Difference between revisions of "1963 IMO Problems/Problem 5"
(New page: ==Problem== Prove that <math>\cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=\frac{1}{2}</math>. ==Solution== {{solution}} ==See Also== {{IMO box|year=1963|num-b=4|num-a=6...) |
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==Solution== | ==Solution== | ||
{{solution}} | {{solution}} | ||
| + | We have S | ||
| + | = cos(π/7) - cos(2π/7) + cos(3π/7) | ||
| + | = cos(π/7) + cos(3π/7) + cos(5π/7) | ||
| + | |||
| + | Then, product-sum formulae, we have | ||
| + | S * 2* sin(π/7) = sin(2π/7) + sin(4π/7) - sin(2π/7) + sin(6π/7) - sin(4π/7) | ||
| + | = sin(6π/7) = sin(π/7) | ||
| + | |||
| + | Thus S = 1/2 | ||
| + | |||
==See Also== | ==See Also== | ||
{{IMO box|year=1963|num-b=4|num-a=6}} | {{IMO box|year=1963|num-b=4|num-a=6}} | ||
Revision as of 13:46, 24 November 2009
Problem
Prove that 
.
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it. We have S = cos(π/7) - cos(2π/7) + cos(3π/7) = cos(π/7) + cos(3π/7) + cos(5π/7)
Then, product-sum formulae, we have S * 2* sin(π/7) = sin(2π/7) + sin(4π/7) - sin(2π/7) + sin(6π/7) - sin(4π/7) = sin(6π/7) = sin(π/7)
Thus S = 1/2
See Also
| 1963 IMO (Problems) • Resources | ||
| Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
| All IMO Problems and Solutions | ||
