Sin2x Formula with Solved Example

Sin2x Formula

Sin2x is one of the formulae for double angles in trigonometry.

Using this formula, we can get the sine of the doubled value of an angle. Sin is defined as the ratio of the length of the opposing side (of the angle) to the length of the hypotenuse in a right-angled triangle. Sin is one of the main trigonometric ratios. Various Sin2x-related formulae may be proved using fundamental trigonometric formulas. As the range of the sin function is [-1, 1], so is the range of Sin2x.

In the remainder of this article, we will also examine Sin2x (sin squared x) and its formula. We shall get the formula for Sin2x and Sin2x by expressing them in terms of various trigonometric functions using various trigonometric formulas.

What exactly is Sin2x?

Sin2x is a trigonometric formula used to solve many trigonometric, integration, and differentiation issues. It is used to simplify trigonometric expressions. The Sin2x formula may be represented in several ways using various trigonometric formulae. The most frequent form of Sin2x is Sin2x = 2 sinx cosx, which represents the product of the sine function and cosine function. We may also express Sin2x using the tangent function.

Sin2x Formula

Sin2x is the double angle identity used in trigonometry for the sine function. The study of the relationship that exists between the three sides and angles of a right triangle is known as trigonometry. There are two fundamental expressions for Sin2x:

Sin2x = 2 sin x cos x 

Sin2x = (2tan x)​/(1 + tan2x)

These are the primary Sin2x formulae. Using the trigonometric equation Sin2x + cos2x = 1, this expression may be expressed in terms of sin x (or cos x) alone. This trigonometric identity allows us to write sinx =  √(1 – cos2x) and cosx =.  √(1 – Sin2x). Therefore, the formula for Sin2x given cos and sin are:

Sin2x = 2 √(1 – cos2x) cos x 

Sin2x = 2 sin x √(1 – Sin2x)

Solved Examples

Example: Determine sin2A if cos A = 3/5 and A is in quadrant I given that cos A = 3/5.

Solution:

Pythagorean identity is ours.

sin2A + cos2A = 1

sin2A = 1 – cos2A

sin A = ±√(1 − cos2A)

sin A = ±√(1 − (3/5)2)

sin A = ±√(16/25)

sin A = ± 4/5

Since A is located in quadrant I, sin A has a positive value. Thus,

sin A = 4/5

From Sin2x formula, Sin2x = 2 sin x cos x. Based on this,

sin2A equals 2 cos A sin A

= 2 (4/5) (3/5)

= 24/25

Therefore, sin2A is equal to 24/25.

Important formulas: