To find [tex]\(\sin(\theta)\)[/tex] and [tex]\(\cos(\theta)\)[/tex] given [tex]\(\tan(\theta) = 0.6\)[/tex] and knowing that [tex]\(\theta\)[/tex] lies in the third quadrant, we use trigonometric identities and properties specific to this quadrant. Here’s the detailed step-by-step process:
1. Understanding the Third Quadrant:
- In the third quadrant, both sine and cosine are negative.
2. Given Information:
- [tex]\(\tan(\theta) = 0.6\)[/tex]
- [tex]\(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\)[/tex]
3. Express [tex]\(\sin(\theta)\)[/tex] and [tex]\(\cos(\theta)\)[/tex] in terms of [tex]\(\tan(\theta)\)[/tex]:
- Let [tex]\(\sin(\theta) = -a\)[/tex] and [tex]\(\cos(\theta) = -b\)[/tex] (since both are negative in the third quadrant).
- From the identity, [tex]\(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\)[/tex], we have:
[tex]\[
\tan(\theta) = \frac{-a}{-b} = \frac{a}{b}
\][/tex]
- Given [tex]\(\tan(\theta) = 0.6\)[/tex], we get:
[tex]\[
0.6 = \frac{a}{b} \implies a = 0.6b
\][/tex]
4. Use the Pythagorean Identity:
- From [tex]\(\sin^2(\theta) + \cos^2(\theta) = 1\)[/tex]:
[tex]\[
a^2 + b^2 = 1
\][/tex]
- Substitute [tex]\(a = 0.6b\)[/tex]:
[tex]\[
(0.6b)^2 + b^2 = 1
\][/tex]
[tex]\[
0.36b^2 + b^2 = 1
\][/tex]
[tex]\[
1.36b^2 = 1
\][/tex]
[tex]\[
b^2 = \frac{1}{1.36}
\][/tex]
[tex]\[
b = -\sqrt{\frac{1}{1.36}}
\][/tex]
- Since [tex]\(b\)[/tex] is negative in the third quadrant, we calculate [tex]\(b\)[/tex] as:
[tex]\[
b = -\sqrt{\frac{1}{1.36}} = -\frac{1}{\sqrt{1.36}} \approx -0.8574929257125442
\][/tex]
5. Calculate [tex]\(a\)[/tex]:
- Recall [tex]\(a = 0.6b\)[/tex]:
[tex]\[
a = 0.6 \times (-0.8574929257125442) \approx -0.5144957554275265
\][/tex]
6. Final Values:
- [tex]\(\sin(\theta) = -a \approx -0.5144957554275265\)[/tex]
- [tex]\(\cos(\theta) = -b \approx -0.8574929257125442\)[/tex]
So, the exact values are:
[tex]\[
\sin(\theta) = -0.5144957554275265
\][/tex]
[tex]\[
\cos(\theta) = -0.8574929257125442
\][/tex]
