Difference between revisions of "1987 AHSME Problems/Problem 20"
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== Solution 2 == | == Solution 2 == | ||
| − | We have log(sin(1)/cos(1)*sin(2)/cos(2)*...*sin(89)/cos(89)). However, sin(x) = cos(90 - x). Thus each pair of sin, cos (for example, sin(1), cos(89)) multiplies to 1. thus, we have log(1) => 0 | + | We have log(sin(1)/cos(1)*sin(2)/cos(2)*...*sin(89)/cos(89)). However, sin(x) = cos(90 - x). Thus each pair of sin, cos (for example, sin(1), cos(89)) multiplies to 1. thus, we have log(1) => 0. |
== See also == | == See also == | ||
Revision as of 15:17, 2 February 2018
Contents
Problem
Evaluate
Solution
Because 
, 
, and 
, the answer is 
Solution 2
We have log(sin(1)/cos(1)*sin(2)/cos(2)*...*sin(89)/cos(89)). However, sin(x) = cos(90 - x). Thus each pair of sin, cos (for example, sin(1), cos(89)) multiplies to 1. thus, we have log(1) => 0.
See also
| 1987 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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. 
