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, $2\left(\frac{\cos (45 - A)}{\cos A}\right)\left(\frac{\cos (45 - B)}{\cos B}\right) = 2\frac{\cos 25 \cos 20}{\cos 20 \cos 25} = \box{\text{(B) } 2}$ (Error compiling LaTeX. Unknown error_msg)
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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