Trigonometric Identities: Simplifying Sin(2Θ) With Sin(Θ)

The sine of a double angle is a trigonometric identity, comparing two mathematical expressions that remain valid for any values in a specified range. The formula for the sine of a double angle, sin 2θ, is used to simplify various trigonometry problems and can be expressed in terms of different trigonometric functions, including the sine and cosine, and the tangent function. The formula for sin 2θ is derived from the formula for the sine of a compound angle, sin(θ + φ) = sin(θ)cos(φ) + sin(φ)cos(θ). By substituting φ = θ, we can find the formula for the sine of a double angle: sin 2θ = 2sin(θ)cos(θ). This formula can be used to find the value of sin 2θ when the value of sin(θ) is known.

The double angle formula

> sin 2θ = 2 sin Θ cos Θ

This formula can be derived using the sum formula for sine, i.e., sin(α + β) = sin α cos β + cos α sin β. Substituting α = β = θ, we get:

> sin(θ + θ) = sin θ cos θ + cos θ sin θ

> sin 2θ = 2 sin Θ cos Θ

> sin 2θ = 2 tan Θ / (1 + tan^2 Θ)

This formula can be derived by multiplying and dividing the formula sin 2θ = 2 sin Θ cos Θ by cos Θ, and then substituting the identity sin Θ / cos Θ = tan Θ and cos Θ = 1 / sec Θ. Using the Pythagorean trigonometric identity, sec^2 Θ = 1 + tan^2 Θ, we can substitute to get:

> sin 2θ = 2 tan Θ x (1 / sec^2 Θ)

> sin 2θ = (2 tan Θ) / (1 + tan^2 Θ)

> sin 2x = 2 x (1 / √2) x (1 / √2) = √2 / 2

> (sin Θ + cos Θ)^2 = sin^2 Θ + 2 sin Θ cos Θ + cos^2 Θ

> = (sin^2 Θ + cos^2 Θ) + 2 sin Θ cos Θ

> = 1 + 2 sin Θ cos Θ

> = 1 + sin 2Θ

Thus, we have verified the identity.

Using trigonometric identities and the tangent

The sine of a double angle, sin(2θ), can be calculated using the compound angle formula:

> sin(α + β) = sin(α)cos(β) + sin(β)cos(α)

Substituting α = β = θ, we get:

> sin(2θ) = sin(θ)cos(θ) + cos(θ)sin(θ)

> sin(2θ) = 2sin(θ)cos(θ)

This is the sin(2θ) formula, one of the important double-angle formulas in trigonometry.

We can also use other trigonometric identities to calculate sin(2θ) in various ways. One method involves the Pythagorean identity:

> sin^2(θ) + cos^2(θ) = 1

We can rearrange this equation to express sin(θ) in terms of cos(θ):

> sin(θ) = ±√(1 - cos^2(θ))

Substituting this into the sin(2θ) formula, we get:

> sin(2θ) = 2cos(θ)√(1 - cos^2(θ))

We can further simplify this expression using the Pythagorean identity:

> 1 + tan^2(θ) = sec^2(θ)

Substituting this into the previous equation, we get:

> sin(2θ) = 2tan(θ) / (1 + tan^2(θ))

This is another form of the sin(2θ) formula, expressed in terms of the tangent function.

In summary, the sine of a double angle, sin(2θ), can be calculated using the compound angle formula, which gives the formula sin(2θ) = 2sin(θ)cos(θ). We can also use other trigonometric identities, such as the Pythagorean identity, to derive alternative expressions for sin(2θ). One such expression is sin(2θ) = 2tan(θ) / (1 + tan^2(θ)), which is obtained by rewriting the sin(2θ) formula in terms of the tangent function.

The sine of a compound angle

$$\sin(\theta + \varphi) = \sin(\theta)\cos(\varphi) + \sin(\varphi)\cos(\theta)$$

This formula can be visualised geometrically using triangles. Starting with a triangle with sides labelled according to the values of $\sin(\theta)$, $\cos(\theta)$, and $1$ (the hypotenuse), we can find the values of the sides of two smaller triangles with angle $\theta$. By doing so, we can construct a final triangle that, due to the property of parallel lines, has an angle of $\theta + \varphi$. This allows us to calculate the sine of the compound angle.

The formula for the sine of a compound angle is one of the basic compound angle formulae or trigonometric identities. These identities express trigonometric functions of $(A + B)$ or $(A - B)$ in terms of trigonometric functions of $A$ and $B$. The three fundamental compound angle formulae are:

$$\begin{align*}

\cos(A \pm B) &= \cos(A)\cos(B) \mp \sin(A)\sin(B) \\

\sin(A \pm B) &= \sin(A)\cos(B) \pm \cos(A)\sin(B) \\

\tan(A \pm B) &= \frac{\tan(A) \pm \tan(B)}{1 \mp \tan(A)\tan(B)}

\end{align*}$$

These formulae can be used to simplify various trigonometric expressions and problems. For example, we can use the compound angle formula for sine to derive the formula for the sine of a double angle, $2\theta$. By substituting $\varphi = \theta$ into the compound angle formula, we obtain:

$$\begin{align*}

\sin(2\theta) &= \sin(\theta + \theta) \\

&= \sin(\theta)\cos(\theta) + \sin(\theta)\cos(\theta) \\

&= 2\sin(\theta)\cos(\theta)

\end{align*}$$

This formula can be further manipulated to express the sine of a double angle in terms of the tangent function:

$$\sin(2\theta) = \frac{2\tan(\theta)}{1 + \tan^2(\theta)}$$

The Pythagorean identity

Sin^2(θ) + cos^2(θ) = 1

This equation is derived from the Pythagorean theorem, which states that in a right-angled triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides. By defining the trigonometric ratios for a right-angled triangle, we can express this theorem in terms of sine and cosine.

The sine of an angle θ is defined as the ratio of the side opposite the angle to the hypotenuse:

Sin(θ) = opposite/hypotenuse

The cosine of an angle θ is defined as the ratio of the side adjacent to the angle to the hypotenuse:

Cos(θ) = adjacent/hypotenuse

By squaring both definitions and adding them together, we get:

Sin^2(θ) + cos^2(θ) = opposite^2/hypotenuse^2 + adjacent^2/hypotenuse^2

According to the Pythagorean theorem, the left-hand side of this equation is equal to 1, as it represents the ratio of the sum of the squares of the two legs of the triangle to the square of the hypotenuse. Therefore, we arrive at the Pythagorean identity:

Sin^2(θ) + cos^2(θ) = 1

This identity holds true for all values of θ in the unit circle. It is one of the basic relations between the sine and cosine functions and is used extensively in trigonometry to solve problems and prove other trigonometric identities.

Sec^2(θ) - tan^2(θ) = 1

Similarly, dividing each term by sin^2(θ) yields:

1 + cot^2(θ) = csc^2(θ)

These identities provide relationships between different trigonometric functions and are valuable tools for simplifying trigonometric expressions and solving equations.

The formula for the square of a polynomial

The formula for the sine of a double angle is a trigonometric identity, comparing two mathematical expressions that remain valid for any values in a specified range. To calculate sin 2 theta, we can use the double angle formula, which is a particular case of the compound angle formula:

Sin(α + β) = sin(α)cos(β) + sin(β)cos(α)

Substituting β with α gives us the double angle identity for the sine:

Sin(α + α) = sin(2α) = 2sin(α)cos(α)

We can also calculate sin 2 theta using other trigonometric identities and the tangent. One such identity is the Pythagorean identity:

Sin^2(θ) + cos^2(θ) = 1

Using the formula for the square of a polynomial:

A+b)^2 = a^2+b^2+2ab

We can rewrite the result of the double angle formula as a double product, specifically the square of the sum of sine and cosine of theta:

Sin(θ) + cos(θ))^2 = sin^2(θ) + cos^2(θ) + 2sin(θ)cos(θ)

Substituting the Pythagorean identity, we get:

Sin(θ) + cos(θ))^2 = 1 + 2sin(θ)cos(θ)

Rearranging the order of the factors, we find:

Sin(2θ) = 2sin(θ)cos(θ) = (sin(θ) + cos(θ))^2 - 1

If we introduce the tangent, we can find another way to calculate sin 2 theta. First, rewrite the result of the double angle formula:

2sin(θ)cos(θ) = 2tan(θ)cos^2(θ)

Move the square of the cosine to the denominator and use the Pythagorean identity:

2tan(θ)cos^2(θ) = 2tan(θ)/(sin^2(θ) + cos^2(θ)) = 2tan(θ)/cos^2(θ)

Split the denominator to find the tangent:

2tan(θ)/[tan^2(θ) + 1]

So, the formula for the square of a polynomial is:

A+b)^2 = a^2+2ab+b^2

This formula can be derived by multiplying a polynomial by itself:

Ax^2 + bx + c)^2 = (ax^2 + bx + c) * (ax^2 + bx + c)

Expanding this expression gives:

A^2x^4 + abx^3 + acx^2 + abx^3 + b^2x^2 + bcx + acx^2 + bcx + c^2

Combining like terms:

A^2x^4 + 2abx^3 + (2ac + b^2)x^2 + 2bcx + c^2

Which can be simplified to:

A^2x^4 + 2abx^3 + (2ac + b^2)x^2 + 2bcx + c^2

Frequently asked questions