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

Apply Double Angle Formulas: Use the double angle formulas for sine and cosine: sin(2x) = 2sin(x)cos(x) cos(2x) = cos^2(x) - sin^2(x) tan(2x) = sin(2x)/cos(2x)Find sin(x): Find sin(x) using the Pythagorean identity, sin^2(x) + cos^2(x) = 1: sin^2(x) = 1 - cos^2(x) = 1 - (3/5)^2 = 1 - 9/25 = 16/25 sin(x) = sqrt(16/25) = 4/5 (since x is in quadrant I, sin(x) is positive)Calculate sin(2x) and cos(2x): Calculate sin(2x) and cos(2x) using the values of sin(x) and cos(x): sin(2x) = 2 * (4/5) * (3/5) = 24/25 cos(2x) = (3/5)^2 - (4/5)^2 = 9/25 - 16/25 = -7/25Calculate tan(2x): Calculate tan(2x) using sin(2x) and cos(2x): tan(2x) = (24/25) / (-7/25) = -24/7

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Find derivatives of sine and cosine functions

Full solution

Q. If

\cos x=\frac{3}{5}, x

in quadrant

I

, then find (without finding

x

)

\newline

\begin{array}{l} \sin (2 x)= \\ \cos (2 x)= \\ \tan (2 x)= \end{array}

Apply Double Angle Formulas: Use the double angle formulas for sine and cosine: $\newline$ $\sin(2x) = 2\sin(x)\cos(x)$ $\newline$ $\cos(2x) = \cos^2(x) - \sin^2(x)$ $\newline$ $\tan(2x) = \frac{\sin(2x)}{\cos(2x)}$
Find $\sin(x)$ : Find $\sin(x)$ using the Pythagorean identity, $\sin^2(x) + \cos^2(x) = 1$ : $\sin^2(x) = 1 - \cos^2(x) = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}$ $\sin(x) = \sqrt{\frac{16}{25}} = \frac{4}{5}$ (since $x$ is in quadrant I, $\sin(x)$ is positive)
Calculate $\sin(2x)$ and $\cos(2x)$ : Calculate $\sin(2x)$ and $\cos(2x)$ using the values of $\sin(x)$ and $\cos(x)$ : $\sin(2x) = 2 \times \left(\frac{4}{5}\right) \times \left(\frac{3}{5}\right) = \frac{24}{25}$ $\cos(2x) = \left(\frac{3}{5}\right)^2 - \left(\frac{4}{5}\right)^2 = \frac{9}{25} - \frac{16}{25} = -\frac{7}{25}$
Calculate $\tan(2x)$ : Calculate $\tan(2x)$ using $\sin(2x)$ and $\cos(2x)$ : $\tan(2x) = \frac{24}{25} / \frac{-7}{25} = -\frac{24}{7}$