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

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The discussion centers on proving the trigonometric equality \( \tan 1^\circ + \tan 5^\circ + \tan 9^\circ = \tan 177^\circ - 45 \). It is established that for angles of the form \( \theta = (4k+1)^\circ \) where \( 0 \leq k \leq 44 \), the equation \( \tan(45\theta) = 1 \) holds true. The formula for \( \tan(n\theta) \) is utilized to derive a polynomial equation whose roots sum to 45. Consequently, it is concluded that the sum of the tangents of the specified angles equals 45. This confirms the trigonometric equality as stated.
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prove:

$tan 1^o+tan 5^o+tan 9^o +---------+tan 177^o=45$
 
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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]
 
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Re: trigonometric equality

Opalg said:
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]
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