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If sin x=(2)/(3),x in quadrant I, then find the exact answers for the following (without finding x ): {:[sin(2x)=],[cos(2x)=],[tan(2x)=]:}

If sin⁡x=23,x \sin x=\frac{2}{3}, x in quadrant I I , then find the exact answers for the following (without finding x x ):\newlinesin⁡(2x)=cos⁡(2x)=tan⁡(2x)= \begin{array}{l} \sin (2 x)= \\ \cos (2 x)= \\ \tan (2 x)= \end{array}

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Q. If sin⁡x=23,x \sin x=\frac{2}{3}, x in quadrant I I , then find the exact answers for the following (without finding x x ):\newlinesin⁡(2x)=cos⁡(2x)=tan⁡(2x)= \begin{array}{l} \sin (2 x)= \\ \cos (2 x)= \\ \tan (2 x)= \end{array}
  1. Use double angle formulas: Step 11: Use the double angle formulas for sine and cosine.\newlinesin⁡(2x)=2sin⁡(x)cos⁡(x)\sin(2x) = 2\sin(x)\cos(x)\newlinecos⁡(2x)=cos⁡2(x)−sin⁡2(x)\cos(2x) = \cos^2(x) - \sin^2(x)
  2. Calculate cos⁡(x)\cos(x): Step 22: Calculate cos⁡(x)\cos(x) using the Pythagorean identity since sin⁡(x)=23\sin(x) = \frac{2}{3}.cos⁡(x)=1−sin⁡2(x)=1−(23)2=1−49=59=53\cos(x) = \sqrt{1 - \sin^2(x)} = \sqrt{1 - (\frac{2}{3})^2} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}
  3. Substitute into formulas: Step 33: Substitute sin⁡(x)\sin(x) and cos⁡(x)\cos(x) into the double angle formulas.\newlinesin⁡(2x)=2⋅(23)⋅(53)=459\sin(2x) = 2 \cdot \left(\frac{2}{3}\right) \cdot \left(\frac{\sqrt{5}}{3}\right) = \frac{4\sqrt{5}}{9}\newlinecos⁡(2x)=(53)2−(23)2=59−49=19\cos(2x) = \left(\frac{\sqrt{5}}{3}\right)^2 - \left(\frac{2}{3}\right)^2 = \frac{5}{9} - \frac{4}{9} = \frac{1}{9}
  4. Calculate tan⁡(2x)\tan(2x): Step 44: Calculate tan⁡(2x)\tan(2x) using the formula tan⁡(2x)=sin⁡(2x)cos⁡(2x)\tan(2x) = \frac{\sin(2x)}{\cos(2x)}.\newlinetan⁡(2x)=459/19=45\tan(2x) = \frac{4\sqrt{5}}{9} / \frac{1}{9} = 4\sqrt{5}

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