Difference between revisions of "1969 IMO Problems/Problem 2"
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==Problem== | ==Problem== | ||
Let <math>a_1, a_2,\cdots, a_n</math> be real constants, <math>x</math> a real variable, and <cmath>f(x)=\cos(a_1+x)+\frac{1}{2}\cos(a_2+x)+\frac{1}{4}\cos(a_3+x)+\cdots+\frac{1}{2^{n-1}}\cos(a_n+x).</cmath> Given that <math>f(x_1)=f(x_2)=0,</math> prove that <math>x_2-x_1=m\pi</math> for some integer <math>m.</math> | Let <math>a_1, a_2,\cdots, a_n</math> be real constants, <math>x</math> a real variable, and <cmath>f(x)=\cos(a_1+x)+\frac{1}{2}\cos(a_2+x)+\frac{1}{4}\cos(a_3+x)+\cdots+\frac{1}{2^{n-1}}\cos(a_n+x).</cmath> Given that <math>f(x_1)=f(x_2)=0,</math> prove that <math>x_2-x_1=m\pi</math> for some integer <math>m.</math> | ||
+ | |||
+ | ==Solution== | ||
+ | Because the period of <math>cos(x)</math> is <math>2\pi</math>, the period of <math>f(x)</math> is also <math>2\pi</math>. | ||
+ | <cmath>f(x_1)=f(x_2)=f(x_1+x_2-x_1)</cmath> | ||
+ | We can get <math>x_2-x_1 = 2k\pi</math> for <math>k\in N^*</math>. Thus, <math>x_2-x_1=m\pi</math> for some integer <math>m.</math> | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{IMO box|year=1969|num-b=1|num-a=3}} |
Revision as of 08:30, 12 March 2019
Problem
Let be real constants, a real variable, and Given that prove that for some integer
Solution
Because the period of is , the period of is also . We can get for . Thus, for some integer
See Also
1969 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |