How do you find sin 2x, given tan x = -2 and cos x > 0?

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How do you find sin 2x, given tan x = -2 and cos x > 0?

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Answer 1 · 2015-11-30T14:05:45

Find sin 2x, knowing tan x = -2, and cos x > 0

Ans: $sin 2 x = \frac{4}{5}$ $sin 2x = 4/5$

Explanation:

3 Trig identities to be used:
$1 + {tan}^{2} x = \frac{1}{{cos}^{2} x}$ $1 + tan^2 x = 1/cos^2 x$ (1)
${sin}^{2} x + {cos}^{2} x = 1$ $sin^2 x + cos ^2 x = 1$ (2)
$sin 2 x = 2 sin x . cos x$ $sin 2x = 2sin x.cos x$ (3)
Given tan x = -2. First find cos x and sin x
(1) --> $1 + 4 = \frac{1}{{cos}^{2} x}$ $1 + 4 = 1/(cos^2 x)$ --> ${cos}^{2} x = \frac{1}{5}$ $cos^2 x = 1/5$ --> $cos x = \pm \frac{1}{\sqrt{5}}$ $cos x = +- 1/sqrt5$ .
Since cos x > 0, then $cos x = \frac{1}{\sqrt{5}} .$ $cos x = 1/sqrt5.$
(2) --> ${sin}^{2} x = 1 - {cos}^{2} x = 1 - \frac{1}{5} = \frac{4}{5}$ $sin^2 x = 1 - cos^2 x = 1 - 1/5 = 4/5$ --> $sin x = \pm \frac{2}{\sqrt{5}} .$ $sin x = +- 2/sqrt5.$
Since cos x > 0 then $sin x = \frac{2}{\sqrt{5}}$ $sin x = 2/sqrt5$ .
(3) --> $sin 2 x = 2 sin x . cos x = 2 (\frac{1}{\sqrt{5}}) (\frac{2}{\sqrt{5}}) = \frac{4}{5}$ $sin 2x = 2sin x.cos x = 2(1/sqrt5)(2/sqrt5) = 4/5$

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How do you find sin 2x, given tan x = -2 and cos x > 0?

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