Difference between revisions of "1963 IMO Problems/Problem 5"
Mathboy100 (talk | contribs) (→Solution) |
Mathboy100 (talk | contribs) (→Solution) |
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<cmath>\cos{\frac{2\pi}{7}} + \cos{\frac{4\pi}{7}} + \cos{\frac{6\pi}{7}} = -\frac{1}{2}.</cmath> | <cmath>\cos{\frac{2\pi}{7}} + \cos{\frac{4\pi}{7}} + \cos{\frac{6\pi}{7}} = -\frac{1}{2}.</cmath> | ||
− | Finally, since <math>\cos{x} = -cos{\pi-x}</math>, | + | Finally, since <math>\cos{x} = -\cos{\pi-x}</math>, |
<cmath>\cos{\frac{2\pi}{7}} - \cos{\frac{3\pi}{7}} - \cos{\frac{\pi}{7}} = -\frac{1}{2}</cmath> | <cmath>\cos{\frac{2\pi}{7}} - \cos{\frac{3\pi}{7}} - \cos{\frac{\pi}{7}} = -\frac{1}{2}</cmath> |
Revision as of 16:26, 8 December 2022
Problem
Prove that .
Solution
Because the sum of the -coordinates of the seventh roots of unity is
, we have
Now, we can apply to obtain
Finally, since ,
~mathboy100
Solution 2
Let . We have
Then, by product-sum formulae, we have
Thus .
Solution 3
Let and
. From the addition formulae, we have
From the Trigonometric Identity, , so
We must prove that . It suffices to show that
.
Now note that . We can find these in terms of
and
:
Therefore . Note that this can be factored:
Clearly , so
. This proves the result.
Solution 4
Let . Thus it suffices to show that
. Now using the fact that
and
, this is equivalent to
But since
is a
th root of unity,
. The answer is then
, as desired.
~yofro
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 |