$\tan{2x}$, $\tan{2\theta}$, $\tan{2A}$ and $\tan{2\alpha}$ are most popular examples for tan double angle function. It is used to expand tan double angle function in terms of tan of angle.

$(1) \,\,\,\,\,\,$ $\tan{2\theta} \,=\, \dfrac{2\tan{\theta}}{1-2\tan^2{\theta}}$

$(2) \,\,\,\,\,\,$ $\tan{2x} \,=\, \dfrac{2\tan{x}}{1-2\tan^2{x}}$

$(3) \,\,\,\,\,\,$ $\tan{2A} \,=\, \dfrac{2\tan{A}}{1-2\tan^2{A}}$

$(4) \,\,\,\,\,\,$ $\tan{2\alpha} \,=\, \dfrac{2\tan{\alpha}}{1-2\tan^2{\alpha}}$

Tan double angle identity in mathematical form is derived in trigonometry by geometrical system.

$\Delta ECD$ is a right angled triangle. Bisect the angle by drawing a line to side $\overline{ED}$ from point $C$ and it intersect the side $\overline{ED}$ at point $F$. Assume, each angle is denoted by theta symbol.

Draw a perpendicular line to side $\overline{CF}$ from point $F$ and it intersects the side $\overline{CE}$ at point $G$. Now, 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 $D$.

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 $\overline{ED}$ at point $K$.

$2\theta$ is double angle of $\Delta ICG$. Now, write tan of double angle ($\tan{2\theta}$) in terms of ratio of sides of this triangle.

$\tan{2\theta} \,=\, \dfrac{IG}{IC}$

The length of the side $\overline{IG}$ can be written as the sum of the lengths of the sides $\overline{IJ}$ and $\overline{JG}$. Similarly, the length of the side $\overline{IC}$ as the subtraction of length of side $\overline{CD}$ from length of the side $\overline{ID}$.

In other words, $IC = CD–ID$ and $IG = IJ+JG$

$\implies$ $\tan{2\theta} \,=\, \dfrac{IJ+JG}{CD-ID}$

At point $H$, the side $\overline{GI}$ is split into two sides $\overline{GH}$ and $\overline{HI}$. Therefore, the length of the side $\overline{GI}$ can be written as the sum of the lengths of the sides $\overline{GH}$ and $\overline{HI}$.

$\implies \tan{2\theta} \,=\, \dfrac{GH+HI}{CI}$

At point $I$, the side $\overline{CD}$ is divided into two sides $\overline{CI}$ and $\overline{ID}$. So, the length of the side $\overline{CI}$ can be written as the subtraction of the length of the side $\overline{ID}$ from the length of the side $\overline{CD}$.

$\implies \tan{2\theta} \,=\, \dfrac{GH+HI}{CD-ID}$

The sides $\overline{HI}$ and $\overline{FD}$ are parallel lines and the lengths of them are also equal. Due to this reason, the length of the side $\overline{HI}$ can be replaced by the length of the side $\overline{FD}$. Likewise, the sides $\overline{ID}$ and $\overline{HF}$ are parallel lines and their lengths are also equal. So, replace the length of the side $\overline{ID}$ by the length of the side $\overline{HF}$.

$\implies \tan{2\theta} \,=\, \dfrac{GH+FD}{CD-HF}$

Take the length of the $CD$ common from the denominator.

$\implies \tan{2\theta} \,=\, \dfrac{GH+FD}{CD \Bigg(1-\dfrac{HF}{CD}\Bigg)}$

$\implies \tan{2\theta} \,=\, \dfrac{\dfrac{GH+FD}{CD}}{1-\dfrac{HF}{CD}}$

$\implies \tan{2\theta} \,=\, \dfrac{\dfrac{GH}{CD}+\dfrac{FD}{CD}}{1-\dfrac{HF}{CD}}$

$\overline{FD}$ and $\overline{CD}$ are the sides of the right angled triangle $DCF$. So, consider this triangle and express the ratio of length of the side $\overline{FD}$ to length of the side $\overline{CD}$.

$\tan{\theta} \,=\, \dfrac{FD}{CD}$

Replace the ratio of the sides by the tan function.

$\implies \tan{2\theta} \,=\, \dfrac{\dfrac{GH}{CD}+\tan{\theta}}{1-\dfrac{HF}{CD}}$

03

$\overline{GH}$ is the adjacent side of the triangle $HGF$ and $\overline{CD}$ is adjacent side of the $\Delta DCF$. Geometrically, $\Delta HGF$ and $\Delta DCF$ are similar triangles.

Express the lengths of adjacent sides $\overline{GH}$ and $\overline{CD}$ in alternative form.

According to $\Delta HGF$.

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

According to $\Delta DCF$.

$\cos{\theta} \,=\, \dfrac{CD}{CF}$

The two ratios are same. So, make them equal.

$\dfrac{GH}{FG} \,=\, \dfrac{CD}{CF}$

$\implies \dfrac{GH}{CD} \,=\, \dfrac{GF}{CF}$

Replace the ratio of length of the side $\overline{GH}$ to length of the side $\overline{CD}$ by the ratio of length of the side $\overline{CD}$ to length of the side $\overline{CF}$ in the expansion of the cos of double angle formula.

$\implies \tan{2\theta} \,=\, \dfrac{\dfrac{GF}{CF}+\tan{\theta}}{1-\dfrac{HF}{CD}}$

04

$\overline{GF}$ and $\overline{CF}$ are two sides of the right angled triangle $FCG$.

The ratio of length of the side $\overline{GF}$ to length of the side $\overline{CF}$ is tan of angle theta.

$\tan{\theta} \,=\, \dfrac{GF}{CF}$

Substitute the ratio of length of the side GF to length of the side CF by its equal value.

$\implies \tan{2\theta} \,=\, \dfrac{\tan{\theta}+\tan{\theta}}{1-\dfrac{HF}{CD}}$

$\implies \tan{2\theta} \,=\, \dfrac{2\tan{\theta}}{1-\dfrac{HF}{CD}}$

05

Once again, use the property of similar triangles of $\Delta HGF$ and $\Delta DCF$.

According to $\Delta HGF$.

$\sin{\theta} \,=\, \dfrac{HF}{GF}$

According to $\Delta DCF$.

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

The two expressions are same. Therefore, make them equal.

$\dfrac{HF}{GF} \,=\, \dfrac{DF}{CF}$

$\implies HF \,=\, \dfrac{DF \times GF}{CF}$

Replace the length of the side $\overline{HF}$ by its equivalent value in the expansion of the tan of double angle law.

$\implies \tan{2\theta} \,=\, \dfrac{2\tan{\theta}}{1-\dfrac{\dfrac{DF \times GF}{CF}}{CD}}$

$\implies \tan{2\theta} \,=\, \dfrac{2\tan{\theta}}{1-\dfrac{DF \times GF}{CD \times CF}}$

$\implies \tan{2\theta} \,=\, \dfrac{2\tan{\theta}}{1-\dfrac{DF}{CD} \times \dfrac{GF}{CF}}$

06

$\overline{DF}$ and $\overline{CD}$ are the sides of the right angled triangle $DCF$.

Express the ratio of the length of the side $\overline{DF}$ to length of the side $\overline{CD}$ and it is the equal to tan of the angle theta.

$\tan{\theta} \,=\, \dfrac{DF}{CD}$

Replace the value of the ratio of the sides in the formula.

$\implies \tan{2\theta} \,=\, \dfrac{2\tan{\theta}}{1-\tan{\theta} \times \dfrac{GF}{CF}}$

07

$\overline{GF}$ and $\overline{CF}$ are two sides of the right angled triangle $FCG$. So, the ratio of length of the side $\overline{GF}$ to length of the side $\overline{CF}$ in trigonometric function form.

$\tan{\theta} \,=\, \dfrac{GF}{CF}$

Write the ratio of the sides as the tan of angle theta to complete the proof of tan of double angle rule geometrically.

$\implies \tan{2\theta} \,=\, \dfrac{2\tan{\theta}}{1-\tan{\theta} \times \tan{\theta}}$

$\therefore \,\,\,\,\,\, \tan{2\theta} \,=\, \dfrac{2\tan{\theta}}{1-\tan^2{\theta}}$

The trigonometric formula is mainly used to expand the tan of double angle in terms of tan of angle and can also use in reverse operation.

Take $\theta \,=\, 30^\circ$ and check both sides of the formula.

$\tan{(2 \times 30^\circ)}$

$=\,\,\, \tan{(60^\circ)}$

$=\,\,\, \sqrt{3}$

Now, evaluate the expansion of the tan of the double angle.

$\dfrac{2\tan{(30^\circ})}{1-\tan^2{(30^\circ)}}$

$=\,\,\, \dfrac{2 \times \dfrac{1}{\sqrt{3}}}{1-{\Bigg(\dfrac{1}{\sqrt{3}}\Bigg)}^2}$

$=\,\,\, \dfrac{\dfrac{2}{\sqrt{3}}}{1-\dfrac{1}{3}}$

$=\,\,\, \dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{3-1}{3}}$

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

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

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

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

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

$=\,\,\, \sqrt{3}$

Compare both results and they both are equal. Hence, the tan of double angle formula is called as a double angle trigonometric identity.

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