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# Find the value of $\cos 480^{\circ}\sin 150^{\circ} +\sin 600^{\circ}\cos 390^{\circ}$.A. 0B. 1C. $\dfrac{1}{2}$D. $-1$

Last updated date: 16th Mar 2023
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Hint:We explain the process of finding values for associated angles. We find the rotation and the position of the angle for $\cos 480^{\circ},\sin 150^{\circ},\sin 600^{\circ},\cos 390^{\circ}$. We explain the changes that are required for that angle. Depending on those things we find the solution.

We need to find the ratio values for $\cos 480^{\circ},\sin 150^{\circ},\sin 600^{\circ},\cos 390^{\circ}$. For general form, we need to convert the value of x into the closest multiple of $\dfrac{\pi }{2}$ and add or subtract a certain value $\alpha$ from that multiple of $\dfrac{\pi }{2}$ to make it equal to $x$.

Let’s assume $x=k\times \dfrac{\pi }{2}+\alpha$, $k\in \mathbb{Z}$. Here we took the addition of $\alpha$. We also need to remember that $\left| \alpha \right|\le \dfrac{\pi }{2}$. Now we take the value of $k$. For ratio cos, if it’s even then keep the ratio as cos and if it’s odd then the ratio changes to sin ratio from cos. For ratio sin, if it’s even then keep the ratio as sin and if it’s odd then the ratio changes to cos ratio from sin.

Then we find the position of the given angle as a quadrant value measured in counter clockwise movement from the origin and the positive side of X-axis. If for ratio cos the angel falls in the first or fourth quadrant then the sign remains positive but if it falls in the second or third quadrant then the sign becomes negative. For ratios sin the angel falls in the first or second quadrant then the sign remains positive but if it falls in the third or fourth quadrant then the sign becomes negative. The final form becomes,
$\cos 480^{\circ}=\cos \left( 5\times \dfrac{\pi }{2}+30 \right)=-\sin \left( 30 \right)=-\dfrac{1}{2}$
$\Rightarrow \sin 150^{\circ}=\sin \left( 1\times \dfrac{\pi }{2}+60 \right)=\cos \left( 60 \right)=\dfrac{1}{2}$
$\Rightarrow \sin 600^{\circ}=\sin \left( 6\times \dfrac{\pi }{2}+60 \right)=-\sin \left( 60 \right)=-\dfrac{\sqrt{3}}{2}$
$\Rightarrow \cos 390^{\circ}=\cos \left( 4\times \dfrac{\pi }{2}+30 \right)=\cos \left( 30 \right)=\dfrac{\sqrt{3}}{2}$
$\therefore \cos 480^{\circ}\sin 150^{\circ}+\sin 600^{\circ}\cos 390^{\circ}=\left( -\dfrac{1}{2} \right)\left( \dfrac{1}{2} \right)+\left( -\dfrac{\sqrt{3}}{2} \right)\left( \dfrac{\sqrt{3}}{2} \right)=-1$

Therefore, the value of $\cos 480^{\circ}\sin 150^{\circ}+\sin 600^{\circ}\cos 390^{\circ}$ is $-1$. The correct option is D.

Note:We need to remember that the easiest way to avoid the change of ratio thing is to form the multiple of $\pi$ instead of $\dfrac{\pi }{2}$. It makes the multiplied number always even. In that case we don’t have to change the ratio. If $x=k\times \pi +\alpha =2k\times \dfrac{\pi }{2}+\alpha$. Value of $2k$ is always even.