May the tangent be with you

Geometry Level 3

If $P$ is the fundamental period of the function $f(a) = \dfrac{ 2\tan(4a)}{1 + \tan^2(4a)}$ , find the value of $\dfrac {16}\pi P$ .

16 4 8 2

1 solution

Akhil Bansal
Nov 29, 2015

$f(a) = \dfrac{ 2\tan(4a)}{1 + \tan^2(4a)}$ $f(a) = \dfrac{ 2\left(\dfrac{\sin(4a)}{\cos(4a)}\right) }{\sec^2(4a)}$ $f(a) = 2\sin(4a)\cos(4a) = \sin(8a)$ Fundamental period of $\sin(8a)$ is $\dfrac{2\pi}{8} = P$ $\dfrac{16}{\pi}P = \dfrac{16}{\pi} \times \dfrac{\pi}{4} = \boxed{ 4}$

May the tangent be with you

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