sin 120^{circ} cos 150^{circ} - cos 240^{circ | vedclass

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

Question

(B) We use the values of trigonometric functions for allied angles:$\sin 120^{\circ} = \sin(180^{\circ} - 60^{\circ}) = \sin 60^{\circ} = \frac{\sqrt{

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(B) We use the values of trigonometric functions for allied angles:$\sin 120^{\circ} = \sin(180^{\circ} - 60^{\circ}) = \sin 60^{\circ} = \frac{\sqrt{3}}{2}$$\cos 150^{\circ} = \cos(180^{\circ} - 30^{\circ}) = -\cos 30^{\circ} = -\frac{\sqrt{3}}{2}$$\cos 240^{\circ} = \cos(180^{\circ} + 60^{\circ}) = -\cos 60^{\circ} = -\frac{1}{2}$$\sin 330^{\circ} = \sin(360^{\circ} - 30^{\circ}) = -\sin 30^{\circ} = -\frac{1}{2}$Substituting these values into the expression:$\left(\frac{\sqrt{3}}{2}\right) \left(-\frac{\sqrt{3}}{2}\right) - \left(-\frac{1}{2}\right) \left(-\frac{1}{2}\right)$$= -\frac{3}{4} - \frac{1}{4}$$= -\frac{4}{4} = -1$

$\sin 120^{\circ} \cos 150^{\circ} - \cos 240^{\circ} \sin 330^{\circ}$ is equal to :

Similar Questions

From a disc of mass $M$ and radius $R$,a circular hole of diameter $R$ is cut such that its rim passes through the center of the disc. The moment of inertia of the remaining part of the disc about a perpendicular axis passing through the center of the original disc is

Ovulation is induced under the influence of which hormone?

$X$-rays of wavelength $0.140 \ nm$ are scattered from a block of carbon. What will be the wavelength of $X$-rays scattered at $90^{\circ}$ (in $nm$)?

Nitrogen fixation in the root nodules of $Alnus$ is performed by:

$A$ weak acid $HA$ has a $K_a$ of $1.00 \times 10^{-5}$. If $0.100 \ mol$ of this acid is dissolved in one litre of water,the percentage of acid dissociated at equilibrium is closest to.....$\%$

Vedclass Test Series

Exam Paper Generator

Online Exam Module