use the given information to find the exact value of each of the following. a. sin 2θ b. cos 2θ c. tan 2θ…

Explanation:

Step1: Find cosθ

Since $\sin^{2}\theta+\cos^{2}\theta = 1$, then $\cos\theta=-\sqrt{1 - \sin^{2}\theta}$ (because $\theta$ is in quadrant II where cosine is negative). Given $\sin\theta=\frac{2}{3}$, we have $\cos\theta=-\sqrt{1 - (\frac{2}{3})^{2}}=-\sqrt{1-\frac{4}{9}}=-\sqrt{\frac{5}{9}}=-\frac{\sqrt{5}}{3}$.

Step2: Find sin2θ

Use the double - angle formula $\sin2\theta = 2\sin\theta\cos\theta$. Substitute $\sin\theta=\frac{2}{3}$ and $\cos\theta = -\frac{\sqrt{5}}{3}$ into it, we get $\sin2\theta=2\times\frac{2}{3}\times(-\frac{\sqrt{5}}{3})=-\frac{4\sqrt{5}}{9}$.

Step3: Find cos2θ

Use the double - angle formula $\cos2\theta=\cos^{2}\theta-\sin^{2}\theta$. Substitute $\sin\theta=\frac{2}{3}$ and $\cos\theta = -\frac{\sqrt{5}}{3}$ into it. $\cos2\theta=(-\frac{\sqrt{5}}{3})^{2}-(\frac{2}{3})^{2}=\frac{5}{9}-\frac{4}{9}=\frac{1}{9}$.

Step4: Find tan2θ

Use the formula $\tan2\theta=\frac{\sin2\theta}{\cos2\theta}$. Substitute $\sin2\theta = -\frac{4\sqrt{5}}{9}$ and $\cos2\theta=\frac{1}{9}$ into it, we get $\tan2\theta=- 4\sqrt{5}$.

Answer:

a. $-\frac{4\sqrt{5}}{9}$ b. $\frac{1}{9}$ c. $-4\sqrt{5}$