Cos Triple angle formula

Formula

$\large \cos{3\theta}$ $\,=\,$ $4\cos^3{\theta}-3\cos{\theta}$

Other forms

Cos triple angle formula is usually written in three standard forms.

Angle Triple angle Formula
$x$ $3x$ $\cos{3x}$ $\,=\,$ $4\cos^3{x}-3\cos{x}$
$A$ $3A$ $\cos{3A}$ $\,=\,$ $4\cos^3{A}-3\cos{A}$
$\alpha$ $3\alpha$ $\cos{3\alpha}$ $\,=\,$ $4\cos^3{\alpha}-3\cos{\alpha}$

The angle of the right angled triangle can be denoted by any symbol but the triple angle is equal to thrice the symbol. However, the expansion of cosine of triple angle rule is in same form.

Proof

Assume $\theta$ is an angle of the right angled triangle and triple angle is written as $3\theta$. Cosine of triple angle is written as $\cos{3\theta}$ in trigonometry and its expansion can be derived in geometrical method as follows.

01

According to ΔJDH

triple angle triangle

The angle of the right angled triangle $JDH$ is $3\theta$ and express cosine of triple angle in ratio form by this triangle.

$\cos{3\theta} \,=\, \dfrac{DJ}{DH}$

The side $\overline{DE}$ is divided as sides $\overline{DJ}$ and $\overline{JE}$ by the perpendicular side $\overline{HJ}$ at point $J$. Due to this, the length of the side $\overline{DJ}$ can be written as the subtraction of the length of the side $\overline{JE}$ from the length of the side $\overline{DE}$.

$DJ \,=\, DE\,–\,JE$

Replace the length of the side $\overline{DJ}$ in ratio form of the cosine of triple angle formula.

$\implies$ $\cos{3\theta} \,=\, \dfrac{DE\,-\,JE}{DH}$

02

According to ΔEDG

right angled triangle EDG

$\overline{DE}$ is adjacent side of the right angled triangle $EDG$ and the angle of the $\Delta EDG$ is $2\theta$.

On the basis of $\Delta EDG$, express the length of the side $\overline{DE}$ in its alternative form.

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

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

The mathematical relation expresses that the length of the side $\overline{DE}$ can be written as the product of the length of the side $\overline{DG}$ and cosine of double angle theta.

$\implies$ $\cos{3\theta} \,=\, \dfrac{DG \cos{2\theta} \,-\,JE}{DH}$

03

Relation between JE and KG

JE and KG parallel lines

The sides $\overline{JE}$ and $\overline{KG}$ are parallel lines geometrically and their lengths are also same.

$JE \,=\, KG$

Hence, the length of the side $\overline{JE}$ can be replaced by the length of the side $\overline{KG}$ in the expansion of the cos of triple angle law.

$\implies$ $\cos{3\theta} \,=\, \dfrac{DG \cos{2\theta} \,-\,KG}{DH}$

04

According to ΔKHG

right angled triangle KHG

$\overline{KG}$ is opposite side of the right angled triangle $KHG$. As per this triangle, write the length of the side $\overline{KG}$ in its equivalent form. Geometrically, the angle of this triangle is also $\theta$.

$\sin{2\theta} \,=\, \dfrac{KG}{GH}$

$\implies KG \,=\, GH \sin{2\theta}$

So, the length of the side $\overline{KG}$ can be substituted by the product of the length of the side $\overline{GH}$ and sine of double angle theta.

$\implies$ $\cos{3\theta} \,=\, \dfrac{DG \cos{2\theta} \,-\, GH \sin{2\theta}}{DH}$

05

According to ΔGDH

$\implies$ $\cos{3\theta}$ $\,=\,$ $\dfrac{DG}{DH} \times \cos{2\theta}$ $\,-\,$ $\dfrac{GH}{DH} \times \sin{2\theta}$

right angled triangle GDH

$\overline{DG}$, $\overline{DH}$ and $\overline{GH}$ are sides of the right angled triangle $GDH$. According to this triangle, the ratio of the sides can be expressed in trigonometric functions form.

$\sin{\theta} \,=\, \dfrac{GH}{DH}$

$\cos{\theta} \,=\, \dfrac{DG}{DH}$

As per these two trigonometric expressions, transform the expansion of the cos triple angle rule.

$\implies$ $\cos{3\theta}$ $\,=\,$ $\cos{\theta} \times \cos{2\theta}$ $\,-\,$ $\sin{\theta} \times \sin{2\theta}$

$\implies$ $\cos{3\theta}$ $\,=\,$ $\cos{\theta} \cos{2\theta}$ $\,-\,$ $\sin{\theta} \sin{2\theta}$

06

Application of Double angle rules

The cos triple angle formula is derived in terms of sine and cosine functions with angle and double angle. Apply double angle formulas to transform the expansion in terms of angle of the triangle purely.

Cosine of double angle formula can be expanded in terms of cosine of angle.

$\cos{2\theta} \,=\, 2\cos^2{\theta}-1$

Similarly, the sine of double angle formula can also be expanded in terms of sine of angle.

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

Transform the entire expansion in terms of angle of the triangle purely by these two trigonometric identities.

$\implies$ $\cos{3\theta}$ $\,=\,$ $\cos{\theta}(2\cos^2{\theta}-1)$ $\,-\,$ $\sin{\theta}(2\sin{\theta}\cos{\theta})$

$\implies$ $\cos{3\theta}$ $\,=\,$ $2\cos^3{\theta}-\cos{\theta}$ $\,-\,$ $2\sin^2{\theta}\cos{\theta}$

$\implies$ $\cos{3\theta}$ $\,=\,$ $2\cos^3{\theta}-\cos{\theta}$ $\,-\,$ $2\cos{\theta}\sin^2{\theta}$

The square of sine of angle can be converted in terms of square of cosine of angle by the Pythagorean identity of sine and cosine functions.

$\implies$ $\cos{3\theta}$ $\,=\,$ $2\cos^3{\theta}-\cos{\theta}$ $\,-\,$ $2\cos{\theta}(1-\cos^2{\theta})$

$\implies$ $\cos{3\theta}$ $\,=\,$ $2\cos^3{\theta}-\cos{\theta}$ $\,-\,$ $2\cos{\theta}+2\cos^3{\theta}$

$\therefore \,\,\,\,\,\,$ $\cos{3\theta}$ $\,=\,$ $4\cos^3{\theta}-3\cos{\theta}$

Therefore, it is derived that cosine of triple angle is equal to the subtraction of the thrice cosine of angle from the four times cube of the cosine of angle. It is usually used to expand the cosine of triple angle in terms of cosine of angle and vice-versa.

Verification

For example, consider $\theta = 30^\circ$. and find the value of both sides of the formula.

$\cos{(3 \times 30^\circ)}$

$=\,\,\, \cos{(90^\circ)}$

$=\,\,\, 0$

This time, evaluate the expansion of the cos triple angle property.

$=\,\,\, 4\cos^3{(30^\circ)}-3\cos{(30^\circ)}$

$=\,\,\, 4 \times {\Bigg(\dfrac{\sqrt{3}}{2}\Bigg)}^3-3\Bigg(\dfrac{\sqrt{3}}{2}\Bigg)$

$=\,\,\, 4 \times \Bigg(\dfrac{3\sqrt{3}}{8}\Bigg)-\dfrac{3\sqrt{3}}{2}$

$=\,\,\, \Bigg(4 \times \dfrac{3\sqrt{3}}{8}\Bigg)-\dfrac{3\sqrt{3}}{2}$

$\require{cancel} =\,\,\, \dfrac{\cancel{4} \times 3\sqrt{3}}{\cancel{8}}-\dfrac{3\sqrt{3}}{2}$

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

$=\,\,\, 0$

Observe the values of both sides of this formula.

$\cos{(3 \times 30^\circ)}$ $\,=\,$ $4\cos^3{(30^\circ)}-3\cos{(30^\circ)}$ $\,=\,$ $0$

They both are same and equal to zero. It is proved that cos triple angle formula is true for all the angles and it is called as a triple angle trigonometric identity.

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