MHB Prove Trigonometric Equality: $tan 1^o+tan 5^o+tan 9^o = tan 177^o-45$

[Albert1]

prove:

$tan 1^o+tan 5^o+tan 9^o +---------+tan 177^o=45$

 

[Opalg]

Re: trigonometric equality

Outline solution:
[sp]For $0\leqslant k\leqslant 44$, the angles $\theta = (4k+1)^\circ$ satisfy $\tan(45\theta) = 1.$

The formula for $\tan(n\theta)$ gives $\tan(45\theta) = \dfrac{{45\choose1}t - {45\choose3}t^3 + \ldots -{45\choose43}t^{43} + t^{45}}{1 - {45\choose2}t^2 - \ldots + {45\choose44}t^{44}} = \dfrac{45t -\ldots + t^{45}}{1-\ldots + 45t^{44}},$ where $t = \tan\theta.$ So the equation $\tan(45\theta) = 1$ (for $\theta$) corresponds to the equation $\dfrac{45t -\ldots + t^{45}}{1-\ldots + 45t^{44}} = 1$ (for $t$), or equivalently $t^{45} - 45t^{44} - \ldots -1=0.$ The sum of the roots of that equation is $45.$ Therefore $$\sum_{k=0}^{44}\tan(4k+1)^\circ = 45.$$[/sp]

 

[Albert1]

Re: trigonometric equality

Outline solution:
[sp]For $0\leqslant k\leqslant 44$, the angles $\theta = (4k+1)^\circ$ satisfy $\tan(45\theta) = 1.$

The formula for $\tan(n\theta)$ gives $\tan(45\theta) = \dfrac{{45\choose1}t - {45\choose3}t^3 + \ldots -{45\choose43}t^{43} + t^{45}}{1 - {45\choose2}t^2 - \ldots - {45\choose44}t^{44}} = \dfrac{45t -\ldots + t^{45}}{1-\ldots + 45t^{44}},$ where $t = \tan\theta.$ So the equation $\tan(45\theta) = 1$ (for $\theta$) corresponds to the equation $\dfrac{45t -\ldots + t^{45}}{1-\ldots + 45t^{44}} = 1$ (for $t$), or equivalently $t^{45} - 45t^{44} - \ldots -1=0.$ The sum of the roots of that equation is $45.$ Therefore $$\sum_{k=0}^{44}\tan(4k+1)^\circ = 45.$$[/sp]

perfect (Yes)