$\large \sin{2\theta} \,=\, \dfrac{2\tan{\theta}}{1+\tan^2{\theta}}$

The formula for sine of double angle is occasionally written in any one of the following three forms.

Angle | Double angle | Formula |
---|---|---|

$x$ | $2x$ | $\sin{2x}$ $\,=\,$ $\dfrac{2\tan{x}}{1+\tan^2{x}}$ |

$A$ | $2A$ | $\sin{2A}$ $\,=\,$ $\dfrac{2\tan{A}}{1+\tan^2{A}}$ |

$\alpha$ | $2\alpha$ | $\sin{2\alpha}$ $\,=\,$ $\dfrac{2\tan{\alpha}}{1+\tan^2{\alpha}}$ |

The angle of a right angled triangle can be represented by any symbol and twice that symbol is known as double angle. However, the expansion of sin double angle rule in terms of tan of angle is in the above form.

If theta is an angle of a right angled triangle, then the double angle is $2\theta$. The expansion of sine of double angle is written in terms of sine and cosine of angle.

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

01

Sine of angle is a multiplying factor in the expansion of the sine of double angle and transform it in terms of the tangent of angle.

$\implies \sin{2\theta}$ $\,=\,$ $2\sin{\theta}\cos{\theta} \times 1$

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

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

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

$\implies \sin{2\theta}$ $\,=\,$ $2 \times \dfrac{\sin{\theta}}{\cos{\theta}} \times \cos^2{\theta}$

The ratio of sine of angle to cosine of angle is tangent of angle as per quotient rule of sine and cosine functions.

$\implies \sin{2\theta}$ $\,=\,$ $2 \times \tan{\theta} \times \cos^2{\theta}$

$\implies \sin{2\theta}$ $\,=\,$ $2\tan{\theta} \times \cos^2{\theta}$

02

As per reciprocal relation of cosine and secant, express square of cosine of angle theta as reciprocal of square of the secant of the angle.

$\implies \sin{2\theta}$ $\,=\,$ $2\tan{\theta} \times \dfrac{1}{\sec^2{\theta}}$

$\implies \sin{2\theta}$ $\,=\,$ $\dfrac{2\tan{\theta}}{\sec^2{\theta}}$

Write the square of secant of angle in terms of tan of angle by applying the Pythagorean identity of tangent and secant of angle.

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

Take $\theta = 30^\circ$ and observe results of both sides of this trigonometry rule.

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

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

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

Now, find the value of the expansion for the angle $30$ degrees.

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

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

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

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

$=\,\,\, \dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{4}{3}}$

$=\,\,\, \dfrac{2}{\sqrt{3}} \times \dfrac{3}{4}$

$=\,\,\, \dfrac{2 \times 3}{\sqrt{3} \times 4}$

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

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

Compare the results of the sine double angle with result of its expansion in terms of tan. They both are equal. Hence, it is called as a double angle trigonometric identity in trigonometry.

$\sin{(2 \times 30^\circ)}$ $\,=\,$ $\dfrac{2\tan{(30^\circ)}}{1+\tan^2{(30^\circ)}}$ $\,=\, \dfrac{\sqrt{3}}{2}$

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