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

  • Context: MHB 
  • Thread starter Thread starter Albert1
  • Start date Start date
  • Tags Tags
    Trigonometric
Click For Summary
SUMMARY

The trigonometric equality \( \tan 1^\circ + \tan 5^\circ + \tan 9^\circ = \tan 177^\circ - 45 \) is proven by analyzing the angles \( \theta = (4k+1)^\circ \) for \( 0 \leq k \leq 44 \), which satisfy \( \tan(45\theta) = 1 \). Using the formula for \( \tan(n\theta) \), it is established that the roots of the equation \( t^{45} - 45t^{44} - \ldots - 1 = 0 \) sum to 45. Consequently, the sum \( \sum_{k=0}^{44} \tan(4k+1)^\circ = 45 \) confirms the equality.

PREREQUISITES
  • Understanding of trigonometric functions, specifically tangent.
  • Familiarity with polynomial equations and their roots.
  • Knowledge of combinatorial coefficients, denoted as \( {n \choose k} \).
  • Basic grasp of angle transformations in trigonometry.
NEXT STEPS
  • Study the properties of tangent functions and their periodicity.
  • Explore the derivation and applications of the formula for \( \tan(n\theta) \).
  • Learn about polynomial root-finding techniques and their implications in trigonometry.
  • Investigate the significance of combinatorial coefficients in trigonometric identities.
USEFUL FOR

Mathematicians, students studying trigonometry, and educators looking to deepen their understanding of trigonometric identities and polynomial equations.

Albert1
Messages
1,221
Reaction score
0
prove:

$tan 1^o+tan 5^o+tan 9^o +---------+tan 177^o=45$
 
Mathematics news on Phys.org
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]
 
Last edited:
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]
perfect (Yes)
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K