Tan Triple angle formula

Formula

$\large \tan{3\theta} \,=\, \dfrac{3\tan{\theta}-\tan^3{\theta}}{1-3\tan^2{\theta}}$

Other forms

Tangent of triple angle formula is written in the following three standard forms alternatively.

Angle Triple angle Formula
$x$ $3x$ $\tan{3x} \,=\, \dfrac{3\tan{x}-\tan^3{x}}{1-3\tan^2{x}}$
$A$ $3A$ $\tan{3A} \,=\, \dfrac{3\tan{A}-\tan^3{A}}{1-3\tan^2{A}}$
$\alpha$ $3\alpha$ $\tan{3\alpha} \,=\, \dfrac{3\tan{\alpha}-\tan^3{\alpha}}{1-3\tan^2{\alpha}}$

The angle of a right angled triangle can be denoted by symbol but the triple angle should be thrice of that symbol. Likewise, the expansion of tan of the triple angle will be in the same form.

Proof

Assume theta is an angle of the right angled triangle and the triple angle is written as $3\theta$. Tan of triple angle is written as $\tan{3\theta}$ and it can be expanded in terms of tan of angle theta by deriving a proof in geometrical method.

01

According to ΔHDJ

triple angle triangle

In right angled triangle $HDJ$, the angle is $3\theta$ and it represents triple angle. Tan triple angle is written as $\tan{3\theta}$ and express tan of triple angle in terms of ratio of the sides to write it in mathematical form.

$\tan{3\theta} \,=\, \dfrac{HJ}{DJ}$

The side $\overline{HJ}$ is split as sides $\overline{HK}$ and $\overline{KJ}$ at point $K$ by the perpendicular line $\overline{KG}$. Hence, the length of the side $\overline{HJ}$ can be written as sum of the length of the sides $\overline{HK}$ and $\overline{KJ}$.

$HJ \,=\, HK + KJ$

Similarly, the side $\overline{DE}$ is split by the perpendicular side $\overline{HJ}$ at point $J$. So, the length of the side $\overline{DJ}$ can be written as subtraction of the side $\overline{JE}$ from the side $\overline{DE}$.

$DJ = DE\,–\,JE$

Replace the length of the sides $\overline{HJ}$ and $\overline{DJ}$ by these relations in tan triple angle expression.

$\implies \tan{3\theta} \,=\, \dfrac{HK+KJ}{DE-JE}$

02

Relation of side KJ with EG

parallel lines KJ and GE

Geometrically, the side $\overline{KJ}$ is parallel to the side $\overline{EG}$ and the lengths of them are also equal. Therefore, the length of the side $\overline{KJ}$ can be replaced by the length of the side $\overline{EG}$.

KJ = EG

Replace the length of the side $\overline{KJ}$ by the length of the side $\overline{EG}$ in the tan triple angle rule.

$\implies \tan{3\theta} \,=\, \dfrac{HK+EG}{DE-JE}$

03

According to ΔGDE

right angled triangle EDG

$\overline{EG}$ is a side of the right angled triangle $GDE$ and its angle is $2\theta$. Express length of the side $\overline{GE}$ in terms of tan of double angle.

$\tan{2\theta} \,=\, \dfrac{EG}{DE}$

$\implies EG \,=\, DE \tan{2\theta}$

Replace the length of the side $\overline{EG}$ by the product of the length of the side $\overline{DE}$ and tan of double angle in the law of tan of triple angle.

$\implies \tan{3\theta} \,=\, \dfrac{HK+DE\tan{2\theta}}{DE-JE}$

Take the length of the side $\overline{DE}$ common from both numerator and denominator.

$\implies \tan{3\theta} \,=\, \dfrac{DE\Bigg(\dfrac{HK}{DE}+\tan{2\theta \Bigg)}}{DE \Bigg(1-\dfrac{JE}{DE}\Bigg)}$

$\implies \require{cancel} \tan{3\theta} \,=\, \dfrac{\cancel{DE}\Bigg(\dfrac{HK}{DE}+\tan{2\theta \Bigg)}}{\cancel{DE}\Bigg(1-\dfrac{JE}{DE}\Bigg)}$

$\implies \tan{3\theta} \,=\, \dfrac{\dfrac{HK}{DE}+\tan{2\theta}}{1-\dfrac{JE}{DE}}$

04

According to ΔKHG and ΔEDG

right angled triangle KHG and EDG

$\overline{HK}$ and $\overline{DE}$ are adjacent sides of $\Delta KHG$ and $\Delta EDG$ respectively but the ratio of them is interrupted the derivation because they cannot be expressed in terms of tan functions. Hence, they should be expressed in their equivalent value.

Geometrically, $\Delta KHG$ and $\Delta EDG$ are similar triangles. Hence, their angles are also same and the ratios of any two should be equal.

As per $\Delta KHG$

$\cos{2\theta} \,=\, \dfrac{HK}{HG}$

As per $\Delta EDG$

$\cos{2\theta} \,=\, \dfrac{DE}{DG}$

They are equal in value.

$\dfrac{HK}{HG} \,=\, \dfrac{DE}{DG}$

$\implies \dfrac{HK}{DE} \,=\, \dfrac{HG}{DG}$

Use the relation between ratios of sides and replace the ratio of length of the side HK to DE by the ratio of length of the side HG to DG in the property of the tangent triple angle.

$\implies \tan{3\theta} \,=\, \dfrac{\dfrac{HG}{DG}+\tan{2\theta}}{1-\dfrac{JE}{DE}}$

05

According to ΔGDH

right angled triangle GDH

$\overline{HG}$ and $\overline{DG}$ are sides of the right angled triangle $GDH$. So, consider this triangle and express the ratio of length of the side $\overline{HG}$ to length of the side $\overline{DG}$ in trigonometric function form.

$\tan{\theta} \,=\, \dfrac{HG}{DG}$

Now, replace the ratio in the expansion of the tan of triple angle identity.

$\implies \tan{3\theta} \,=\, \dfrac{\tan{\theta}+\tan{2\theta}}{1-\dfrac{JE}{DE}}$

06

Property of side JE with KG

JE and KG parallel lines

$\overline{JE}$ and $\overline{KG}$ are two sides but the side $\overline{JE}$ is parallel to the side $\overline{KG}$ and their lengths are also equal.

$JE \,=\, KG$

Replace the length of the side $\overline{JE}$ by the length of the side $\overline{KG}$ in the expansion of the tangent triple angle rule.

$\implies \tan{3\theta} \,=\, \dfrac{\tan{\theta}+\tan{2\theta}}{1-\dfrac{KG}{DE}}$

07

According to ΔKHG

right angled triangle KHG

$\overline{KG}$ is opposite side of the right angled triangle $KHG$ and angle of this triangle is $2\theta$. Express the relation between them in trigonometric function form.

$\tan{2\theta} \,=\, \dfrac{KG}{HK}$

$\implies KG \,=\, HK \tan{2\theta}$

Replace the length of the side $\overline{KG}$ by the product of the length of the side $\overline{HK}$ and tan of double angle.

$\implies \tan{3\theta} \,=\, \dfrac{\tan{\theta}+\tan{2\theta}}{1-\dfrac{HK \tan{2\theta}}{DE}}$

$\implies \tan{3\theta} \,=\, \dfrac{\tan{\theta}+\tan{2\theta}}{1-\dfrac{HK}{DE} \times \tan{2\theta}}$

08

According to ΔKHG and ΔEDG

right angled triangle KHG and EDG

Just like in step $4$.

$\cos{2\theta} \,=\, \dfrac{HK}{HG} \,=\, \dfrac{DE}{DG}$

$\implies \dfrac{HK}{DE} \,=\, \dfrac{HG}{DG}$

Therefore, use this relation and change the ratio by its equivalent value in the tan triple angle expansion.

$\implies \tan{3\theta} \,=\, \dfrac{\tan{\theta}+\tan{2\theta}}{1-\dfrac{HG}{DG} \times \tan{2\theta}}$

09

According to ΔGDH

right angled triangle GDH

$\overline{HG}$ and $\overline{DG}$ are opposite and adjacent sides of the right angled triangle $GDH$ and the angle of this triangle is theta. Hence, express relation between three of them in mathematical form.

$\tan{\theta} \,=\, \dfrac{HG}{DG}$

Now, replace the ratio in the trigonometric function form in the tangent triple angle formula to obtain its expansion in terms of tan functions.

$\implies \tan{3\theta} \,=\, \dfrac{\tan{\theta}+\tan{2\theta}}{1-\tan{\theta} \times \tan{2\theta}}$

$\implies \tan{3\theta} \,=\, \dfrac{\tan{\theta}+\tan{2\theta}}{1-\tan{\theta} \tan{2\theta}}$

10

Expansion in terms of Tan of angle

Tan of triple angle formula is expanded in terms of tan of angle and tan of double. It can be expressed purely in terms of tan of angle by substituting tan of double angle by its expansion in terms of tan of angle.

According to the expansion of the Tan double angle formula.

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

Now, transform the tan of triple angle formula by this identity.

$\implies$ $\tan{3\theta}$ $\,=\,$ $\dfrac{\tan{\theta}+\dfrac{2\tan{\theta}}{1-\tan^2{\theta}}}{1-\tan{\theta} \times \dfrac{2\tan{\theta}}{1-\tan^2{\theta}}}$

$\implies$ $\tan{3\theta}$ $\,=\,$ $\dfrac{\dfrac{\tan{\theta}(1-\tan^2{\theta}) + 2\tan{\theta}}{1-\tan^2{\theta}}}{1-\dfrac{2\tan^2{\theta}}{1-\tan^2{\theta}}}$

$\implies$ $\tan{3\theta}$ $\,=\,$ $\dfrac{\dfrac{\tan{\theta}-\tan^3{\theta}+ 2\tan{\theta}}{1-\tan^2{\theta}}}{\dfrac{1(1-\tan^2{\theta})-2\tan^2{\theta}}{1-\tan^2{\theta}}}$

$\implies$ $\tan{3\theta}$ $\,=\,$ $\dfrac{\dfrac{3\tan{\theta}-\tan^3{\theta}}{1-\tan^2{\theta}}}{\dfrac{1-\tan^2{\theta}-2\tan^2{\theta}}{1-\tan^2{\theta}}}$

$\implies$ $\tan{3\theta}$ $\,=\,$ $\dfrac{\dfrac{3\tan{\theta}-\tan^3{\theta}}{1-\tan^2{\theta}}}{\dfrac{1-3\tan^2{\theta}}{1-\tan^2{\theta}}}$

$\implies$ $\tan{3\theta}$ $\,=\,$ $\dfrac{3\tan{\theta}-\tan^3{\theta}}{1-\tan^2{\theta}}$ $\times$ $\dfrac{1-\tan^2{\theta}}{1-3\tan^2{\theta}}$

$\implies$ $\tan{3\theta}$ $\,=\,$ $\dfrac{(3\tan{\theta}-\tan^3{\theta}) \times (1-\tan^2{\theta})}{(1-\tan^2{\theta}) \times (1-3\tan^2{\theta})}$

$\implies$ $\tan{3\theta}$ $\,=\,$ $\require{cancel} \dfrac{(3\tan{\theta}-\tan^3{\theta}) \times \cancel{(1-\tan^2{\theta})}}{\cancel{(1-\tan^2{\theta})} \times (1-3\tan^2{\theta})}$

$\therefore \,\,\,\,\,\,$ $\tan{3\theta}$ $\,=\,$ $\dfrac{3\tan{\theta}-\tan^3{\theta}}{1-3\tan^2{\theta}}$

Verification

Take $\theta = 45^\circ$ and evaluate both values of the both sides of this formula. Then compare them.

$\tan{(3 \times 45^\circ)}$

$=\,\,\, \tan{(135^\circ)}$

$=\,\,\, \tan{(90^\circ +45^\circ)}$

$=\,\,\, -\cot{45^\circ}$

$=\,\,\, -1$

Now, find the value of the expansion for angle $45^\circ$.

$=\,\,\, \dfrac{3\tan{45^\circ}-\tan^3{45^\circ}}{1-3\tan^2{45^\circ}}$

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

$=\,\,\, \dfrac{3-1}{1-3 \times 1}$

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

$=\,\,\, \dfrac{2}{-2}$

$\require{cancel} =\,\,\, \dfrac{\cancel{2}}{\cancel{-2}}$

$=\,\,\, \dfrac{1}{-1}$

$=\,\,\, -1$

Compare both results and they are equal in value.

$\tan{(3 \times 45^\circ)}$ $\,=\,$ $\dfrac{3\tan{45^\circ}-\tan^3{45^\circ}}{1-3\tan^2{45^\circ}}$ $\,=\, -1$

Hence, the tan triple angle formula is called a triple angle identity of the trigonometric functions.

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