Answer :
Certainly! Let's solve the problem step by step.
Given:
[tex]\[ \cos \theta = -\frac{3}{5}, \theta \text{ is in quadrant III.} \][/tex]
### Finding [tex]\(\sin \theta\)[/tex]
In trigonometry, one of the fundamental identities is:
[tex]\[ \cos^2 \theta + \sin^2 \theta = 1 \][/tex]
We substitute [tex]\(\cos \theta = -\frac{3}{5}\)[/tex] into the identity:
[tex]\[ \left( -\frac{3}{5} \right)^2 + \sin^2 \theta = 1 \][/tex]
[tex]\[ \frac{9}{25} + \sin^2 \theta = 1 \][/tex]
[tex]\[ \sin^2 \theta = 1 - \frac{9}{25} \][/tex]
[tex]\[ \sin^2 \theta = \frac{25}{25} - \frac{9}{25} \][/tex]
[tex]\[ \sin^2 \theta = \frac{16}{25} \][/tex]
Taking the square root of both sides:
[tex]\[ \sin \theta = \pm \frac{4}{5} \][/tex]
Because [tex]\( \theta \)[/tex] is in quadrant III, [tex]\(\sin \theta\)[/tex] is negative. Hence:
[tex]\[ \sin \theta = -\frac{4}{5} \][/tex]
### Finding [tex]\(\sin 2\theta\)[/tex]
The double-angle identity for sine is:
[tex]\[ \sin 2\theta = 2 \sin \theta \cos \theta \][/tex]
Substitute the values we have found:
[tex]\[ \sin 2\theta = 2 \left( -\frac{4}{5} \right) \left( -\frac{3}{5} \right) \][/tex]
[tex]\[ \sin 2\theta = 2 \cdot \frac{12}{25} \][/tex]
[tex]\[ \sin 2\theta = \frac{24}{25} \][/tex]
### Finding [tex]\(\cos 2\theta\)[/tex]
The double-angle identity for cosine is:
[tex]\[ \cos 2\theta = \cos^2 \theta - \sin^2 \theta \][/tex]
Substitute the values we have found:
[tex]\[ \cos 2\theta = \left( -\frac{3}{5} \right)^2 - \left( -\frac{4}{5} \right)^2 \][/tex]
[tex]\[ \cos 2\theta = \frac{9}{25} - \frac{16}{25} \][/tex]
[tex]\[ \cos 2\theta = -\frac{7}{25} \][/tex]
### Finding [tex]\(\tan 2\theta\)[/tex]
The identity for tangent of a double-angle is:
[tex]\[ \tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} \][/tex]
Substitute the values we have found:
[tex]\[ \tan 2\theta = \frac{\frac{24}{25}}{-\frac{7}{25}} \][/tex]
[tex]\[ \tan 2\theta = \frac{24}{-7} \][/tex]
[tex]\[ \tan 2\theta = -\frac{24}{7} \][/tex]
### Determining the Quadrant of [tex]\(2 \theta\)[/tex]
Since [tex]\( \theta \)[/tex] is in quadrant III, and angles in quadrant III are between 180° - 270° or [tex]\( \pi \leq \theta < \frac{3\pi}{2} \)[/tex]:
[tex]\[ 360^\circ \leq 2\theta < 540^\circ \text{ or } 2\pi \leq 2\theta < 3\pi \][/tex]
This range includes angles that lie in both Quadrant II (270° - 360° or [tex]\( \frac{3\pi}{2} \)[/tex] to [tex]\( 2\pi \)[/tex]) and Quadrant III (360° - 450° or [tex]\( 2\pi \)[/tex] to [tex]\( \frac{5\pi}{2} \)[/tex]). Since [tex]\( \theta \)[/tex] was specifically in the latter part of quadrant III, it means [tex]\( 2\theta \)[/tex] is strictly in Quadrant III.
Therefore, the final results:
- [tex]\( \sin 2\theta = \frac{24}{25} \)[/tex]
- [tex]\( \cos 2\theta = -\frac{7}{25} \)[/tex]
- [tex]\( \tan 2\theta = -\frac{24}{7} \)[/tex]
- [tex]\( 2\theta \)[/tex] is in Quadrant III
Given:
[tex]\[ \cos \theta = -\frac{3}{5}, \theta \text{ is in quadrant III.} \][/tex]
### Finding [tex]\(\sin \theta\)[/tex]
In trigonometry, one of the fundamental identities is:
[tex]\[ \cos^2 \theta + \sin^2 \theta = 1 \][/tex]
We substitute [tex]\(\cos \theta = -\frac{3}{5}\)[/tex] into the identity:
[tex]\[ \left( -\frac{3}{5} \right)^2 + \sin^2 \theta = 1 \][/tex]
[tex]\[ \frac{9}{25} + \sin^2 \theta = 1 \][/tex]
[tex]\[ \sin^2 \theta = 1 - \frac{9}{25} \][/tex]
[tex]\[ \sin^2 \theta = \frac{25}{25} - \frac{9}{25} \][/tex]
[tex]\[ \sin^2 \theta = \frac{16}{25} \][/tex]
Taking the square root of both sides:
[tex]\[ \sin \theta = \pm \frac{4}{5} \][/tex]
Because [tex]\( \theta \)[/tex] is in quadrant III, [tex]\(\sin \theta\)[/tex] is negative. Hence:
[tex]\[ \sin \theta = -\frac{4}{5} \][/tex]
### Finding [tex]\(\sin 2\theta\)[/tex]
The double-angle identity for sine is:
[tex]\[ \sin 2\theta = 2 \sin \theta \cos \theta \][/tex]
Substitute the values we have found:
[tex]\[ \sin 2\theta = 2 \left( -\frac{4}{5} \right) \left( -\frac{3}{5} \right) \][/tex]
[tex]\[ \sin 2\theta = 2 \cdot \frac{12}{25} \][/tex]
[tex]\[ \sin 2\theta = \frac{24}{25} \][/tex]
### Finding [tex]\(\cos 2\theta\)[/tex]
The double-angle identity for cosine is:
[tex]\[ \cos 2\theta = \cos^2 \theta - \sin^2 \theta \][/tex]
Substitute the values we have found:
[tex]\[ \cos 2\theta = \left( -\frac{3}{5} \right)^2 - \left( -\frac{4}{5} \right)^2 \][/tex]
[tex]\[ \cos 2\theta = \frac{9}{25} - \frac{16}{25} \][/tex]
[tex]\[ \cos 2\theta = -\frac{7}{25} \][/tex]
### Finding [tex]\(\tan 2\theta\)[/tex]
The identity for tangent of a double-angle is:
[tex]\[ \tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} \][/tex]
Substitute the values we have found:
[tex]\[ \tan 2\theta = \frac{\frac{24}{25}}{-\frac{7}{25}} \][/tex]
[tex]\[ \tan 2\theta = \frac{24}{-7} \][/tex]
[tex]\[ \tan 2\theta = -\frac{24}{7} \][/tex]
### Determining the Quadrant of [tex]\(2 \theta\)[/tex]
Since [tex]\( \theta \)[/tex] is in quadrant III, and angles in quadrant III are between 180° - 270° or [tex]\( \pi \leq \theta < \frac{3\pi}{2} \)[/tex]:
[tex]\[ 360^\circ \leq 2\theta < 540^\circ \text{ or } 2\pi \leq 2\theta < 3\pi \][/tex]
This range includes angles that lie in both Quadrant II (270° - 360° or [tex]\( \frac{3\pi}{2} \)[/tex] to [tex]\( 2\pi \)[/tex]) and Quadrant III (360° - 450° or [tex]\( 2\pi \)[/tex] to [tex]\( \frac{5\pi}{2} \)[/tex]). Since [tex]\( \theta \)[/tex] was specifically in the latter part of quadrant III, it means [tex]\( 2\theta \)[/tex] is strictly in Quadrant III.
Therefore, the final results:
- [tex]\( \sin 2\theta = \frac{24}{25} \)[/tex]
- [tex]\( \cos 2\theta = -\frac{7}{25} \)[/tex]
- [tex]\( \tan 2\theta = -\frac{24}{7} \)[/tex]
- [tex]\( 2\theta \)[/tex] is in Quadrant III
