1985 AHSME Problems/Problem 16
Problem
If and , then the value of is
Solution 1
Noting that , we apply the angle sum formula giving so Hence
Solution 2
Expanding in terms of sines and cosines, we obtain
Recalling the angle sum identities this reduces to
Now, using the product-to-sum formula we can simplify the denominator, yielding
Finally, since , we have , so
Remark: Notice that we only used the fact that , so we have in fact shown that not just for and , but also for all such that for integers .
Solution 3
As in Solution 2, we rewrite the expression as and hence as Using the angle sum identities we obtain Therefore the expression becomes
Solution 4
As in Solutions 2 and 3, the expression becomes
Now, using the identity and the double-angle identity , we observe that
Since and are acute, we have , so and . Hence, taking the positive square root of both sides in the above identity, the expression becomes Recalling the further identity , together with the half-angle identity we finally obtain
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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