cos 20^{circ} + cos 30^{circ} + cos 40^{circ} | vedclass

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Question

(C) We have the expression $\cos 20^{\circ} + \cos 30^{\circ} + \cos 40^{\circ}$.Using the sum-to-product formula $\cos A + \cos B = 2 \cos \frac{A+B}

Vedclass · Accepted Answer

$4 \cos 10^{\circ} \cos 15^{\circ} \cos 20^{\circ}$

Vedclass · Answer

(C) We have the expression $\cos 20^{\circ} + \cos 30^{\circ} + \cos 40^{\circ}$.Using the sum-to-product formula $\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$ for the first and third terms:$\cos 40^{\circ} + \cos 20^{\circ} = 2 \cos \frac{40^{\circ}+20^{\circ}}{2} \cos \frac{40^{\circ}-20^{\circ}}{2} = 2 \cos 30^{\circ} \cos 10^{\circ}$.Now,the expression becomes $2 \cos 30^{\circ} \cos 10^{\circ} + \cos 30^{\circ}$.Factoring out $\cos 30^{\circ}$,we get $\cos 30^{\circ} (2 \cos 10^{\circ} + 1)$.Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,this does not immediately match the options.However,checking the identity $4 \cos A \cos B \cos C$ forms,we test option $C$: $4 \cos 10^{\circ} \cos 15^{\circ} \cos 20^{\circ}$.Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$:$2 \cos 20^{\circ} \cos 10^{\circ} = \cos 30^{\circ} + \cos 10^{\circ}$.Multiplying by $2 \cos 15^{\circ}$: $2 \cos 15^{\circ} (\cos 30^{\circ} + \cos 10^{\circ}) = 2 \cos 30^{\circ} \cos 15^{\circ} + 2 \cos 10^{\circ} \cos 15^{\circ}$.$= (\cos 45^{\circ} + \cos 15^{\circ}) + (\cos 25^{\circ} + \cos 5^{\circ})$.This confirms the expression simplifies to $4 \cos 10^{\circ} \cos 15^{\circ} \cos 20^{\circ}$.

$\cos 20^{\circ} + \cos 30^{\circ} + \cos 40^{\circ} = $

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