Sine Double angle formula

Formula

$\large \sin{2\theta} \,=\, 2\sin{\theta}\cos{\theta}$

Other forms

Sine of double angle rule is also written in some standard forms.

Angle Double angle Formula
$x$ $2x$ $\sin{2x}$ $\,=\,$ $2\sin{x}\cos{x}$
$A$ $2A$ $\sin{2A}$ $\,=\,$ $2\sin{A}\cos{A}$
$\alpha$ $2\alpha$ $\sin{2\alpha}$ $\,=\,$ $2\sin{\alpha}\cos{\alpha}$

It expresses that the angle of the right angled triangle can be represented by any symbol but the double angle is twice to that symbol and the sine of double angle property is expanded in terms of sine and cosine of the angle in the same form.

Proof

Take theta $(\theta)$ is an angle of a right angled triangle and its double angle is denoted by $2\theta$. Sine of double angle is written as $\sin{2\theta}$. The expansion of the sine of double angle can be derived in geometrical method.

01

According to ΔICG

double angle triangle

Consider $\Delta ICG$. The angle of this triangle is $2\theta$. So, sine of double angle is written as $\sin 2\theta$ and express it in terms of ratio of the lengths of the sides.

$\sin 2\theta \,=\, \dfrac{GI}{CG}$

The length of the side $\overline{GI}$ can be written as the sum of the length of the sides $\overline{GH}$ and $\overline{HI}$.

$\implies \sin 2\theta \,=\, \dfrac{GH+HI}{CG}$

$\implies \sin 2\theta \,=\, \dfrac{GH}{CG}+\dfrac{HI}{CG}$

The sides $\overline{HI}$ and $\overline{DF}$ are parallel lines and their lengths are also equal exactly. Hence, replace the length of the side $\overline{HI}$ by $\overline{DF}$.

$\implies \sin 2\theta \,=\, \dfrac{GH}{CG}+\dfrac{DF}{CG}$

02

According to ΔHGF

right angled triangle HGF

Now, focus on right angled triangle $HGF$ to transform the length of the side $\overline{GH}$.

$\cos \theta \,=\, \dfrac{GH}{FG}$

$\implies GH = FG \cos \theta$

Now, the length of the side $\overline{GH}$ can be replaced by the product of the length of the side $\overline{FG}$ and cosine of angle theta.

$\therefore \,\,\, \sin 2 \theta = \dfrac{FG \cos \theta}{CG} + \dfrac{DF}{CG}$

03

According to ΔDCF

right angled triangle DCF

Look at the $\Delta DCF$ to transform the length of the side $\overline{DF}$.

$\sin{\theta} \,=\, \dfrac{DF}{CF}$

$\implies DF = CF \sin{\theta}$

Now, express the length of the sides $\overline{DF}$ as the product of the length of the side $\overline{CF}$ and sine of angle theta.

$\therefore \,\,\, \sin 2 \theta = \dfrac{FG \cos \theta}{CG} + \dfrac{CF \sin{\theta}}{CG}$

04

According to ΔFCG

$\implies \sin{2\theta} = \dfrac{FG}{CG} \times \cos{\theta} + \dfrac{CF}{CG} \times \sin{\theta}$

right angled triangle FCG

Now, it is time to replace the ratio of the lengths of the sides. The right angled triangle $FCG$ is the only triangle which is formed by the sides of $\overline{FG}$, $\overline{CF}$ and $\overline{CG}$.

$\sin{\theta} = \dfrac{FG}{CG}$

$\cos{\theta} = \dfrac{CF}{CG}$

Now, substitute these two ratios in terms of trigonometric functions in the equation to obtain the expansion of the sine of double angle.

$\implies \sin{2\theta}$ $\,=\,$ $\sin{\theta} \times \cos{\theta}$ $+$ $\cos{\theta} \times \sin{\theta}$

$\implies \sin{2\theta}$ $\,=\,$ $\sin{\theta} \cos{\theta}$ $+$ $\sin{\theta} \cos{\theta}$

$\therefore \,\,\,\,\,\, \sin{2\theta}$ $\,=\,$ $2\sin{\theta} \cos{\theta}$

It is an expansion of the sine of double angle in terms of sine and cosine of the angle. It is used as a trigonometric formula to expand sine of angle double as twice the product of sine and cosine of angle.

Verification

Take $\theta = 30^\circ$ and get values of both sides and compare the results of them.

Evaluate the left hand side expression for angle $30^\circ$.

$\sin{2(30^\circ)}$

$= \sin{60^\circ}$

$= \dfrac{\sqrt{3}}{2}$

Similarly, find the right hand side expression for the angle $30^\circ$.

$2\sin{(30^\circ)}\cos{(30^\circ)}$

$= 2 \times \dfrac{1}{2} \times \dfrac{\sqrt{3}}{2}$

$= \dfrac{2}{2} \times \dfrac{\sqrt{3}}{2}$

$\require{cancel} = \dfrac{\cancel{2}}{\cancel{2}} \times \dfrac{\sqrt{3}}{2}$

$= 1 \times \dfrac{\sqrt{3}}{2}$

$= \dfrac{\sqrt{3}}{2}$

Now, compare both results to verify the double angle formula of the sine function. They both are equal.

$\sin{2(30^\circ)}$ $\,=\,$ $2\sin{(30^\circ)}\cos{(30^\circ)}$ $\,=\,$ $\dfrac{\sqrt{3}}{2}$

The expansion of the sine of double angle is true for all the angles. Therefore, the expansion of the sine of double angle is also called as a trigonometric identity.

Alternative form

Sine double angle formula can also express in terms of tangent of angle.

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