MATHEMATICS

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.

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, 30^o, 45^o, 60^o and 90^oare given in the table below.

Angle A	0	30^o	45^o	60^o	90^o
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, 45^o and 90^o 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 30^o, sin 45^o, sin 60^o and sin 90^o 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 30^o, cos 45^o, cos 60^o and cos 90^o 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.

sin²A + cos²A = 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 180^o, i.e., A + B + C = 180^o. For a right-angled triangle, angle B is 90^o and A + C = 90^o. On using 90^o - C = A, the following are readily obtained:
- sin (90^o - A) = sin C = c / b = cos A
- cos (90^o - A) = cos C = a / b = sin A
- tan (90^o - 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 $e i x = cos x + i sin x$ , 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  may be interpreted as the radian measure of an angle as follows: If  , think of wrapping a length  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  , 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  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  (in radians) generated in the process will satisfy the following relation:
  
  Observe that the length s of string gives the measure of the angle  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  radians and also 360 degrees (or  ), that is:
  
  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:
  
  
  EXAMPLE 1 Express the angle  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:
  
  Therefore,  .
  
  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)
  where  is any real number (of course  measures an angle).
  Example. Show that
  Answer. By definitions of the trigonometric functions we have
  Hence we have
  Using the magic identity we get
  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
  Answer. We have by definition of the trigonometric functions
  Hence
  Using the magic identity we get
  Putting stuff together we get
  This gives
  Using the magic identity we get
  Therefore we have
  
  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
  and
  Of course, we used the fact that
  
  Example. verify the identity
  
  Answer. We have
  which gives
  But
  and since
  and  , we get finally
  
  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
  or we may take  . In this case we have
  
  Example. Find the exact value of
  
  Answer. We have
  Hence, using the additions formulas for the cosine function we get
  Since
  we get
  
  Example. Find the exact value for
  
  Answer. We have
  Since
  we get
  Finally we have
  
  Remark. Using the addition formulas, we generate the following identities
  
  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
  
  Example. Check the identities
  Answer. We will check the first one. the second one is left to the reader as an exercise. We have
  Hence
  which implies
  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  as an expression involving the trigonometric functions with their first power.
  Answer. We have
  Hence
  Since  , we get
  or
  
  Example. Verify the identity
  Answer.We have
  Using the Double-Angle formulas we get
  Putting stuff together we get
  From the Double-Angle formulas, one may generate easily the Half-Angle formulas
  In particular, we have
  
  Example. Use the Half-Angle formulas to find
  Answer. Set  . Then
  Using the above formulas, we get
  Since  , then  is a positive number. Therefore, we have
  Same arguments lead to
  
  Example. Check the identities
  Answer. First note that
  which falls from the identity  . So we need to verify only one identity. For example, let us verify that
  using the Half-Angle formulas, we get
  which reduces to
  
  Product and Sum Formulas
  From the Addition Formulas, we derive the following trigonometric formulas (or identities)
  
  Remark. It is clear that the third formula and the fourth are identical (use the property  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  as a sum of trigonometric functions.
  Answer. We have
  which gives
  Note that the above formulas may be used to transform a sum into a product via the identities
  
  Example. Express  as a product.
  Answer. We have
  Note that we used  .
  
  Example. Verify the formula
  Answer. We have
  and
  Hence
  which clearly implies
  
  Example. Find the real number x such that  and
  Answer. Many ways may be used to tackle this problem. Let us use the above formulas. We have
  Hence
  Since  , the equation  gives  and the equation  gives  . Therefore, the solutions to the equation
  are
  
  Example. Verify the identity
  Answer. We have
  Using the above formulas we get
  Hence
  which implies
  Since  , we get
  
  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):
  
  EXAMPLE 1: Find all solutions of the equation  .
  Solution: We can graphically visualize all the angles u which satisfy the equation by noticing that  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  that satisfy the equation:  and  . Since the period of the sine function is  , it follows that all solutions of the original equation are:
  
  Table of Trigonometric Identities
  Reciprocal identities
  
  Pythagorean Identities
  
  Quotient Identities
  
  Co-Function Identities
  
  Even-Odd Identities
  
  Sum-Difference Formulas
  
  Double Angle Formulas
  
  Power-Reducing/Half Angle Formulas
  
  Sum-to-Product Formulas
  
  Product-to-Sum Formulas
  
  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:
  
  Here are their graphs: the (pronounce: "kosh") is pictured in red, the function (rhymes with the "Grinch") is depicted in blue.
  
  As their trigonometric counterparts, the function is even, while the function is odd.
  Their most important property is their version of the Pythagorean Theorem.
  The verification is straightforward:
  While , , parametrizes the unit circle, the hyperbolic functions , , parametrize the standard hyperbola , x>1.
  In the picture below, the standard hyperbola is depicted in red, while the point 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:
  
  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:
  Start with the right side and multiply out:
  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  (also denoted by  ) by reflecting the graph of about the line y=x:
  
  Since  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  , and try to solve for x:
  This is a quadratic equation with  instead of x as the variable. y will be considered a constant.
  So using the quadratic formula, we obtain
  Since  for all x, and since  for all y, we have to discard the solution with the minus sign, so
  and consequently
  Read that last sentence again slowly!
  We have found out that
  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 70^o = tan 20^o + 2 tan 50^o.
  Answer:
  
  Trigonometry Solutions:Question 10
  
  Question:
  Answer:   Here is the Trigonometry Solution , By data:
  
  By componendo and dividendo