The sin double angle identity is a double angle trigonometric identity and it is used as a formula in trigonometry to expand sin double angle functions such as $\sin{2x}$, $\sin{2\theta}$, $\sin{2A}$, $\sin{2\alpha}$ and etc. The expansion of sin double angle formula is actually derived in geometric system. Now, it is your time to learn how to derive sin double angle identity geometrically.

The sin of double angle identity is derived geometrically by constructing a right triangle with double angle. In this case, the right triangle (represented by $\Delta DCE$) with double angle is constructed geometrically by the following four steps.

- Draw a line from point $\small C$ to side $\small \overline{DE}$ to bisect the $\small \angle DCE$ and it intersects the side $\small \overline{DE}$ at point $\small F$. Take, each angle is represented by $x$.
- Draw a line to side $\small \overline{CD}$ from point $\small F$ but it should be perpendicular to side $\small \overline{CF}$ and take that it intersects the side $\small \overline{CD}$ at point $\small G$.
- Draw a perpendicular line to side $\small \overline{CE}$ from point $\small G$. It intersects the side $\small \overline{CF}$ at point $\small H$ and perpendicularly intersects side $\small \overline{CE}$ at point $\small I$.
- Draw a perpendicular line to side $\small \overline{GI}$ from point $\small F$ and it intersects the side $\small \overline{GI}$ at point $\small J$. Similarly, draw a perpendicular line to side $\small \overline{DE}$ from point $\small H$ and it intersects the side $\small \overline{DE}$ at point $\small K$.

A right triangle with double angle is constructed geometrically and it is used to derive the expansion of sin double angle identity in trigonometric form.

The side $\overline{CF}$ bisects the $\angle DCE$ as two equal angles $\angle ECF$ and $\angle FCD$. If $\angle ECF$ is denoted by $x$, then $\angle FCD$ is also equal to $x$.

$\angle DCE = \angle ECF + \angle FCD$

$\implies \angle DCE = x + x$

$\implies \angle DCE = 2x$

The $\angle DCE$ and $\angle ICG$ are congruent. Therefore, $\angle ICG = \angle DCE = 2x$.

Now, write sin of double angle ($\sin{2x}$) in its trigonometric ratio form as per $\Delta ICG$.

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

The side $\overline{JF}$ splits the side $\overline{GI}$ as two sides $\overline{GJ}$ and $\overline{JI}$. So, the length of side $\overline{GI}$ can be written as the sum of them mathematically.

$GI = GJ + JI$

Now, replace the length of side $\overline{GI}$ by its equivalent value in $\sin{(2x)}$ expansion.

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

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

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 $\overline{EF}$. Therefore, $JI = EF$.

$\implies$ $\sin{2x} \,=\, \dfrac{GJ}{CG}+\dfrac{EF}{CG}$

The $\sin{(2x)}$ is expanded in terms of the ratios between the sides. Now, let’s try to express each ratio in the form a trigonometric function.

The side $\overline{EF}$ is opposite side of the $\Delta ECF$ and its angle is $x$. Try to write the length of side $\overline{EF}$ in terms of a trigonometric function.

$\sin{x} = \dfrac{EF}{CF}$

$\implies EF = {CF}\sin{x}$

Now, substitute the length of perpendicular $\overline{EF}$ by its equivalent value in the expansion of $\sin{2x}$.

$\implies$ $\sin{2x} \,=\, \dfrac{GJ}{CG}+\dfrac{{CF}\sin{x}}{CG}$

$\implies$ $\sin{2x} \,=\, \dfrac{GJ}{CG}+\dfrac{CF}{CG} \times \sin{x}$

The sides $\overline{CF}$ and $\overline{CG}$ are sides of the right triangle $GCF$ and the angle of this triangle is also $x$. According to trigonometry, the ratio of them can be represented by $\cos{x}$.

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

Now, replace the ratio of lengths of sides $\overline{CF}$ to $\overline{CG}$ by its equivalent trigonometric function in the expansion of sin of double angle function.

$\implies$ $\sin{2x} \,=\, \dfrac{GJ}{CG}+\cos{x} \times \sin{x}$

$\implies$ $\sin{2x} \,=\, \dfrac{GJ}{CG}+\cos{x}\sin{x}$

$\implies$ $\sin{2x} \,=\, \dfrac{GJ}{CG}+\sin{x}\cos{x}$

The side $\overline{GJ}$ is one of the sides of the $\Delta JGF$ but its angle is unknown. So, it is not possible to express the length of the side $\overline{GJ}$ in terms of a trigonometric function. Hence, let’s find the angle of this triangle geometrically.

The sides $\overline{CE}$ and $\overline{JF}$ are parallel lines and $\overline{CF}$ is their transversal. $\angle ECF$ and $\angle JFC$ are interior alternate angles and they are equal geometrically.

$\angle JFC = \angle ECF = x$

The side $\overline{FG}$ is a perpendicular line to side $\overline{CF}$. So, the $\angle CFG = 90^°$. but the side $\overline{JF}$ splits the $\angle CFG$ as $\angle JFC$ and $\angle JFG$.

$\angle JFC + \angle JFG = 90^°$

It is proved about that $\angle JFC = x$.

$\implies x + \angle JFG = 90^°$

$\implies \angle JFG = 90^°-x$

$\Delta JGF$ is a right triangle in which $\angle GJF = 90^°$ and $\angle JFG = 90^°-x$. The $\angle JGF$ can be calculated by the sum of angles rule of a triangle.

$\angle JGF + \angle GJF + \angle JFG$ $=$ $180^°$

$\implies \angle JGF + 90^° + 90^°-x$ $=$ $180^°$

$\implies \angle JGF + 180^°-x$ $=$ $180^°$

$\implies \angle JGF $ $=$ $180^°-180^°+x$

$\,\,\, \therefore \,\,\,\,\,\, \angle JGF = x$

It is proved that the $\angle JGF$ is also equal to $x$ because it is a similar triangle.

It is already derived that

$\sin{2x} \,=\, \dfrac{GJ}{CG}+\sin{x}\cos{x}$

Now, let’s try to write the ratio between sides $\overline{GJ}$ and $\overline{CG}$ in terms of trigonometric functions to get the expansion of sin double angle identity.

$\overline{GJ}$ is adjacent side of $\Delta JGF$ and its angle is derived as $x$.

$\cos{x} = \dfrac{GJ}{GF}$

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

Substitute the length of $\overline{GJ}$ by its equivalent value in the expansion of $\sin{(2x)}$ function.

$\sin{2x}$ $\,=\,$ $\dfrac{{GF}\cos{x}}{CG}+\sin{x}\cos{x}$

$\implies \sin{2x}$ $\,=\,$ $\dfrac{GF}{CG} \times \cos{x}+\sin{x}\cos{x}$

The sides $\overline{GF}$ and $\overline{CG}$ are sides of $\Delta GFC$. So, consider right triangle $GFC$ once more time.

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

Therefore, replace the ratio of lengths of sides $\overline{GF}$ to $\overline{CG}$ by trigonometric function $\sin{x}$ in the $\sin{2x}$ expansion.

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

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

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

Therefore, it is proved geometrically that the sin of double angle function can be expanded as two times the product of sin of angle and cos of angle.

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