Proof of Sin double angle formula

Sine of double angle is a trigonometric identity and it is used as a rule to expand sine of double functions such as $\sin{2x}$, $\sin{2\theta}$, $\sin{2A}$, $\sin{2\alpha}$ and etc. in terms of sine of angle and cosine of angle.

$(1) \,\,\,\,\,\,$ $\sin{2\theta} \,=\, 2\sin{\theta}\cos{\theta}$

$(2) \,\,\,\,\,\,$ $\sin{2x} \,=\, 2\sin{x}\cos{x}$

$(3) \,\,\,\,\,\,$ $\sin{2A} \,=\, 2\sin{A}\cos{A}$

$(4) \,\,\,\,\,\,$ $\sin{2\alpha} \,=\, 2\sin{\alpha}\cos{\alpha}$

Proof

Sine of double angle rule is actually derived mathematically in geometrical system. So, learn how to derive the sine of double angle identity geometrically to use it as a trigonometric formula for expanding sine double angle functions easily in terms of sine and cosine of angles.

Construction of a Triangle with double angle

animation for constructing double angle triangle

$\Delta DCE$ is a right angled triangle. Draw an angle bisector from point $C$ and extend it until it intersects the side $\overline{DE}$ at point $F$. Take each angle as $\theta$.

Draw a line from point $F$ (perpendicular line to $\overline{CF}$) and extend it until it intersects the side $\overline{CE}$ at point $G$.

Draw a perpendicular line to side $\overline{CD}$ from point $G$. It intersects the side $\overline{CF}$ at point $H$ and also intersects the side $\overline{CD}$ at point $I$ perpendicularly.

Finally, draw a perpendicular line to side $\overline{GI}$ from point $F$ and it intersects the side $\overline{GI}$ at point $J$. Similarly, draw a perpendicular line to side $\overline{DE}$ from point $H$ and it intersects the side $\overline{DE}$ at point $K$.

Expressing sine double angle in terms of ratio of sides

Consider $\Delta ICG$ and its angle is $2\theta$. Write, sine of double angle ($\sin{2\theta}$) in terms of ratio of the sides.

double angle triangle

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

The length of side $\overline{GI}$ can be written as sum of the lengths of the sides $\overline{GJ}$ and $\overline{JI}$.

$\implies$ $\sin{2\theta} \,=\, \dfrac{GJ+JI}{CG}$

$\implies$ $\sin{2\theta} \,=\, \dfrac{GJ}{CG}+\dfrac{JI}{CG}$

The sides $\overline{GI}$ and $\overline{DE}$ are parallel lines. So, the length of the side $\overline{JI}$ is exactly equal to the length of the side $\overline{FD}$. Therefore, $JI = FD$.

$\implies$ $\sin{2\theta} \,=\, \dfrac{GJ}{CG}+\dfrac{FD}{CG}$

Transforming the equation in terms of cos of angle

$\Delta DCF$ and $\Delta KHF$ are similar triangles. So, the $\angle KHF$ is same as $\angle DCF$, which means $\angle KHF = \theta$. Geometrically, the $\angle KHF$ and $\angle HFJ$ are interior alternative angles. Thus, the $\angle HFJ = \theta$.

double angle triangle

Geometrically, $\angle GFH$ is a right angle. So, $\angle GFH = 90^\circ$ but $\angle HFJ = \theta$. Hence, $\angle GFJ = \angle GFH – \angle HFJ$.

$\,\,\, \therefore \,\,\,\,\,\,$ $\angle GFJ = 90^\circ –\theta$

$\Delta JGF$ is a right angled triangle. In this triangle, the $\angle GJF = 90^\circ$ and $\angle GFJ = 90^\circ -\theta$ but $\angle JGF$ is unknown but it can be evaluated by applying sum of angles of a triangle rule.

$\angle GJF + \angle GFJ+ \angle JGF = 180^\circ$

$\implies$ $90^\circ + 90^\circ-\theta + \angle JGF = 180^\circ$

$\implies$ $180^\circ -\theta + \angle JGF = 180^\circ$

$\implies$ $\angle JGF = 180^\circ -180^\circ +\theta$

$\implies$ $\angle JGF = \theta$

Now, comeback to sine double angle equation.

$\implies$ $\sin{2\theta} \,=\, \dfrac{GJ}{CG}+\dfrac{FD}{CG}$

The length of side $\overline{GJ}$ can be expressed in another form as per the $\Delta JGF$.

$\cos{\theta} \,=\, \dfrac{GJ}{GF}$

$\implies GJ = GF \cos{\theta}$

$\,\,\, \therefore \,\,\,\,\,\,$ $\sin{2\theta} \,=\, \dfrac{GF \cos{\theta}}{CG}+\dfrac{FD}{CG}$

Transforming the equation in terms of sin of angle

double angle triangle

According to $\Delta DCF$

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

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

Now, transform the sine of double angle equation by the above rule.

$\implies$ $\sin{2\theta} \,=\, \dfrac{GF \cos{\theta}}{CG}+\dfrac{CF \sin{\theta}}{CG}$

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

Obtaining the expansion of sin of double angle

double angle triangle

$\overline{GF}$, $\overline{CF}$ and $\overline{CG}$ are the sides of the $\Delta FCG$. So, transform the ratios of them by respective trigonometric functions as per this triangle.

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

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

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

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

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

Geometrically, it has proved that sine of double angle is expanded as twice the product of sine of angle and cosine of angle.



Follow us
Email subscription