1975 AHSME Problems/Problem 30
Contents
[hide]Problem 30
Let . Then equals
Solution
Using the difference to product identity, we find that is equivalent to Since sine is an odd function, we find that , and thus . Using the property , we find We multiply the entire expression by and use the double angle identity of sine twice to find Using the property , we find Substituting this back into the equation, we have Dividing both sides by , we have
Solution 2
We observe that is twice of , so evaluating and using the double angle identify is enough to find
We first let . Notice that and . Rewriting as , we then use the difference of angles identify to find Rearranged to get Using the half-angle identify twice, and noticing that is positive, we get Also,
Thus we can get Squaring both sides and simplifying, this results in From which we get Squaring both sides again, Multiplying both sides by and rearranging, Factoring out from the first two terms and applying difference of squares, We know that which is not equal to , so we divide both sides by Using the rational root test, we find that is a solution of this equation. However, we also know that is not equal to , so we divide both sides by to get Using the quadratic formula, we find that We know must be positive, which gives us
Thus, using the double angle identify, we can find
-SaraLiu24
See Also
1975 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 28 |
Followed by Last Problem | |
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.