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Difference between revisions of "1987 AHSME Problems/Problem 20"

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== Solution 2 ==
 
== Solution 2 ==
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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.
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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

Problem

Evaluate $\log_{10}(\tan 1^{\circ})+\log_{10}(\tan 2^{\circ})+\log_{10}(\tan 3^{\circ})+\cdots+\log_{10}(\tan 88^{\circ})+\log_{10}(\tan 89^{\circ}).$

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \frac{1}{2}\log_{10}(\frac{\sqrt{3}}{2}) \qquad \textbf{(C)}\ \frac{1}{2}\log_{10}2\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ \text{none of these}$

Solution

Because $\tan x \tan (90^\circ - x) = \tan x \cot x = 1$, $\tan 45^\circ = 1$, and $\log a + \log b = \log {ab}$, the answer is $\log_{10} {\tan 1^\circ \tan 2^\circ \dots \tan 89^\circ} = \log_{10} 1 = 0.$ $\boxed{\textbf{(A)}}.$

Solution 2

We have $\log{(\frac{\sin 1^\circ}{\cos 1^\circ} \times \frac{\sin 2^\circ}{\cos 2^\circ} \times ... \times \frac{\sin 89^\circ}{\cos 89^\circ})}$. However, $\sin{(x)} = \cos{(90^\circ - x)}$. Thus each pair of $\sin, \cos$ (for example, $\sin{1^\circ}, \cos{89^\circ}$) multiplies to $1$. Hence 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

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