Difference between revisions of "1985 AHSME Problems/Problem 16"
Freddylukai (talk | contribs) (→Solution) |
|||
Line 66: | Line 66: | ||
==See Also== | ==See Also== | ||
{{AHSME box|year=1985|num-b=15|num-a=17}} | {{AHSME box|year=1985|num-b=15|num-a=17}} | ||
+ | {{MAA Notice}} |
Revision as of 12:01, 5 July 2013
Contents
[hide]Problem
If and , then the value of is
Solution
Solution 1
First, let's leave everything in variables and see if we can simplify .
We can write everything in terms of sine and cosine to get .
We can multiply out the numerator to get .
It may seem at first that we've made everything more complicated, however, we can recognize the numerator from the angle sum formulas:
Therefore, our fraction is equal to .
We can also use the product-to-sum formula
to simplify the denominator:
.
But now we seem stuck. However, we can note that since , we have , so we get
Note that we only used the fact that , so we have in fact not just shown that for and , but for all such that , for integer .
Solution 2
We can see that . We also know that . First, let us expand .
We get .
Now, let us look at .
By the sum formula, we know that
Then, since , we can see that
Then
Thus, the sum become and the answer is
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.