Difference between revisions of "1987 AHSME Problems/Problem 20"
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==Solution== | ==Solution== | ||
Because <math>\tan x \tan (90^\circ - x) = \tan x \cot x = 1</math>, <math>\tan 45^\circ = 1</math>, and <math>\log a + \log b = \log {ab}</math>, the answer is <math>\log_{10} {\tan 1^\circ \tan 2^\circ \dots \tan 89^\circ} = \log_{10} 1 = 0.</math> <math>\boxed{\textbf{(A)}}.</math> | Because <math>\tan x \tan (90^\circ - x) = \tan x \cot x = 1</math>, <math>\tan 45^\circ = 1</math>, and <math>\log a + \log b = \log {ab}</math>, the answer is <math>\log_{10} {\tan 1^\circ \tan 2^\circ \dots \tan 89^\circ} = \log_{10} 1 = 0.</math> <math>\boxed{\textbf{(A)}}.</math> | ||
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| + | == Solution 2 == | ||
| + | We have <math>\log{(\frac{\sin 1^\circ}{\cos 1^\circ} \times \frac{\sin 2^\circ}{\cos 2^\circ} \times ... \times \frac{\sin 89^\circ}{\cos 89^\circ})}</math>. However, <math>\sin{(x)} = \cos{(90^\circ - x)}</math>. Thus each pair of <math>\sin, \cos</math> (for example, <math>\sin{1^\circ}, \cos{89^\circ}</math>) multiplies to <math>1</math>. Hence we have <math>\log{1} = 0</math>. | ||
== See also == | == See also == | ||
Latest revision as of 11:42, 31 March 2018
Contents
Problem
Evaluate
Solution
Because 
, 
, and 
, the answer is 
Solution 2
We have 
. However, 
. Thus each pair of 
(for example, 
) multiplies to 
. Hence we have 
.
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. 
