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

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

Angle | Triple angle | Formula |
---|---|---|

$x$ | $3x$ | $\sin{3x}$ $\,=\,$ $3\sin{x}-4\sin^3{x}$ |

$A$ | $3A$ | $\sin{3A}$ $\,=\,$ $3\sin{A}-4\sin^3{A}$ |

$\alpha$ | $3\alpha$ | $\sin{3\alpha}$ $\,=\,$ $3\sin{\alpha}-4\sin^3{\alpha}$ |

The angle can be represented by any symbol and the triple angle is equal to thrice the symbol but the expansion is in the same form.

Theta is an angle of the triangle and $2\theta$ is another angle of the triangle. The sum of them is $3\theta$ and it is called as triple angle. Sine of triple angle is written as $\sin{3\theta}$ in mathematics. The $\sin{3\theta}$ can be expanded in terms of sine of angle by the geometrical method.

01

Focus on the right angled triangle $JDH$. The angle of it is $3\theta$.

Express sine of triple angle in terms of the ratio of length of the opposite side to length of the hypotenuse of the $\Delta JDH$.

$\sin{3\theta} \,=\, \dfrac{HJ}{DH}$

The side $\overline{HJ}$ is divided into two sides $\overline{HK}$ and $\overline{KJ}$ by the perpendicular line $\overline{KG}$ at point $K$. Therefore, express the length of the side $\overline{HJ}$ as the sum of the lengths of the sides $\overline{HK}$ and $\overline{KJ}$.

$\implies$ $\sin{3\theta} \,=\, \dfrac{HK+KJ}{DH}$

$\implies$ $\sin{3\theta} \,=\, \dfrac{HK}{DH}+\dfrac{KJ}{DH}$

02

$\overline{HK}$ is adjacent side of the $\Delta KHG$ and consider this triangle to express the length of the side $\overline{HK}$ in another form.

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

$\implies HK \,=\, GH \cos{2\theta}$

Replace the length of the side $\overline{HK}$ in expansion of the sine triple angle rule.

$\implies$ $\sin{3\theta} \,=\, \dfrac{GH \cos{2\theta}}{DH}+\dfrac{KJ}{DH}$

03

$\overline{KJ}$ and $\overline{GE}$ are parallel lines and their lengths are same too.

$KJ \,=\, GE$

Therefore, the length of the side $\overline{KJ}$ can be replaced by the length of the side $\overline{GE}$.

$\implies$ $\sin{3\theta} \,=\, \dfrac{GH \cos{2\theta}}{DH}+\dfrac{GE}{DH}$

04

$\overline{GE}$ is a side of the right angled triangle $EDG$. Consider this triangle and transform the length of the side $\overline{GE}$ in its alternative form.

$\sin{2\theta} \,=\, \dfrac{GE}{DG}$

$\implies GE \,=\, DG \sin{2\theta}$

The length of the side $\overline{GE}$ can be substituted by the product of length of the side $\overline{DG}$ and sine of double angle theta.

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

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

05

$\overline{GH}$, $\overline{DG}$ and $\overline{DH}$ are the sides of the right angled triangle $GDH$. Transform the each ratio of two sides in trigonometric functions form to obtain the expansion of the sine of triple angle law.

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

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

Replace the both ratios by the associated trigonometric functions.

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

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

06

The expansion of the sine of triple angle is in terms of angle and double angle. Hence, use double angle trigonometric identities to obtain the expansion of sine of triple angle formula in terms of angle purely.

According to the sin double angle formula, it can be expanded in product form of sine and cosine of angle.

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

As per the cos double angle formula, it can be expanded in terms of the sine of angle.

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

Use both expansions of sine and cosine of double angle identities to replace them in the expansion of the sine triple angle rule.

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

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

Apply Pythagorean trigonometric identity, square of cosine can be transformed in terms of square of sine of angle.

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

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

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

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

It is proved that the sine of triple angle is expanded as the subtraction of the four times cube of sine of angle from the thrice sine of angle. The rule is actually applied in mathematics to expand the sine of triple angle in terms of sine of angle and also used in opposite operation.

For example, take $\theta = 30^\circ$ and evaluate the value of sin of triple angle.

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

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

$=\,\,\, 1$

Now, find the value of the expansion of the sine triple angle identity.

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

$=\,\,\, 3 \times \Bigg(\dfrac{1}{2}\Bigg)-4 \times {\Bigg(\dfrac{1}{2}\Bigg)}^3$

$=\,\,\, \dfrac{3}{2}-4 \times \Bigg(\dfrac{1}{8}\Bigg)$

$=\,\,\, \dfrac{3}{2}-\dfrac{4}{8}$

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

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

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

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

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

$=\,\,\, 1$

Compare both sides of the results and the values of both sides are equal. Hence, the sin triple angle formula is also called as triple angle trigonometric identity in trigonometry mathematics.

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