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We have already proved that Sin(A + B) = SinA.CosB + CosA.SinB. Now are going to prove another important trigonometry formula. Let us assume A and B are two different angles.

Sin(A – B) = SinA.CosB – CosA.SinB

Let us consider A and B are two different angles. This is trigonometry formula is most useful formula. We use this formula in problem solving.

Sin(A + B) = SinA.CosB + CosA.SinB

Sine is first trigonometry function, which gives relationship between opposite side and hypotenuse. This trigonometry function always depends on angle of triangle, which could be in either degrees or radians. Sine always gives values between zero and one. In other words, the minimum and maximum values of Sine are zero and one.

Mathematically we can define sine trigonometry function as

SinΘ = Opposite/ Hypotenuse

Sine is an odd function because sine gives negative values for negative angles. This first trigonometry function is positive in only first and second quadrants.

One of our site visitors posted a problem today and the person name is Yemisi. Now, i would like to solve this problem for that person. Here is the one of problems.

Given that, Tan A=m/n and A is acute angle.

1. sin(A+90)
2. cos(A+45)
3. sec(A+90)

We know that Tan A=m/n. Now, we have to find SinA and CosA values to solve this problem easily.

Let us consider, Opposite side is m and adjacent side is n as per tangent definition. So, the hypotenuse is the √(m² + n²)

Now we can write Sine and Cosine functions as stated below.

SinA = m/√(m² + n²)

CosA = n/√(m² + n²)

Today, i received a comment in matzic. One of our mathematical students is requested me to solve five problems. I would like to solve those problems for him. The below trigonometry function is the given by that person. The person name is Yemisi. It is my pleasure to solve this problem.

[Sin30° - 3Cos90° - Cos45°{(1/Sin45°)+(1/Sin60°)}]/(2.tan60°)

First of we have to know the values of Sine, tangent and Cosine trigonometry functions values for respective angles.

We have know that

Sin30° = 1/2

Cos90° = 0

Sin45° = Cos45° = 1/√2

Sin60° = (√3)/2

Tan60° = √3

We have successfully learned Hyperbolic functions and its properties. Now, we going to learn one useful formula in hyperbolic functions. This formula looks like trigonometry formula Sin2x.

Our main objective is to derive Sinh2x formula in terms of Sinhx and Coshx.

According to Hyperbolic Function Sinhx definition, we can write Sinhx as stated below

Sinhx = (e^x – e^-x)/2

Similarly, we can also write Sinh2x as stated below.

Sinh2x = (e^2x – e^-2x)/2

The exponential function looks like a² – b² form. So, we can write it as (a + b).(a – b), which means,

Trigonometry is one mathematical concept. Trigonometry plays significant role in almost all fields especially Engineering. The word trigonometry derived with the help of two words ‘trigon’ and ‘metron’ which means measuring the slides of a triangle.

Trigonometry concepts developed with six trigonometry functions. These are the list of trigonometry functions.

1. Sine
2. Cosine
3. Tangent
4. Co-tangent
5. Secant
6. Co-secant

Trigonometry formulas

We have learned properties of quadratic equation when discriminant is not equal to zero. Now let us consider, the discriminant of a quadratic equation is grater than zero. In other words, b² – 4ac > 0. In this case, the roots α and β are two different real numbers.

We can take  x² + 4x + 3 = 0 as an example

The discriminant b² – 4ac = (4)² – 4.1.3 = 16 – 12 = 4 > 0

Now you understood. So, this quadratic equation should have two different roots. Similarly, this quadratic equation’s roots are two real numbers. Let us see how it is different practically.

Let us assume, the discriminant of a quadratic equation is not equal to zero. In other words, b² – 4ac ≠ 0. In this case, the roots α and β are two different roots.

For example, consider  x² + 4x + 3 = 0 is an quadratic equation

The discriminant b² – 4ac = (4)² – 4.1.3 = 16 – 12 = 4 ≠ 0

Now you understood. So, this quadratic equation should have two different roots. Let us see how it is different practically.

α = [- b + √(b² - 4.a.c)]/2.a = [- 4 + √4]/2.1 = [- 4 + 2]/2 = – 1

β = [- b - √(b² - 4.a.c)]/2.a = [- 4 - √4]/2.1 = [- 4 - 2]/2 = – 3

-1 and -3 are not equal.

So, we can finally say that the quadratic equation has two different two roots when the discriminant is not equal to zero.

We have successfully derived two roots α and β of a quadratic equation previously. Now, we are going to discuss an important factor which is called as Discriminant. Discussing about Discriminant helps us to understand the properties of roots.

As we already know that the general solution of ax² + bx + c = 0 quadratic equation is

x = [-b ± √(b² – 4ac)]/2a

In this general solution, b² – 4ac is called as discriminant. Disciminant looks like very silly thing but it plays very important role in quadratic equation. That’s why we are discussing here about discriminant of quadratic equation separately.