Question

The arithmetic mean of the numbers $2\sin 2^{\circ },4\sin 4^{\circ },6\sin 6^{\circ },\dots ,180\sin {180}^{\circ }$ is $\tan k^{\circ }$. Then the least positive integral value of $k$ is

Answer 1

Let $\frac{2\sin 2^{\circ }+4\sin 4^{\circ }+6\sin 6^{\circ }+\cdots +180\sin {180}^{\circ }}{90}=x$

Multiplying both sides by $\sin 1^{\circ },$ we get
$2\sin 2^{\circ }\sin 1^{\circ }+2\left(2\sin 4^{\circ }\sin 1^{\circ }\right)+\cdots +89\left(2\sin {178}^{\circ }\sin 1^{\circ }\right)=90x\sin 1^{\circ }$

Using the identity, $2\sin A\cos B=\cos (A-B)-\cos (A+B)$ we get
$(\cos 1^{\circ }-\cos 3^{\circ })+2(\cos 3^{\circ }-\cos 5^{\circ })+\cdots +89(\cos {177}^{\circ }-\cos {179}^{\circ })=90x\sin 1^{\circ }$
$\Rightarrow \cos 1^{\circ }+\cos 3^{\circ }+\cos 5^{\circ }+\cdots +\cos {175}^{\circ }+\cos {177}^{\circ }-89\cos {179}^{\circ }=90x\sin 1^{\circ }$
$\Rightarrow \cos 1^{\circ }+(\cos 3^{\circ }+\cos {177}^{\circ })+\cdots +(\cos {89}^{\circ }+\cos {91}^{\circ })-89\cos {179}^{\circ }=90x\sin 1^{\circ }$
$\Rightarrow \cos 1^{\circ }+0+\cdots +0+89\cos 1^{\circ }=90x\sin 1^{\circ }$
$\Rightarrow 90\cos 1^{\circ }=90x\sin 1^{\circ }$
$\Rightarrow x=\cot 1^{\circ }$
$\Rightarrow x=\tan {89}^{\circ }$
$\Rightarrow k=89$