Answer :
To find the exact values of \(\cos \alpha\) and \(\cos(2\alpha)\) given that \(\sin \alpha = \frac{9}{13}\) and \(\alpha\) is in Quadrant II, we need to follow these steps:
1. Determine \(\cos \alpha\) using the Pythagorean identity.
We start with the Pythagorean identity:
[tex]\[ \sin^2 \alpha + \cos^2 \alpha = 1 \][/tex]
Plugging in the given value \(\sin \alpha = \frac{9}{13}\):
[tex]\[ \left( \frac{9}{13} \right)^2 + \cos^2 \alpha = 1 \][/tex]
[tex]\[ \frac{81}{169} + \cos^2 \alpha = 1 \][/tex]
Solving for \(\cos^2 \alpha\):
[tex]\[ \cos^2 \alpha = 1 - \frac{81}{169} \][/tex]
[tex]\[ \cos^2 \alpha = \frac{169}{169} - \frac{81}{169} \][/tex]
[tex]\[ \cos^2 \alpha = \frac{88}{169} \][/tex]
Taking the square root of both sides to find \(\cos \alpha\):
[tex]\[ \cos \alpha = \pm \sqrt{\frac{88}{169}} \][/tex]
[tex]\[ \cos \alpha = \pm \frac{\sqrt{88}}{13} \][/tex]
Simplifying \(\sqrt{88}\):
[tex]\[ \sqrt{88} = \sqrt{4 \times 22} = 2 \sqrt{22} \][/tex]
[tex]\[ \cos \alpha = \pm \frac{2 \sqrt{22}}{13} \][/tex]
Since \(\alpha\) is in Quadrant II, \(\cos \alpha\) is negative:
[tex]\[ \cos \alpha = -\frac{2 \sqrt{22}}{13} \][/tex]
2. Calculate \(\cos(2\alpha)\) using the double-angle formula for cosine.
The double-angle formula for cosine is:
[tex]\[ \cos(2\alpha) = 2 \cos^2 \alpha - 1 \][/tex]
Substituting \(\cos \alpha = -\frac{2 \sqrt{22}}{13}\):
[tex]\[ \cos^2 \alpha = \left( -\frac{2 \sqrt{22}}{13} \right)^2 = \frac{88}{169} \][/tex]
Now, apply this to the double-angle formula:
[tex]\[ \cos(2\alpha) = 2 \left( \frac{88}{169} \right) - 1 \][/tex]
[tex]\[ \cos(2\alpha) = \frac{176}{169} - 1 \][/tex]
[tex]\[ \cos(2\alpha) = \frac{176}{169} - \frac{169}{169} \][/tex]
[tex]\[ \cos(2\alpha) = \frac{7}{169} \][/tex]
This fraction can be converted into a decimal form if needed:
[tex]\[ \cos(2\alpha) \approx 0.0414 \][/tex]
Thus, the value of [tex]\(\cos(2 \alpha)\)[/tex] is approximately [tex]\(0.0414\)[/tex].
1. Determine \(\cos \alpha\) using the Pythagorean identity.
We start with the Pythagorean identity:
[tex]\[ \sin^2 \alpha + \cos^2 \alpha = 1 \][/tex]
Plugging in the given value \(\sin \alpha = \frac{9}{13}\):
[tex]\[ \left( \frac{9}{13} \right)^2 + \cos^2 \alpha = 1 \][/tex]
[tex]\[ \frac{81}{169} + \cos^2 \alpha = 1 \][/tex]
Solving for \(\cos^2 \alpha\):
[tex]\[ \cos^2 \alpha = 1 - \frac{81}{169} \][/tex]
[tex]\[ \cos^2 \alpha = \frac{169}{169} - \frac{81}{169} \][/tex]
[tex]\[ \cos^2 \alpha = \frac{88}{169} \][/tex]
Taking the square root of both sides to find \(\cos \alpha\):
[tex]\[ \cos \alpha = \pm \sqrt{\frac{88}{169}} \][/tex]
[tex]\[ \cos \alpha = \pm \frac{\sqrt{88}}{13} \][/tex]
Simplifying \(\sqrt{88}\):
[tex]\[ \sqrt{88} = \sqrt{4 \times 22} = 2 \sqrt{22} \][/tex]
[tex]\[ \cos \alpha = \pm \frac{2 \sqrt{22}}{13} \][/tex]
Since \(\alpha\) is in Quadrant II, \(\cos \alpha\) is negative:
[tex]\[ \cos \alpha = -\frac{2 \sqrt{22}}{13} \][/tex]
2. Calculate \(\cos(2\alpha)\) using the double-angle formula for cosine.
The double-angle formula for cosine is:
[tex]\[ \cos(2\alpha) = 2 \cos^2 \alpha - 1 \][/tex]
Substituting \(\cos \alpha = -\frac{2 \sqrt{22}}{13}\):
[tex]\[ \cos^2 \alpha = \left( -\frac{2 \sqrt{22}}{13} \right)^2 = \frac{88}{169} \][/tex]
Now, apply this to the double-angle formula:
[tex]\[ \cos(2\alpha) = 2 \left( \frac{88}{169} \right) - 1 \][/tex]
[tex]\[ \cos(2\alpha) = \frac{176}{169} - 1 \][/tex]
[tex]\[ \cos(2\alpha) = \frac{176}{169} - \frac{169}{169} \][/tex]
[tex]\[ \cos(2\alpha) = \frac{7}{169} \][/tex]
This fraction can be converted into a decimal form if needed:
[tex]\[ \cos(2\alpha) \approx 0.0414 \][/tex]
Thus, the value of [tex]\(\cos(2 \alpha)\)[/tex] is approximately [tex]\(0.0414\)[/tex].
