1}{2}\right) \\ && ... ...\ x &\approx &0.174532925\ \mbox{ radians }\\ && \\ && \\ && \end{eqnarray*}

      We know that the $\sin $e function is positive in the first and the second quadrant. Therefore two of the solutions are the angle 3x that terminates in the first quadrant and the angle $\pi -3x$ that terminates in the second quadrant. We have already solved for 3x.
      \begin{eqnarray*}&& \\ \sin \left( \pi -3x\right) &=&\displaystyle \frac{1}{2} ... ...-3x &=&\sin ^{-1}\left( \displaystyle \frac{1}{2}\right) \\ && \end{eqnarray*}
      \begin{eqnarray*}&& \\ 3x &=&\pi -\sin ^{-1}\left( \displaystyle \frac{1}{2}\ri... ... \\ && \\ x &\approx &0.872665\ \mbox{ radians } \\ && \\ && \end{eqnarray*}
      The solutions are $x=\displaystyle \frac{1}{3}\sin ^{-1}\left( \displaystyle \frac{1}{2}\right) $ and $ x=\displaystyle \frac{\pi }{3}-\displaystyle \frac{1}{3}\sin ^{-1}\left( \displaystyle \frac{1}{2}\right) .\bigskip \bigskip\bigskip $
      The period of the $\sin\left( 3x\right) $ function is $\displaystyle \frac{2\pi }{3}.$This means that the values will repeat every $\displaystyle \frac{2\pi }{3}$ radians in both directions. Therefore, the exact solutions are $x=\displaystyle \frac{1}{3}\sin ^{-1}\left( \displaystyle \frac{1}{2}\right) \pm n\left( \displaystyle \frac{2\pi }{3}\right) $ and$x= \displaystyle \frac{\pi }{3}-\displaystyle \frac{1}{3}\sin ^{-1}\left( \displaystyle \frac{1}{2}\right) \pm n\left( \displaystyle \frac{2\pi }{3}\right) $ where n is an integer.

      The approximate solutions are $x=0.174532925\pm n\left( \displaystyle \frac{2\pi }{3} \right) $ and $x\approx 0.872665\pm n\left( \displaystyle \frac{2\pi }{3}\right) $ where n is an integer.


      These solutions may or may not be the answers to the original problem. You much check them, either numerically or graphically, with the original equation.

      Numerical Check:

      Check the answer x=0.174532925

      • Left Side: $\qquad 2\sin\left( 3x\right) -1=2\sin\left( 3\left( 0.174532925\right) \right) -1=0.000000000362\approx 0\bigskip $
      • Right Side:        $0\bigskip $
      Since the left side equals the right side when you substitute 0.174532925for x, then 0.174532925 is a solution.



      Check the answer x=0.872665

      • Left Side: $\qquad 2\sin\left( 3x\right) -1=2\sin\left( 3\left( x\approx 0.872665\right) \right) -1=-0.0000019433\approx 0\bigskip $
      • Right Side:        $0\bigskip $
      Since the left side equals the right side when you substitute $x\approx 0.872665$ for x, then $x\approx 0.872665$ is a solution.


      Graphical Check:

      Graph the equation
      $f(x)=2\sin (3x)-1.$
      Note that the graph crosses the x-axis many times indicating many solutions. You can see that the graph crosses at 0.174532925. Since the period is $\displaystyle \frac{2\pi }{3}\approx 2.094395$, it crosses again at0.174532925+2.094395=2.2689 and at 0.174532925+2(2.094395)=4.3633, etc. The graph crosses at 0.872665.

      Since the period is $\displaystyle \frac{2\pi }{3}\approx 2.094395$, it will cross again at $0.872665+\left( 2.094395\right) =2.9671$ and at 0.872665+2(2.094395)=5.061455, etc $.\bigskip\bigskip $

      The Derivatives of Trigonometric Functions

      \begin{displaymath}\begin{array}{cc}\sin^\prime x=\cos x&\cos^\prime x=-\cos x\\... ...\prime x=\sec x \tan x&\csc^\prime x=-\csc x \cot x \end{array}\end{displaymath}
      Trigonometric functions are useful in our practical lives in diverse areas such as astronomy, physics, surveying, carpentry etc. How can we find the derivatives of the trigonometric functions?
      Our starting point is the following limit: 
      \begin{displaymath}\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1\cdot\end{displaymath}

      Using the derivative language, this limit means that $\sin'(0) = 1$. This limit may also be used to give a related one which is of equal importance: 
      \begin{displaymath}\lim_{x \rightarrow 0} \frac{\cos(x)-1}{x} = 0\end{displaymath}

      To see why, it is enough to rewrite the expression involving the cosine as 
      \begin{displaymath}\frac{\cos(x)-1}{x} = \frac{(\cos(x)-1)(\cos(x) + 1)}{x(\cos(x) + 1)} = \frac{(\cos^2(x)-1)}{x(\cos(x) + 1)}\end{displaymath}

      But $\cos^2(x)-1 = -\sin^2(x)$, so we have 
      \begin{displaymath}\lim_{x \rightarrow 0} \frac{\cos(x)-1}{x} = \lim_{x \rightar... ... \rightarrow 0} x \frac{-\sin^2(x)}{x^2(\cos(x) + 1)} = 0 \cdot\end{displaymath}

      This limit equals $\cos'(0)$ and thus $\cos'(0) = 0$.
      In fact, we may use these limits to find the derivative of $\sin(x)$ and $\cos(x)$ at any point x=a. Indeed, using the addition formula for the sine function, we have 
      \begin{displaymath}\sin(a + h) = \sin(a) \cos(h) + \sin(h) \cos(a) \cdot\end{displaymath}

      So 
      \begin{displaymath}\frac{\sin(a + h) - \sin(a)}{h} = \sin(a)\frac{1 - \cos(h)}{h} + \cos(a) \frac{\sin(h)}{h}\end{displaymath}

      which implies 
      \begin{displaymath}\lim_{h \rightarrow 0} \frac{\sin(a + h) - \sin(a)}{h} = \cos(a) \cdot\end{displaymath}

      So we have proved that $\sin'(a)$ exists and $\sin'(a) = \cos(a)$.
      Similarly, we obtain that $\cos'(a)$ exists and that $\cos'(a) = -\sin(a)$.
      Since $\tan(x)$, $\cot(x)$, $\sec(x)$, and $\csc(x)$ are all quotients of the functions $\sin(x)$ and $\cos(x)$, we can compute their derivatives with the help of the quotient rule:

      \begin{displaymath}\begin{array}{llll} \displaystyle \frac{d}{dx} (\tan(x)) = \s... ...style \frac{d}{dx} (\csc(x)) = -\csc(x) \cot(x) \\ \end{array}\end{displaymath}

      It is quite interesting to see the close relationship between $\tan(x)$ and $\sec(x)$ (and also between $\cot(x)$ and $\csc(x)$).
      From the above results we get

      \begin{displaymath}\sin''(x) = - \sin(x)\;\;\mbox{and}\;\; \cos''(x) = - \cos(x)\cdot\end{displaymath}

      These two results are very useful in solving some differential equations.


      Example 1. Let $f(x) = \sin(2 x)$. Using the double angle formula for the sine function, we can rewrite 
      \begin{displaymath}\sin(2 x) = 2 \sin(x) \cos(x)\cdot\end{displaymath}

      So using the product rule, we get 
      \begin{displaymath}\frac{d}{dx}\Big(\sin(2x)\Big) = 2 \Big( \cos(x) \cos(x) - \sin(x) \sin(x) \Big) = 2 \Big( \cos^2(x) - \sin^2(x) \Big)\end{displaymath}

      which implies, using trigonometric identities, 
      \begin{displaymath}\frac{d}{dx}\Big(\sin(2x)\Big) = 2 \cos(2x)\cdot\end{displaymath}


      Hyperbolic Functions

      The hyperbolic functions enjoy properties similar to the trigonometric functions; their definitions, though, are much more straightforward:
      displaymath121

      displaymath122
      Here are their graphs: the tex2html_wrap_inline125 (pronounce: "kosh") is pictured in red, the tex2html_wrap_inline127 function (rhymes with the "Grinch") is depicted in blue.


      As their trigonometric counterparts, the tex2html_wrap_inline125 function is even, while the tex2html_wrap_inline127 function is odd.
      Their most important property is their version of the Pythagorean Theorem.

      • tex2html_wrap_inline133
      The verification is straightforward:eqnarray18
      While tex2html_wrap_inline135 , tex2html_wrap_inline137 , parametrizes the unit circle, the hyperbolic functions tex2html_wrap_inline139 , tex2html_wrap_inline141 , parametrize the standard hyperbola tex2html_wrap_inline143 , x>1.
      In the picture below, the standard hyperbola is depicted in red, while the point tex2html_wrap_inline139 for various values of the parameter t is pictured in blue.


      The other hyperbolic functions are defined the same way, the rest of the trigonometric functions is defined:
      eqnarray36


      tanh x
      coth x
      sech x
      csch x

      For every formula for the trigonometric functions, there is a similar (not necessary identical) formula for the hyperbolic functions:
      Let's consider for example the addition formula for the hyperbolic cosine function:
      • tex2html_wrap_inline151
      Start with the right side and multiply out:eqnarray61



      Inverse Hyperbolic Functions

      The hyperbolic sine function is a one-to-one function, and thus has an inverse. As usual, we obtain the graph of the inverse hyperbolic sine function tex2html_wrap_inline53 (also denoted by tex2html_wrap_inline55 ) by reflecting the graph oftex2html_wrap_inline57 about the line y=x:
      Since tex2html_wrap_inline61 is defined in terms of the exponential function, you should not be surprised that its inverse function can be expressed in terms of the logarithmic function:
      Let's set tex2html_wrap_inline63 , and try to solve for x:
      eqnarray14
      This is a quadratic equation with tex2html_wrap_inline67 instead of x as the variable. y will be considered a constant.
      So using the quadratic formula, we obtain
      displaymath47
      Since tex2html_wrap_inline73 for all x, and since tex2html_wrap_inline77 for all y, we have to discard the solution with the minus sign, so
      displaymath48
      and consequently
      displaymath49
      Read that last sentence again slowly!
      We have found out that
      • tex2html_wrap_inline81

      Here is some problems which have Trigonometry Solutions with detail explanation step by step .


      Question 1


      Question:   

      Answer:    

      Question 2


      Question:   
      Answer:    

      Question 3


      Question:   
      Answer:    

      Question 4


      Question:   
      Answer:    


      Question 5


      Question:   

      Answer:    

      Question 6


      Question:   If A+B=45, prove that (1 + tan A) (1 + tan B) = 2.

      Answer:    





      Question 7


      Question:   
      Answer:    

      Question 8


      Question:   
      Answer:    





      Question 9


      Question:   Prove that tan 70o = tan 20o + 2 tan 50o.
      Answer:    








      Trigonometry Solutions:Question 10


      Question:   
      Answer:   Here is the Trigonometry Solution ,  By data: 


      By componendo and dividendo





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3 comments:

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November 27, 2019 at 1:31 AM
Ram Dev said...

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January 5, 2021 at 6:24 AM
Easy Shiksha said...

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May 30, 2021 at 11:01 PM

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