To determine [tex]\(\sin 2x\)[/tex], [tex]\(\cos 2x\)[/tex], and [tex]\(\tan 2x\)[/tex] given that [tex]\(\cos x = -\frac{1}{\sqrt{10}}\)[/tex] and [tex]\(x\)[/tex] is in quadrant III, let's go through the steps systematically.
### Step 1: Find the value of [tex]\(\sin x\)[/tex]
Given:
[tex]\[
\cos x = -\frac{1}{\sqrt{10}}
\][/tex]
In quadrant III, both sine and cosine are negative. We use the Pythagorean identity for sine and cosine:
[tex]\[
\sin^2 x + \cos^2 x = 1
\][/tex]
Substitute [tex]\(\cos x\)[/tex]:
[tex]\[
\sin^2 x + \left( -\frac{1}{\sqrt{10}} \right)^2 = 1
\][/tex]
[tex]\[
\sin^2 x + \frac{1}{10} = 1
\][/tex]
[tex]\[
\sin^2 x = 1 - \frac{1}{10}
\][/tex]
[tex]\[
\sin^2 x = \frac{10}{10} - \frac{1}{10}
\][/tex]
[tex]\[
\sin^2 x = \frac{9}{10}
\][/tex]
[tex]\[
\sin x = -\sqrt{\frac{9}{10}} \quad (\text{since } x \text{ is in quadrant III})
\][/tex]
[tex]\[
\sin x = -\frac{3}{\sqrt{10}}
\][/tex]
[tex]\[
\sin x = -0.9486832980505138
\][/tex]
### Step 2: Calculate [tex]\(\sin 2x\)[/tex]
Using the double angle formula for sine:
[tex]\[
\sin 2x = 2 \sin x \cos x
\][/tex]
Substitute the known values:
[tex]\[
\sin 2x = 2 \left( -\frac{3}{\sqrt{10}} \right) \left( -\frac{1}{\sqrt{10}} \right)
\][/tex]
[tex]\[
\sin 2x = 2 \left( \frac{3 \cdot 1}{10} \right)
\][/tex]
[tex]\[
\sin 2x = 2 \cdot \frac{3}{10}
\][/tex]
[tex]\[
\sin 2x = \frac{6}{10}
\][/tex]
[tex]\[
\sin 2x = 0.6
\][/tex]
### Step 3: Calculate [tex]\(\cos 2x\)[/tex]
Using the double angle formula for cosine:
[tex]\[
\cos 2x = 2 \cos^2 x - 1
\][/tex]
Substitute [tex]\(\cos x = -\frac{1}{\sqrt{10}}\)[/tex]:
[tex]\[
\cos 2x = 2 \left( -\frac{1}{\sqrt{10}} \right)^2 - 1
\][/tex]
[tex]\[
\cos 2x = 2 \left( \frac{1}{10} \right) - 1
\][/tex]
[tex]\[
\cos 2x = \frac{2}{10} - 1
\][/tex]
[tex]\[
\cos 2x = 0.2 - 1
\][/tex]
[tex]\[
\cos 2x = -0.8
\][/tex]
### Step 4: Calculate [tex]\(\tan 2x\)[/tex]
Using the double angle formula for tangent:
[tex]\[
\tan 2x = \frac{\sin 2x}{\cos 2x}
\][/tex]
Substitute:
[tex]\[
\tan 2x = \frac{0.6}{-0.8}
\][/tex]
[tex]\[
\tan 2x = -\frac{6}{8}
\][/tex]
[tex]\[
\tan 2x = -\frac{3}{4}
\][/tex]
[tex]\[
\tan 2x = -0.75
\][/tex]
### Final Results
[tex]\[
\sin 2x = 0.6
\][/tex]
[tex]\[
\cos 2x = -0.8
\][/tex]
[tex]\[
\tan 2x = -0.75
\][/tex]
