If cos theta = frac{3}{5} and cos phi = frac{ | vedclass

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(B) Given $\cos 	heta = \frac{3}{5}$ and $\cos \phi = \frac{4}{5}$.Since $	heta$ and $\phi$ are acute angles,$\sin 	heta = \sqrt{1 - \cos^2 	heta}

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$\frac{7}{5\sqrt{2}}$

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(B) Given $\cos 	heta = \frac{3}{5}$ and $\cos \phi = \frac{4}{5}$.Since $	heta$ and $\phi$ are acute angles,$\sin 	heta = \sqrt{1 - \cos^2 	heta} = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$ and $\sin \phi = \sqrt{1 - \cos^2 \phi} = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$.Using the formula $\cos(	heta - \phi) = \cos 	heta \cos \phi + \sin 	heta \sin \phi$,we get:$\cos(	heta - \phi) = \left(\frac{3}{5}ight)\left(\frac{4}{5}ight) + \left(\frac{4}{5}ight)\left(\frac{3}{5}ight) = \frac{12}{25} + \frac{12}{25} = \frac{24}{25}$.We know that $2\cos^2\left(\frac{	heta - \phi}{2}ight) = 1 + \cos(	heta - \phi)$.Substituting the value,$2\cos^2\left(\frac{	heta - \phi}{2}ight) = 1 + \frac{24}{25} = \frac{49}{25}$.Therefore,$\cos^2\left(\frac{	heta - \phi}{2}ight) = \frac{49}{50}$.Taking the square root,$\cos\left(\frac{	heta - \phi}{2}ight) = \sqrt{\frac{49}{50}} = \frac{7}{5\sqrt{2}}$.

If $\cos \theta = \frac{3}{5}$ and $\cos \phi = \frac{4}{5},$ where $\theta$ and $\phi$ are positive acute angles,then $\cos \frac{\theta - \phi}{2} = $

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