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,$\sin 	heta = \sqrt{1 - \cos^2 	heta} = \sqrt{1 - (\frac{3}{5})^2} = \frac{4}{5}$ and $\sin \phi = \sqrt{1 - \cos^2 \phi} = \sqrt{1 - (\frac{4}{5})^2} = \frac{3}{5}$.Using the formula $\cos(	heta - \phi) = \cos 	heta \cos \phi + \sin 	heta \sin \phi$,we get:$\cos(	heta - \phi) = (\frac{3}{5})(\frac{4}{5}) + (\frac{4}{5})(\frac{3}{5}) = \frac{12}{25} + \frac{12}{25} = \frac{24}{25}$.Using the half-angle identity $2\cos^2(\frac{	heta - \phi}{2}) = 1 + \cos(	heta - \phi)$:$2\cos^2(\frac{	heta - \phi}{2}) = 1 + \frac{24}{25} = \frac{49}{25}$.$\cos^2(\frac{	heta - \phi}{2}) = \frac{49}{50}$.Taking the square root,$\cos(\frac{	heta - \phi}{2}) = \sqrt{\frac{49}{50}} = \frac{7}{5\sqrt{2}}$.

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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