Trigonometry is an interesting as well as an important branch of Mathematics. It has many identities that are very useful for learning and deriving the many equations and formulas in science. This article will look at some specific kinds of trigonometric formulae which are popular as the double angle formulae. These formulae are possible with all 6 kinds of trigonometry ratios. Here we will see the Sin 2X formula with the concept, derivation, and examples. Such formulae are popular as they involve trigonometric functions of double angles. Let us learn it!

## Concept of Sin 2x

We will take the right-angled triangle. In this triangle, we have three sides namely â€“ Hypotenuse, opposite side (Perpendicular) and Adjacent side (Height). The largest side is the hypotenuse, the side opposite to the angle is opposite and the side where both hypotenuse and opposite rests is the adjacent side. There are six fundamental ratios which are the core of trigonometry. These are,

- Sine (sin)
- Cosine (cos)
- Tangent (tan)
- Secant (sec)
- Cosecant (csc)
- Cotangent (cot)

Double angle identities and formulae are useful for solving certain integration problems where a double formula may make things much simpler to solve. Therefore in mathematics as well as in physics, such formulae are useful for deriving many important identities.

The trigonometric formulas like Sin2x, Cos 2x, Tan 2x are popular as double angle formulae, because they have double angles in their trigonometric functions. For solving many problems we may use these widely.

The Sin 2x formula is:

\(Sin 2x = 2 sin x cos x\)

Where x is the angle.

Source: en.wikipedia.org

## Derivation of the Formula

It is clear that Sin value for the double angle is in the form of a product of sin and Cos values of a single angle. We can easily derive this formula using the addition formula for Sin angles.

We know that the addition formula for sin is given as:

\(Sin (X + Y) = Sin X Cos Y + Cos X Sin Y,\)

Where X and Y are the two angles.

In the above formula replace Y by X, with the assumption that both angles X and Y are equal. Thus,

\(Sin (X+X) = Sin X Cos X + Cos X Sin Y\)

Hence Sin 2x = 2 Sin x Cos x

## Solved Examples for Sin 2x Formula

Q.1: Find the value of Calculate \(sin75 ^{\circ}sin15 ^{\circ}\).

Solution: As given,

\(sin75 ^{\circ}Â sin15 ^{\circ}\\\)

\(= sin(90 ^{\circ}-15 ^{\circ}) sin15 ^{\circ} \\\)

\(= cos15 ^{\circ} sin15 ^{\circ} [Â as \ cos x = sin (90 ^{\circ}- x ) ]\\\)

i.e., \(= \frac{1}{2} sin 30^{\circ}Â [ applying \ double\ angle\ formula \ sin 2x = 2 sin x cos x ]\\\)

So, \(= \frac {1}{2} \times \frac{1}{2} [ as\ sin 30^{\circ}Â = \frac{1}{2} ]\\= \frac {1}{4}\)

Thus \(sin75 ^{\circ}Â sin15 ^{\circ}\) will be \(\frac{1}{4}\)

Q.2: Find value of \(sin 90 ^{\circ}\) using its double angle formula.

Solution: We know that double angle formula for sin is : \(Sin 2x = 2 Sin x Cos x\\\)

Putting \(\ x = 45^{\circ} \\\)

\(Sin (2 \times 45^{\circ}) = 2 \ Sin 45^{\circ} Cos 45^{\circ} \\\)

Since \(\ Sin 45^{\circ} =\frac{1}{\sqrt{2}} \\\)

\(Cos 45^{\circ} = \frac{1}{\sqrt{2}} \\\)

Therefore, by substituting we get:

\(Sin 90^{\circ} = 2 \times \frac{1}{\sqrt{2}}\\ \times \frac{1}{\sqrt{2}}\\\)

\(= 2 \times \frac{1}{2}\\ = 1\)

Thus value of is \(Sin 90^{\circ} 1\).

I get a different answer for first example.

I got Q1 as 20.5

median 23 and

Q3 26