Sum and Difference Formulas in Trigonometry

Sum and Difference Formulas for Sine and Cosine
The sum Formulas for Sine and Cosine are:
Sin(A+ B) = SinA.CosB + CosA.SinB
Cos(A+B) = CosA.CosB – SinA.SinB

The difference Formulas for Sine and Cosine are:
Sin(A – B) = SinA.CosB  - CosA.SinB
Cos(A – B) = CosA.CosB + SinA.SinB

Trigonometric Sum and Difference Formulas
Sin(x+y) = sin(x).cos(y) + cos(x).sin(y)
Cos(x+y) = cos(x).cos(y) – sin(x).sin(y)
Tan(x+y) = [tan(x) + tan(y)]/[1- tan(x).tan(y)]

Sin(x-y) = sin(x).cos(y) – cos(x).sin(y)
Cos(x-y) = cos(x).cos(y) + sin(x).sin(y)
Tan(x-y) = [tan(x) – tan(y)]/[1+ tan(x).tan(y)]

Let us solve some of trigonometric problems using trig sum and difference formulas
Solve, cos(30 degrees)cos(15 degrees) – sin(30 degrees)sin(15 degrees) without actually solving. The given trigonometric expression is in the form cos(x).cos(y) – sin(x).sin(y) which is equal to cos(x+y) a trig sum formula. Comparing the terms we get, x = 30 degrees and y = 15 degrees and hence x+ y = 30 + 15 = 45 degrees. Finally we get, cos(x+y) = cos(30+15) = cos(45) = sqrt(2)/2

Sum and Difference Formulas Trig functions sine, cosine and tangent are given as follows:
Sin(alpha+ beta) = sin(alpha).cos(beta) + cos(alpha).sin(beta)
solve sin(75 degrees)
75 degrees is not special angle, but we can split 75 to give 45 + 30, we know both 45 and 30 degrees are special angles. So, we can re-write sin(75 degrees) = sin(45+30) applying the sum formula of sine, we get
Sin(45).cos(30) + cos(45).sin(30) = (1/2)(1/sqrt2) + (sqrt3/2) (1/sqrt2) = sqrt(2)[sqrt(3) +1]/4

Cos(alpha+beta) = cos(alpha).cos(beta) – sin(alpha).cos(beta)
Solve cos(5 pi/12) = cos(pi/4 + pi/6)= cos(pi/4).cos(pi/6) – sin(pi/4).sin(pi/6) = [sqrt(2)/2 ].[sqrt(3)/2] – [sqrt(2)/2. ½]= [sqrt(6) – sqrt(2)]/4

Tan(alpha+beta)
= sin(alpha+beta)/cos(alpha+beta)
= [sin(alpha)cos(beta) + cos(alpha)sin(beta)]/[cos(alpha)cos(beta) – sin(alpha).sin(beta)]
= {[sin(alpha)cos(beta)/cos(alpha)cos(beta)] +[ cos(alpha)sin(beta)/cos(alpha)cos(beta)]}
 Divided by [cos(alpha)cos(beta)/ cos(alpha)cos(beta)] – [sin(alpha).sin(beta)/ cos(alpha)cos(beta)]
= [tan(alpha) + tan(beta)]/[1- tan(alpha)tan(beta)]
tan(alpha+beta) =  [tan(alpha) + tan(beta)]/[1- tan(alpha)tan(beta)]

Sin(alpha- beta) = sin(alpha).cos(beta) – cos(alpha).sin(beta)
Solve sin(15) = sin(45- 30) = sin(45).cos(30) – cos(45).sin(30)
                  = [sqrt(2)/2].[sqrt(3)/2] – [sqrt(2)/2].[1/2] = [sqrt(6) – sqrt(2)]/4
Cos(alpha – beta) = cos(alpha).cos(beta) + sin(alpha).sin(beta)
Verify cos(alpha – pi) = - cos(alpha). Using the above difference formula for cosine we get,
Cos(alpha – pi) = cos(alpha).cos(pi) + sin(alpha).sin(pi) we know that cos(pi) = -1 and sin(pi) = 0
Substituting the values, we get, cos(alpha – pi) = - cos(alpha) + 0 = - cos(alpha) [verified]

tan(alpha – beta)
 = sin(alpha – beta)/cos(alpha – beta)
= [sin(alpha)cos(beta) - cos(alpha)sin(beta)]/[cos(alpha)cos(beta) + sin(alpha).sin(beta)]
= {[sin(alpha)cos(beta)/cos(alpha)cos(beta)] -[ cos(alpha)sin(beta)/cos(alpha)cos(beta)]}
 Divided by [cos(alpha)cos(beta)/ cos(alpha)cos(beta)] + [sin(alpha).sin(beta)/ cos(alpha)cos(beta)]
= [tan(alpha) - tan(beta)]/[1 +tan(alpha)tan(beta)]
tan(alpha – beta)= = [tan(alpha) - tan(beta)]/[1 +tan(alpha)tan(beta)]

Derivative table

Derivative means differentiation of any function. In other words calculating a slope of a function. Here we have to discuss about derivative table. Derivative table is a collection of all differentiation formulas in summarized form in a single table. It can be different table for different function. In derivative table first type is power of X, under this type we differentiate a constant term (d/dx C=0), function X (d/dx X=1) and function x^n (d/dx x^n=n *x^n-1).

Second type in derivative table is exponential and logarithmic functions. In this type we differentiate exponential function d/dx  e^x=  e^x, logarithmic function d/dx logx= 1/x and function like b^x which is equal to bxln(b). third type in derivative table is trigonometric function, its is also know as trig derivative table, in this type we differentiate sin function d/dx sinx=cosx, cosine function, tangent function, cosec function , sec function, cotangent function. By these differential formulas we can make trig derivatives table, we can also proof trig function by algebraic method, in this method we first proof the sin function then with the help of sin proof, we can proof cosine function and in last with the help of both sin and cosine function proof we can proof tangent function.

Forth type in derivative table is inverse trigonometric function, calculus part mostly used in this type. In this type we differentiate inverse sin function d/dx sin^-1x= 1/(√1-x^2), inverse cos function, d/dx cos^-1x = -1/(√1-x^2), inverse tan function d/dxtan^-1x = 1/(1+x^2), inverse sec function d/dx sec^-1x = 1/(x*√x^2 -1), inverse cot function d/dx cot ^-1x =-1/(1+x^2). Fifth and the last type in derivative table is hyperbolic function. In this type we differentiate hyperbolic sin d/dx sinhx= coshx, hyperbolic cos function d/dx coshx= sinhx, hyberbolic sec function d/dx sechx= -tanhx*sechx, hyperbolic cosec function d/dxcosechx = -cothx* cosechx, hyperbolic tan function d/dxtanhx= 1-tanh2x, hyperbolic cot function d/dxcothx= 1-coth2x.

Double integrals in polar coordinates, in double integration we have region (D) in every trigonometry problem, region D is much easier to describe in Cartesian form. But some region such as disk, ring or any portion of disk or ring, if we use Cartesian coordinate it will be very difficult to differentiate these functions. So for these functions we use polar coordinate. Suppose in any problem disk is given with radius 2 then in Cartesian it is difficult to differentiate with limit, but when we change limits in polar coordinate in θ and r  form  then it become simple and we can integrate  it easily. Double integrals over rectangles means integration of a area made by curve, which cut x axis and y axis at equal points.

About the Newton Raphson method

Consider the following equation:
Y = x^2 + 2x – 3
If you are asked to find the roots of that equation, it is a pretty simple process of factoring the equation and then using the zero product rule to solve for possible values of x.

Illustrated as follows:
Y = x^2 + 2x – 3
 = (x+3)(x-1)
For finding zeros, y = 0
So, (x+3)(x-1) = 0
Using the zero product rule,
(x+3) = 0 ; (x-1) = 0
Thus, x = -3 or x = 1. Therefore these are the two zeros of the given equation.

Now consider the following equation:
Y = 2x^2 + x+ 5
Since in this case the left hand side polynomial is prime, we can simply use the quadratic formula to find the real roots of this equation (if any).

But now suppose if the function is like this:
Y = 3x^5 + x^3 – 2x^2  - x + 7
There are no simple methods or formulae for finding the roots (or zeros) of such a function. This is where we use some method of approximation. One such method is the Newton – Raphson method named after the mathematicians  Issac Newton and Joseph Raphson who invented the method. The idea here is to start with some approximate zero of the function and then through iterative process come to closer and closer approximations of the zeros of the polynomial function.  For a function of one variable the Newton -

Raphson algorithm can be stated as follows:
1. For a function f, we guess the zero = x0 to begin our iterations.
2. For the next best approximation of the zero of f we use the following formula:
X1 = x0 – f(x0)/f’(x0). Thus f(x1) is closer to 0 than f(x0).
3. The next best approximation would be x2 given by the formula:
X2 = x1 – f(x1)/f’(x1). Thus f(x2) is closer to 0 than f(x1)
4. The above iterative process can be continued for as many number of times as we like. Each time the zero we get would be closer and closer to the actual zero of the polynomial.
5. The general formula for the (n+1)th approximation is like this:
X(x+1) = x(n) – f(x(n))/f’(x(n))

For any newton – raphson example, we begin by guessing a zero. For functions with smaller coefficients, we can conveniently assume 0 as an approximate zero. For other functions, we may graph the function using few test values to come to some approximate first guess.

Graphing and solving logarithmic functions

A logarithmic function is the inverse of an exponential function. The general form of an exponential function would be y = b^x, where b is the base of the exponent and x is the exponent or the index. b belongs to positive real numbers and x is any real number. Therefore we see that the domain of an exponential function is all real numbers; whereas the range of this exponential function would be all positive real numbers.

The general form of a logarithmic function would be y = log_b?x, where b is the base of the logarithm. It is the same base that we used above in the exponential function. The x and y have switched places. Therefore, if we were to write the exponential function y = b^x in logarithmic form it would be log_b?y = x.

Graphs of logarithmic functions: To graph logarithmic functions online could be easy, but to graph them manually is not difficult either. Since we know that the logarithmic function is the inverse of an exponential function, we can make a table of values and plot the points, and then join the points with a curve. Let us try to see that with an example.

Example 1: Graph the function y = log_2?x.

Tabulating the results we have:
X (1/4) (1/2) 1 2 4 8
y -2 -1 0 1 2 3

Now we plot those ordered pairs on a graph sheet and run a curve through it. See picture below:

The blue curve above is the graph of the logarithmic function y = log_2?x. The red curve is the graph of the inverse exponential function y = 2^x.
Solving logarithmic functions:  There are various methods to solve logarithmic functions. One way is to convert the logarithmic function to the corresponding exponential function and then solve.
Example 2: Solve log_5?125 = y.
Solution: Converting to exponential form we have, 5^y = 125
5^y = 5^3
Since in the above equation the bases are equal, the exponents would also be equal.
Therefore, y = 3 is the answer.

Linear approximation Formulae

Introduction: Sometimes we can approximate complicated functions with simpler ones that give the accuracy we want for specific applications and are easier to work with. The approximating functions are called linearization. They are based on tangents. We introduce new variables dx and dy and define them in a way that gives new meaning to the Leibniz notation dy/dx. We will use dy to estimate error in measurement and sensitivity to change.

Linear approximation: If we see the graph of y = x^2 and y = 2x – 1, the tangent to a curve y = f(x) lies close to the curve near the point of tangency. For a brief interval to either side, the y-values along the tangent line give a good approximation to the y-values on the curve.

The more we magnify the graph of a function near a point where the function is differentiable, the flatter the graph becomes and the more it resembles its tangent. In the graph of the function y = f(x), the tangent line passes through the point (a, f(a)), so its point slope equation is y = f(a) + f ’(a) (x – a). Thus, the tangent is the graph of the function L(x) = f(a) + f ’ (a) (x – a). For as long as the line remains close to the graph of f, L(x) gives a good approximation to f(x).

According to the definition, Linear approximation formula can be given as below: If f is differentiable at x = a, then the approximating function L (x) = f(a) + f ’ (a) (x – a)is the linearization of f at a. The approximation f (x) ˜ L (x) of f by L is the standard linear approximate of ‘f’ at a” and is also known as Local Linear Approximation .The point x = a is the centre of the approximation. Let us take some Linear approximation examples

Example: Find the linearization of f(x) = sqrt(1 + x) at x = 0.
Solution: With f ’ (x) = ½ (1 + x)^-1/2,
We have f(0) = 1, f ’ (0) = ½, and
L (x) = f (a) + f ’ (a) (x – a) = 1 + ½ (x – 0) = 1 + x/2.
The approximation sqrt(1 + x) ˜ 1 + x/2 gives
Sqrt(1.2) ˜ 1 + 0.2/2 = 1.10
Sqrt(1.05) ˜ 1 + 0.05/2 = 1.025
Sqrt(1.005) ˜ 1 + 0.005/2 = 1.00250

A linear approximate normally loses accuracy away from its centre. The approximation sqrt(1 + x) ˜ 1 + x/2 will probably be too crude to be useful near x = 3. There, we need linearization at x = 3.