Sure, let's find [tex]\(\sin 2x\)[/tex], [tex]\(\cos 2x\)[/tex], and [tex]\(\tan 2x\)[/tex] given [tex]\(\sin x = -\frac{1}{\sqrt{10}}\)[/tex] and [tex]\(x\)[/tex] lies in the third quadrant.
### Step-by-Step Solution
#### 1. Determine [tex]\(\cos x\)[/tex]:
To find [tex]\(\cos x\)[/tex], we can use the Pythagorean identity:
[tex]\[ \sin^2 x + \cos^2 x = 1 \][/tex]
Given [tex]\(\sin x = -\frac{1}{\sqrt{10}}\)[/tex]:
[tex]\[ \sin^2 x = \left(-\frac{1}{\sqrt{10}}\right)^2 = \frac{1}{10} \][/tex]
Plug this into the identity:
[tex]\[ \frac{1}{10} + \cos^2 x = 1 \][/tex]
[tex]\[ \cos^2 x = 1 - \frac{1}{10} = \frac{9}{10} \][/tex]
Since [tex]\(x\)[/tex] is in the third quadrant, [tex]\(\cos x\)[/tex] is negative:
[tex]\[ \cos x = -\sqrt{\frac{9}{10}} = -\frac{3}{\sqrt{10}} \][/tex]
#### 2. Calculate [tex]\(\sin 2x\)[/tex]:
We use the double-angle formula for sine:
[tex]\[ \sin 2x = 2 \sin x \cos x \][/tex]
Substitute [tex]\(\sin x\)[/tex] and [tex]\(\cos x\)[/tex]:
[tex]\[ \sin 2x = 2 \left(-\frac{1}{\sqrt{10}}\right) \left(-\frac{3}{\sqrt{10}}\right) \][/tex]
[tex]\[ \sin 2x = 2 \times \frac{1}{\sqrt{10}} \times \frac{3}{\sqrt{10}} \][/tex]
[tex]\[ \sin 2x = 2 \times \frac{3}{10} = \frac{6}{10} = 0.6 \][/tex]
#### 3. Calculate [tex]\(\cos 2x\)[/tex]:
We use the double-angle formula for cosine:
[tex]\[ \cos 2x = \cos^2 x - \sin^2 x \][/tex]
Substitute [tex]\(\cos x\)[/tex] and [tex]\(\sin x\)[/tex]:
[tex]\[ \cos 2x = \left(-\frac{3}{\sqrt{10}}\right)^2 - \left(-\frac{1}{\sqrt{10}}\right)^2 \][/tex]
[tex]\[ \cos 2x = \frac{9}{10} - \frac{1}{10} \][/tex]
[tex]\[ \cos 2x = \frac{8}{10} = 0.8 \][/tex]
#### 4. Calculate [tex]\(\tan 2x\)[/tex]:
We use the identity:
[tex]\[ \tan 2x = \frac{\sin 2x}{\cos 2x} \][/tex]
Substitute [tex]\(\sin 2x\)[/tex] and [tex]\(\cos 2x\)[/tex]:
[tex]\[ \tan 2x = \frac{0.6}{0.8} = 0.75 \][/tex]
#### Final Results
[tex]\[ \sin 2x = 0.6 \][/tex]
[tex]\[ \cos 2x = 0.8 \][/tex]
[tex]\[ \tan 2x = 0.75 \][/tex]
Therefore,
[tex]\[
\begin{array}{ll}
\sin 2 x = 0.6 \\
\cos 2 x = 0.8 \\
\tan 2 x = 0.75
\end{array}
\][/tex]
