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

(Solution)
(Solution 2)
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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 <math> </math>\boxed{\textbf{(A)}}.$
+
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 16:17, 2 February 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(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 (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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