Half Angle Formula at a glance

Half Angle Formula Proof
Half angle for Sine formula is given by, sin(theta/2) = (+/-)sqrt[1/2 (1+ cos(theta))] and
Proof : We know that cos(2theta)=1 – 2 sin^2(theta)
Taking, theta = alpha/2 then 2theta = alpha
The formula would be, cos(alpha) = 1 – 2 sin^2(alpha/2)
Let us now solve for sin(alpha/2)
1 – 2sin^2(alpha/2) = cos(alpha)
2 sin^2(alpha/2) = 1 – cos(alpha)
Sin^2(alpha/2) = [1- cos(alpha]/2
Sin(alpha/2) = (+/-)sqrt[1 – cos(alpha)/2] is the identity for sine of a half angle

The quadrant in which alpha/2 lies decides the sign whether positive or negative for sin(alpha/2)
It is positive if alpha/2 lies in the first or second quadrants and is given by
 sin(alpha/2)= sqrt[1 – cos(alpha)/2]
It is negative if alpha/2 lies in the third or fourth quadrants and is given by
 sin(alpha/2)= - sqrt[1 – cos(alpha)/2]

The Half angle formula for cosine is given by, Cos(alpha/2)=(+/-)sqrt[1/2(1-cos(alpha))]
Proof: We know that  cos(2theta) = 2cos^2(theta) -1
Substituting theta = alpha/2 we get
Cos(alpha) = 2 cos^2(alpha/2) – 1
Solving for cos(alpha/2)
Cos(alpha) = 2cos^2(alpha/2) – 1
2cos^2(alpha/2) = cos(alpha) + 1
Cos^2(alpha/2) = [cos(alpha) + 1]/2
Cos(alpha/2) = (+/-)sqrt[[cos(alpha) + 1]/2] is the Half Angle Formula Cosine

The quadrant in which alpha/2 lies decides the sign whether positive or negative for cosine(alpha/2)
It is positive if alpha/2 lies in the first or fourth quadrants and is given by
cos(alpha/2) = sqrt[[cos(alpha) + 1]/2]
It is negative if alpha/2 lies in the second or third quadrants and is given by
 cos(alpha/2)= - sqrt[1 + cos(alpha)/2]

Half angle Formula Tan is given by, tan(alpha/2) = [1-cos(alpha)]/sin(alpha)
It can also be written as, tan(alpha/2) = sin(alpha)/1+cos(alpha)
Proof: We know that, tan(theta) = sin(theta)/cos(theta)
Taking theta = alpha/2 and substituting, we get
  Tan(alpha/2) = sin(alpha/2)/cos(alpha/2)
Substituting the half angle of sine and cosine we get
Tan(alpha/2) =  (+/-)sqrt[1 – cos(alpha)/2]/ (+/-)sqrt[[cos(alpha) + 1]/2]
        = sqrt[1 – cos(alpha)]/ sqrt[[cos(alpha) + 1]
        = sqrt[1 – cos(alpha)/cos(alpha) + 1]
Rationalizing the denominator, we get
     = sqrt{[1- cos(alpha)]^2/[1+cos(alpha)][1- cos(alpha)]}
     = sqrt{[1- cos(alpha)]^2/1 – cos^2(alpha)}
Using the identity, sin^2(theta) +cos^2(theta) = 1
     = sqrt{[1 – cos(alpha)]^2/sin^2(alpha)}

Finding the square root, we get
  Tan(alpha/2) =(+/-) [1 – cos(alpha)]/sin(alpha) is the identity for Tangent Half Angle

Half Angle Formula for Tangent  is also written as tan(alpha/2) = sin(alpha)/[1+cos(alpha)]
Proof: We have, tan(alpha/2) = [1 – cos(alpha)]/sin(alpha)
Multiplying and dividing with [1+cos(alpha)], we get
Tan(alpha/2) = [1 – cos(alpha)/sin(alpha)]x [1 + cos(alpha)/1+cos(alpha)]
        = 1 – cos^2(alpha)/sin(alpha)[1+cos(alpha)]
Using the identity, sin^2(theta) + cos^2(theta) = 1
      = sin^2(alpha)/sin(alpha)[1+cos(alpha)]
Canceling sin(alpha), we get
Tan(alpha/2) = (+/-) [sin(alpha)/1+cos(alpha)]

Half Angle Formula Tan is given by,
       Tan(alpha/2) =(+/-) [1 – cos(alpha)]/sin(alpha) and also
       Tan(alpha/2) = (+/-) [sin(alpha)/1+cos(alpha)]

Plane Geometry Problems

Introduction to Plane Geometry Problems
Plane geometry deals with shapes drawn in plane like line, circle, square, rectangle, triangle and parallelogram, polygons. A plane is a flat surface. We can easily draw shapes like square, rectangle, parallelogram and triangle etc. The plane geometry problems can be of any type like perimeter, area, angles measure, missing side length etc.

Problems on Area and Perimeter
Perimeter: In simple language perimeter is the distance covered around outside of a plane shape. Perimeter is calculated by simply adding all the sides around outside of the shape. Perimeter is one dimensional quantity.
example:  Calculate the perimeter of triangle of side length 5cm, 7 cm. 9 cm.
In this case Perimeter = 5+7+9 = 21cm.
Word Problems on Plane Geometry

Problem 1: An equilateral triangle has a perimeter of 51cm. How long is each side of the triangle?
Solution:  In equilateral triangle all the sides are equal.
let a be the side
Perimeter  = sum of all the sides of the triangle
      51 =   a+a+a
=   51 =  3a
=   3a = 51
= 3/3a  = 51/3
       a  = 21 cm.
so each side of the triangle is 21cm.

Problem 2: A room is 20ft. long and 15 ft. wide. How many square tiles of side 5ft. required to cover the floor?
Solution:  Room is a rectangle. Area of rectangle = length x width = 20 x15  sq. ft.
               shape of tile is square. Area of square = side x side     =    5 x 5  sq. ft.

Total area of the floor 20 x 15=300
Area of 1 tile 5x5=25


Mixed Problem on Plane Geometry
1. The radius of a circle is 13 centimeters. What is the circle's circumference?
2. Area of a parallelogram is 66sqm. Find the height if the base is 13m.
3. In an isosceles triangle , angle opposite to equal sides are 70 degree each. Find the measure of the third angle?
4. If the area of a square is 81sq.cm. Find the perimeter?
5. There are two circles of radii 5cm. and 3cm and a rectangle of 4cm inside a square of  side 15cm.  Find the area of remaining portion?

First Order Differential Equation

A First Order Differential Equation is involving the unknown function y, its derivative y’ and the variable x. the most general type of  First Order Differential Equation can be written as Y’=f(x,y),or we can write as dy/dx=f(x,y), here f(x,y) is a function of two variables defined on a region in the xy plane. The equations is of first order because it involves only the first derivative y’ or dy/dx.

Now questions is that how do we solve first order differential equation. There are two methods which can be used to solve first order differential equation. First is separation of variables and second is Integrating factor. Again the new question in your mind should be how do we know which method should be used? so here is your answer, first method is separation of variables, this method is use when a differential equation can be written in the form dy/dx = f(y) g(x), where f is the function is y only and g is function of x only. To solve problems by this methods rewrite the problem as 1/f(y) dy= g(x) dx and integrate both side. Use the initial condition to find the constant of integration.

Let’s clear this method by one first order differential equation example. Solve dy/dx=xy with initial condition y(0)=1. So to solve this problem divide through by y we get 1/y dy/dx= x, now integrate both side with respect to x we get ∫ 1/y dy/dx= ∫ xdx, by solving we get ln(y)= x^2/2+c. now make y as the subject of equation so y= Ae^x^2/2 where A= e^c. this is the general solution of this equation.

Now our second method is Integrating factor this method is use when differential equation is give in the form dy/dx + p(x)y = q(x) where p and q are function of x only. Our next point is how do we find Integrating factor∫ The integrating factor is e^∫ p(x) dx. Now to solve problems by this method you need to rearrange the differential equation in to the standard form and find the integrating factor. Multiply through the integrating factor and rewrite the left hand side as the derivative of ye^ ∫ p(x) dx. Integrating both sides gives the general solution. For more information how to solve math problems visit math related expert online sites.

Now we will discus about what is first order nonlinear differential equation and what is first order ordinary differential equation. The first order differential equation is nonlinear if the function or its derivative is multiplied to itself.  One example is du/dx=u^2+1. The next is first order ordinary differential equation, in this differential equation the unknown function is a function of single independent variable.

Calculus Problems with solutions

Calculus is the mathematics of change, of calculating problems that are continually evolving. There are mainly two branches of Calculus, Differential calculus or Differentiation and Integral Calculus or Integration which deals with areas and volumes of complex figures. Let us solve some of the Calculus Problems using the various rules or formulas stated in Calculus.

Calculus question on Limits: Find the limit of lim(x->5) [x^2- 25]/[x^2+x-30]
 First we need to factor both the numerator and the denominator.
That gives us, lim(x->5)[(x+5)(x-5)]/[(x+6)(x-5)].
 Next we cancel the common factors, which is (x-5).
We get, lim(x->5)(x+5)/(x+6). Now, we substitute x = 5 which gives us, 10/11 which is the required limit.
Differentiate the following calculus question, f(x) =x^2/(3x-1)
Using the Quotient rule of differentiation, we get
f’(x)= [(3x-1)d/dx[x^2]- (x^2)[d/dx(3x-1)]/(3x-1)^2
        =[(3x-1)(2x) – (x^2)(3)]/(3x-1)^2  combining the like terms, we get
        = [6x^2-2x-3x^2]/(3x-1)^2 on further simplification
        = [3x^2-2x]/(3x-1)^2  
        = x[3x-2]/(3x-1)^2
   
Calculus Problem on Differentiation of Trigonometry Functions: Differentiate y=2sin(x)-3cos(x)
We use the factor rule of differentiation, d/dx[cf(x) = cd/dx[f(x)].
So, we get, y’= 2 d/dx[sin(x)] – 3d/dx[cos(x)]
 which gives us 2cos(x)-3[-sin(x)]=2cos(x)+3sin(x)

Let us work on a Calculus question on differentiation of a logarithmic function: y = x^x
First let us apply natural logarithm on both sides, log y = log xx = xlog x.
Using the product rule on the right hand side, 1/y[y’]=x(1/x)+(1) ln x= 1+ln(x)
Multiplying y on both sides, y’ = x^x[1+ ln x]

Evaluate arccos [-sqrt(2)/2]. In this calculus question, we first put y =arccos[-sqrt(2)/2]
that gives us cos(y)=- sqrt(2)/2
From the unit circle, we get the angle y as 3(pi)/4

Find the fourth derivative of the given higher order derivatives calculus question, y = cos x
First derivative, y’ = -sin(x)
Second derivative, y’’= -cos(x)
Third derivative, y’’’=-[-sin(x)] = sin(x)
Fourth derivative, y^(4)= cos(x)

Calculus Problems using summation notation: Find the value of Summation(i=0to2)[5+sqrt(4i)] which gives us, =[5+sqrt(4^0)]+ [5+sqrt(4^1)]+ [5+sqrt(4^2)]
  = [5+sqrt(1)]+ [5+sqrt(4)]+ [5+sqrt(4^2)]
=[5+1] +[5+2]+[5+4]
=22

Integrate integral (2+3e^x) is a Calculus question on Integration.
The steps involved are as follows:
Integral(2+3e^x)= integral(2)dx – integral(3e^x)
= integral(2)dx- 3 integral(e^x)
         = 2x- 3e^x+ C

Integration by parts: Let us solve a Calculus question, Integrate integral (xsinx)dx
In this problem, let  u = x and dv = sin x dx
So, we get, du/dx = 1;
du = (1)dx and v = -cos x
Therefore, integral(xsinx)dx= x(-cos x) – integral (-cos x) dx = -x cos(x) + sin(x) + C

A problem solvers solution guide is a Calculus Problem Solver which gives a stepwise explanation of many of the calculus problems. We can even get online calculus question solvers.

Numerical integration methods

Introduction to numerical integration:  First here we will understand the concept   What is Numerical Integration As we know in integration that if ɸ(x) is primitive of f(x) defined on [a, b], then integral a to b f(x) dx = ɸ(b) - ɸ (a). In most of the practical problems we are given a set of numerical values of the function f(x) corresponding to some values of x and in some cases either the primitive does not exist or it cannot be easily determined by elementary means; as a result the computation of the definite integral by the above formula may not be possible. In such circumstances, the numerical methods for integration are needed. 

Numerical integration is the process of computing the value of a definite integral when we are given a set of numerical values of the inte-grand f(x) corresponding to some values of the independent variable x. And also Methods of Numerical Integration  are The trapezoidal rule: Let y = f(x) be a function defined on [a, b] which is divided into n equal su-intervals each of width h so that b – a = nh. Let the values of f(x) for (n + 1) equidistant arguments x₀ = a, x₁ = x₀ + h, x₂ = x₀ + 2h, …., x_n = x₀ + nh = b be y₀, y₁, y₂, …., y_n respectively, then Integral a to b f(x) dx = integral x₀ to x₀ + nh = y dx = h[1/2 (y₀ + y_n) + (y₁ + y₂ + ….. + y_n)].

This rule is known as trapezoidal rule. In the derivation of this formula it is assumed that the y is a linear function of x i.e., the equation of the curve is the form y = a + bx and second rule is Simpson’s one-third rule: Let y = f(x) be a function defined on [a, b] which is divided into n (an even number) equal parts each of width h so that b – a = nh. Suppose the function y = f(x) attains values y₀, y₁, y₂, …., y_n at n + 1 equidistant points x₀ = a, x₁ = x₀ + h, x₂ = x₀ + 2h, …., x_n = x₀ + nh = b respectively. Then Integral a to b f(x) dx = integral x₀ to x₀ + nh = y dx = h[1/2 (y₀ + y_n) + (y₁ + y₂ + ….. + y_n)] = h/3[(y₀ + y_n) + 4(y₁ + y₃ + ….. + y_n-1) + 2(y₂ + y₄ + ….. + y_n-2)] = (one –third of the distance between two consecutive ordinates) × [(sum of the extreme ordinates) + 4(sum of odd ordinates) + 2(sum of even ordinates)] this formula is known as Simpson’s one-third rule.

Its geometric significance is that we replace the graph of the given function by n/2 arcs of second degree polynomials, or parabolas with vertical axes. It is to note here that the interval [a, b] is divided into an even number of subinterval of equal width.