If frac{cos x}{a} = frac{cos (x + theta)}{b} | vedclass

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(C) Let each ratio be equal to $k$,such that $\frac{\cos x}{a} = \frac{\cos (x + 	heta)}{b} = \frac{\cos (x + 2	heta)}{c} = \frac{\cos (x + 3	heta)

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$\frac{b}{c}$

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(C) Let each ratio be equal to $k$,such that $\frac{\cos x}{a} = \frac{\cos (x + 	heta)}{b} = \frac{\cos (x + 2	heta)}{c} = \frac{\cos (x + 3	heta)}{d} = k$.Then $a = \frac{\cos x}{k}$,$b = \frac{\cos (x + 	heta)}{k}$,$c = \frac{\cos (x + 2	heta)}{k}$,and $d = \frac{\cos (x + 3	heta)}{k}$.Consider the expression $\frac{a + c}{b + d} = \frac{\frac{\cos x}{k} + \frac{\cos (x + 2	heta)}{k}}{\frac{\cos (x + 	heta)}{k} + \frac{\cos (x + 3	heta)}{k}}$.$= \frac{\cos x + \cos (x + 2	heta)}{\cos (x + 	heta) + \cos (x + 3	heta)}$.Using the sum-to-product formula $\cos A + \cos B = 2 \cos \left( \frac{A + B}{2} ight) \cos \left( \frac{A - B}{2} ight)$:Numerator: $\cos x + \cos (x + 2	heta) = 2 \cos (x + 	heta) \cos 	heta$.Denominator: $\cos (x + 	heta) + \cos (x + 3	heta) = 2 \cos (x + 2	heta) \cos 	heta$.Thus,$\frac{a + c}{b + d} = \frac{2 \cos (x + 	heta) \cos 	heta}{2 \cos (x + 2	heta) \cos 	heta} = \frac{\cos (x + 	heta)}{\cos (x + 2	heta)}$.Since $\frac{\cos (x + 	heta)}{b} = \frac{\cos (x + 2	heta)}{c} = k$,we have $\frac{\cos (x + 	heta)}{\cos (x + 2	heta)} = \frac{b}{c}$.Therefore,$\frac{a + c}{b + d} = \frac{b}{c}$.

If $\frac{\cos x}{a} = \frac{\cos (x + \theta)}{b} = \frac{\cos (x + 2\theta)}{c} = \frac{\cos (x + 3\theta)}{d}$,then $\left( \frac{a + c}{b + d} \right)$ is equal to:

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