The value of (sin 210^{circ})(sin 585^{circ}) | vedclass

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

Question

(A) We know that $\sin(180^{\circ} + 	heta) = -\sin 	heta$ and $\sin(540^{\circ} + 	heta) = \sin(360^{\circ} + 180^{\circ} + 	heta) = \sin(180^{\c

Vedclass · Accepted Answer

$\frac{1}{2 \sqrt{2}}$

Vedclass · Answer

(A) We know that $\sin(180^{\circ} + 	heta) = -\sin 	heta$ and $\sin(540^{\circ} + 	heta) = \sin(360^{\circ} + 180^{\circ} + 	heta) = \sin(180^{\circ} + 	heta) = -\sin 	heta$. Step $1$: Simplify $\sin 210^{\circ} = \sin(180^{\circ} + 30^{\circ}) = -\sin 30^{\circ} = -\frac{1}{2}$. Step $2$: Simplify $\sin 585^{\circ} = \sin(540^{\circ} + 45^{\circ}) = -\sin 45^{\circ} = -\frac{1}{\sqrt{2}}$. Step $3$: Multiply the values: $(-\frac{1}{2}) 	imes (-\frac{1}{\sqrt{2}}) = \frac{1}{2 \sqrt{2}}$. Thus,the correct option is $A$.

The value of $(\sin 210^{\circ})(\sin 585^{\circ})$ is

Similar Questions

$\tan A = \frac{-60}{11}$ and $A$ does not lie in the $4^{\text{th}}$ quadrant. $\sec B = \frac{41}{9}$ and $B$ does not lie in the $1^{\text{st}}$ quadrant. If $\operatorname{cosec} A + \cot B = K$,then $24K =$

If $\theta$ lies in the third quadrant and $\cos \theta = -\frac{3}{5}$,find the value of $\tan \theta$.

$\tanh (\log x) = $

Find the degree measure of the angle subtended at the centre of a circle of radius $100 \, cm$ by an arc of length $22 \, cm$ (Use $\pi = \frac{22}{7}$).

Find the angle in radians through which a pendulum swings if its length is $75 \, cm$ and the tip describes an arc of length $10 \, cm$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module