Question #eb243

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Question #eb243

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Answer 1 · 2017-03-06T01:53:46

$sin x = \pm \frac{\sqrt{5}}{5}$ $sin x = +- sqrt5/5$

Explanation:

Use trig identity:
${sin}^{2} x = \frac{1}{1 + {cot}^{2} x}$ $sin^2 x = 1/(1 + cot^2 x)$
In this case:
Call x the arctan (1/2)
$tan x = \frac{1}{2}$ $tan x = 1/2$ --> $cot x = \frac{1}{tan} = 2$ $cot x = 1/(tan) = 2$
${sin}^{2} x = \frac{1}{1 + 4} = \frac{1}{5}$ $sin^2 x = 1/(1 + 4) = 1/5$
$sin x = \pm \frac{1}{\sqrt{5}} = \pm \frac{\sqrt{5}}{5}$ $sin x = +- 1/(sqrt5) = +- sqrt5/5$
Because $tan x = \frac{1}{2}$ $tan x = 1/2$ , x could be either in Quadrant 1 or in Quadrant 3.
There fore, sin x could be either positive or negative.

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Question #eb243

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