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If cos x=(3)/(5),x in quadrant I, then find (without finding x ) {:[sin(2x)=],[cos(2x)=],[tan(2x)=]:}

If cos⁡x=35,x \cos x=\frac{3}{5}, x in quadrant I I , then find (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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Full solution

Q. If cos⁡x=35,x \cos x=\frac{3}{5}, x in quadrant I I , then find (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. Apply Double Angle Formulas: 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)\newlinetan⁡(2x)=sin⁡(2x)cos⁡(2x)\tan(2x) = \frac{\sin(2x)}{\cos(2x)}
  2. Find sin⁡(x)\sin(x): Find sin⁡(x)\sin(x) using the Pythagorean identity, sin⁡2(x)+cos⁡2(x)=1\sin^2(x) + \cos^2(x) = 1:sin⁡2(x)=1−cos⁡2(x)=1−(35)2=1−925=1625\sin^2(x) = 1 - \cos^2(x) = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}sin⁡(x)=1625=45\sin(x) = \sqrt{\frac{16}{25}} = \frac{4}{5} (since xx is in quadrant I, sin⁡(x)\sin(x) is positive)
  3. Calculate sin⁡(2x)\sin(2x) and cos⁡(2x)\cos(2x): Calculate sin⁡(2x)\sin(2x) and cos⁡(2x)\cos(2x) using the values of sin⁡(x)\sin(x) and cos⁡(x)\cos(x):sin⁡(2x)=2×(45)×(35)=2425\sin(2x) = 2 \times \left(\frac{4}{5}\right) \times \left(\frac{3}{5}\right) = \frac{24}{25}cos⁡(2x)=(35)2−(45)2=925−1625=−725\cos(2x) = \left(\frac{3}{5}\right)^2 - \left(\frac{4}{5}\right)^2 = \frac{9}{25} - \frac{16}{25} = -\frac{7}{25}
  4. Calculate tan⁡(2x)\tan(2x): Calculate tan⁡(2x)\tan(2x) using sin⁡(2x)\sin(2x) and cos⁡(2x)\cos(2x):tan⁡(2x)=2425/−725=−247\tan(2x) = \frac{24}{25} / \frac{-7}{25} = -\frac{24}{7}

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