Qle, Power-Reducing, and Half- Use the given information to find the exact value of each of the following. $$ \begin{array}{lll} ext { a. } \sin 2 heta & ext { b. } \cos 2 heta & ext { c. } an 2 heta\end{array} $$ $$ \sin heta=\frac{6}{13}, heta $$ lies in quadrant II

Question

Qle, Power-Reducing, and Half- Use the given information to find the exact value of each of the following. $$ \begin{array}{lll}	ext { a. } \sin 2 	heta & 	ext { b. } \cos 2 	heta & 	ext { c. } 	an 2 	heta\end{array} $$ $$ \sin 	heta=\frac{6}{13}, 	heta $$ lies in quadrant II

UpStudy AI · Accepted Answer

To find the exact values of $$\sin 2\theta$$, $$\cos 2\theta$$, and $$\tan 2\theta$$ given that $$\sin \theta = \frac{6}{13}$$ and $$\theta$$ lies in quadrant II, we can use the double angle formulas:

1. **Double Angle Formulas**:
   - $$\sin 2\theta = 2 \sin \theta \cos \theta$$
   - $$\cos 2\theta = \cos^2 \theta - \sin^2 \theta$$
   - $$\tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta}$$

2. **Finding $$\cos \theta$$**:
   Since $$\sin^2 \theta + \cos^2 \theta = 1$$, we can find $$\cos \theta$$:
   $$
   \sin^2 \theta = \left(\frac{6}{13}\right)^2 = \frac{36}{169}
   $$
   $$
   \cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{36}{169} = \frac{169 - 36}{169} = \frac{133}{169}
   $$
   $$
   \cos \theta = -\sqrt{\frac{133}{169}} = -\frac{\sqrt{133}}{13}
   $$
   (We take the negative root because $$\theta$$ is in quadrant II, where cosine is negative.)

3. **Calculating $$\sin 2\theta$$**:
   $$
   \sin 2\theta = 2 \sin \theta \cos \theta = 2 \left(\frac{6}{13}\right) \left(-\frac{\sqrt{133}}{13}\right)
   $$
   $$
   \sin 2\theta = -\frac{12\sqrt{133}}{169}
   $$

4. **Calculating $$\cos 2\theta$$**:
   $$
   \cos 2\theta = \cos^2 \theta - \sin^2 \theta = \left(-\frac{\sqrt{133}}{13}\right)^2 - \left(\frac{6}{13}\right)^2
   $$
   $$
   \cos 2\theta = \frac{133}{169} - \frac{36}{169} = \frac{133 - 36}{169} = \frac{97}{169}
   $$

5. **Calculating $$\tan 2\theta$$**:
   $$
   \tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} = \frac{-\frac{12\sqrt{133}}{169}}{\frac{97}{169}} = -\frac{12\sqrt{133}}{97}
   $$

### Final Results:
- $$ \sin 2\theta = -\frac{12\sqrt{133}}{169} $$
- $$ \cos 2\theta = \frac{97}{169} $$
- $$ \tan 2\theta = -\frac{12\sqrt{133}}{97} $$
- $$ \sin 2\theta = -\frac{12\sqrt{133}}{169} $$ - $$ \cos 2\theta = \frac{97}{169} $$ - $$ \tan 2\theta = -\frac{12\sqrt{133}}{97} $$

Qle, Power-Reducing, and Half-
Use the given information to find the exact value of each of the following.

lies in quadrant II

Solución de inteligencia artificial de Upstudy

Responder

Solución

Extra Insights

preguntas relacionadas

Latest Trigonometry Questions

Qle, Power-Reducing, and Half- Use the given information to find the exact value of each of the following. lies in quadrant II