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Trigonometric equality

Albert

Well-known member
Jan 25, 2013
1,225
prove:

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

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
Re: trigonometric equality

Outline solution:
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 \(\displaystyle \sum_{k=0}^{44}\tan(4k+1)^\circ = 45.\)
 
Last edited:

Albert

Well-known member
Jan 25, 2013
1,225
Re: trigonometric equality

Outline solution:
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 \(\displaystyle \sum_{k=0}^{44}\tan(4k+1)^\circ = 45.\)
perfect (Yes)