Questions: If sin x = 2/3, x in quadrant I, then find (without finding x) sin (2 x)= cos (2 x)= tan (2 x)=

Transcript text: If $\sin x=\frac{2}{3}, x$ in quadrant $I$, then find (without finding $x$ ) \[ \begin{array}{l} \sin (2 x)=\square \\ \cos (2 x)=\square \\ \tan (2 x)=\square \end{array} \]

Solution

Solution Steps

Step 1: Find $\cos x$ using the Pythagorean identity

Given $\sin x = \frac{2}{3}$ and $x$ is in quadrant $I$, we use the Pythagorean identity: \[ \sin^2 x + \cos^2 x = 1 \] Substitute $\sin x = \frac{2}{3}$: \[ \left(\frac{2}{3}\right)^2 + \cos^2 x = 1 \] \[ \frac{4}{9} + \cos^2 x = 1 \] \[ \cos^2 x = 1 - \frac{4}{9} = \frac{5}{9} \] \[ \cos x = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \]

Step 2: Calculate $\sin(2x)$ using the double-angle formula

The double-angle formula for sine is: \[ \sin(2x) = 2 \sin x \cos x \] Substitute $\sin x = \frac{2}{3}$ and $\cos x = \frac{\sqrt{5}}{3}$: \[ \sin(2x) = 2 \cdot \frac{2}{3} \cdot \frac{\sqrt{5}}{3} = \frac{4\sqrt{5}}{9} \]

Step 3: Calculate $\cos(2x)$ using the double-angle formula

The double-angle formula for cosine is: \[ \cos(2x) = \cos^2 x - \sin^2 x \] Substitute $\sin x = \frac{2}{3}$ and $\cos x = \frac{\sqrt{5}}{3}$: \[ \cos(2x) = \left(\frac{\sqrt{5}}{3}\right)^2 - \left(\frac{2}{3}\right)^2 = \frac{5}{9} - \frac{4}{9} = \frac{1}{9} \]