1985 AHSME Problems/Problem 16
Contents
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
Solution 3
Let's write out the expression in terms of sine and cosine, so that we may see that it is equal to Clearly, that is equal to Now, we note that is equal to . Now, we would like to get in the denominator. What springs to mind is the fact that Therefore, we can express the desired value as Because , we see that the fractional part is , and so the sum is , which brings us to the answer .
Solution 4
Similar to above:
Notice that
Thus,
By half angle identity,
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.