If $a = \frac{\pi }{{18}}\,\,{\rm{rad}}$, then $\cos \,\,a + \cos \,\,2a + ... + \cos \,\,18\,a$ is equal to

JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Easy

1 of 1

Application +4 | -1

a = \frac{\pi }{{18}}\,\,\mathrm{rad}

, then

\cos \,\,a + \cos \,\,2a + ... + \cos \,\,18\,a

is equal to

View answer and explanation

Correct option: B

- 1

Option B (

n\pi + {\left( { - 1} \right)^n}\frac{\pi }{4} - \frac{\pi }{4}

) is correct because it is the general solution to the trigonometric equation

\sin \theta + \cos \theta = 1

Practice All Chapter MCQs →

\tan \alpha=\frac{p}{q}

\alpha=6 \beta

\alpha

\frac{1}{2}[p \operatorname{cosec} 2 \beta-q \sec 2 \beta]

\cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ}

\triangle A B C

C

{\sin ^6}\,\,\theta + {\cos ^6}\,\,\theta + 3{\sin ^2}\,\,\theta \,\,{\cos ^2}\,\,\theta

\sin ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}+\sin ^2 \frac{C}{2}=

\sin ^2(A / 2)+\sin ^2(B / 2)-\sin ^2(C / 2)=