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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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} = $

Similar Questions

If $m \tan (\theta - 30^\circ) = n \tan (\theta + 120^\circ)$,then $\frac{m + n}{m - n} = $

The value of $\cos ^{2} 75^{\circ}+\cos ^{2} 45^{\circ}+\cos ^{2} 15^{\circ}-\cos ^{2} 30^{\circ}-\cos ^{2} 60^{\circ}$ is

$\frac{\tan 52^{\circ} - \tan 38^{\circ}}{\tan 14^{\circ}} = $

If $\tan \alpha = \frac{m}{m + 1}$ and $\tan \beta = \frac{1}{2m + 1}$,then $\alpha + \beta = $

If $\theta+\phi=\frac{\pi}{4}$,then $(1+\tan \theta)(1+\tan \phi)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module