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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Thread 'Area of Overlapping Squares'

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?
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