If sin 600^{circ} cos 30^{circ} + cos 120^{ci | vedclass

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

Question

(C) Given expression: $k = \sin 600^{\circ} \cos 30^{\circ} + \cos 120^{\circ} \sin 150^{\circ}$.First,simplify the trigonometric angles:$\sin 600^{\c

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(C) Given expression: $k = \sin 600^{\circ} \cos 30^{\circ} + \cos 120^{\circ} \sin 150^{\circ}$.First,simplify the trigonometric angles:$\sin 600^{\circ} = \sin(360^{\circ} + 240^{\circ}) = \sin 240^{\circ} = \sin(180^{\circ} + 60^{\circ}) = -\sin 60^{\circ} = -\frac{\sqrt{3}}{2}$.$\cos 120^{\circ} = \cos(180^{\circ} - 60^{\circ}) = -\cos 60^{\circ} = -\frac{1}{2}$.$\sin 150^{\circ} = \sin(180^{\circ} - 30^{\circ}) = \sin 30^{\circ} = \frac{1}{2}$.$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$.Substitute these values into the expression:$k = \left(-\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(-\frac{1}{2}\right) \left(\frac{1}{2}\right)$.$k = -\frac{3}{4} - \frac{1}{4}$.$k = -\frac{4}{4} = -1$.

If $\sin 600^{\circ} \cos 30^{\circ} + \cos 120^{\circ} \sin 150^{\circ} = k$,then $k =$

Similar Questions

$\sin ^4 \frac{\pi}{8} + \sin ^4 \frac{3\pi}{8} + \sin ^4 \frac{5\pi}{8} + \sin ^4 \frac{7\pi}{8} = $

Which of the following is correct?

$\frac{\sin 3\theta + \sin 5\theta + \sin 7\theta + \sin 9\theta}{\cos 3\theta + \cos 5\theta + \cos 7\theta + \cos 9\theta} = $

If $\theta$ is an acute angle and $\tan^{2} \theta + \frac{1}{\tan^{2} \theta} = 2$,then the value of $\theta$ is (in $^{\circ}$)

The value of $\cos 52^\circ + \cos 68^\circ + \cos 172^\circ$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module