Math Doubts

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
Math Doubts
Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers. Know more
Follow us on Social Media
Mobile App for Android users Math Doubts Android App
Math Problems

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.

Learn more