Question

What are the exact values of $\cos 150{}^\circ $ and $\sin 150{}^\circ $?
(a) $-\dfrac{\sqrt{3}}{2},\dfrac{1}{2}$
(b) $\dfrac{\sqrt{3}}{2},\dfrac{1}{2}$
(c) $\dfrac{\sqrt{3}}{2},-\dfrac{1}{2}$
(d) None of these

Answer 1

Hint: In this problem, we are to find out the values of $\cos 150{}^\circ $and $\sin 150{}^\circ $. So, we will start with finding the position of our given and check which quadrant it is into. Thus, we can conclude the signs of the values and also with the help of trigonometric table identities, we get our desired result.

Complete step by step solution:
According to the problem, here, we are trying to find the values of $\cos 150{}^\circ $and $\sin 150{}^\circ $.
So, to start with, we all know, 150 = 180 – 30.
Then, writing in that form in the right place, we get, $\cos 150{}^\circ =\cos \left( 180-30 \right){}^\circ $
But, again, from the trigonometric identities, $\cos \left( 180-\theta \right)=-\cos \theta $ .
Thus, $\cos 150{}^\circ $can be written in this form.
So, we get, $\cos 150{}^\circ =\cos \left( 180-30 \right){}^\circ =-\cos 30{}^\circ $
Again, from trigonometric table identities, we know that the value of $\cos 30{}^\circ $is $\dfrac{\sqrt{3}}{2}$ .
Then, we have, $\cos 150{}^\circ =-\cos 30{}^\circ =-\dfrac{\sqrt{3}}{2}$.
And now, taking $\sin 150{}^\circ $into consideration, we are getting, $\sin 150{}^\circ =\sin \left( 180-30 \right){}^\circ $
But, again, from the trigonometric identities, $\sin \left( 180-\theta \right)=\sin \theta $ .
Thus, $\sin 150{}^\circ $can be written in this form.
So, we get, $\sin 150{}^\circ =\sin \left( 180-30 \right){}^\circ =\sin 30{}^\circ $
Again, from trigonometric table identities, we know that the value of $\sin 30{}^\circ $is $\dfrac{1}{2}$ .
Then, we have, $\sin 150{}^\circ =\sin 30{}^\circ =\dfrac{1}{2}$.
So, the values of $\cos 150{}^\circ $and $\sin 150{}^\circ $are, $-\dfrac{\sqrt{3}}{2}$and $\dfrac{1}{2}$ respectively.

So, the correct answer is “Option a”.

Note: Now, for $\sin 150{}^\circ =\sin 30{}^\circ $, Both are equal because the reference angle for 150 is equal to 30 for the triangle formed in the unit circle. The reference angle is formed when the perpendicular is dropped from the unit circle to the x-axis, which forms a right triangle.
Since, the angle 150 degrees lies on the second quadrant, therefore the value of sin 150 is positive. The internal angle of the triangle is 180 – 150 =30, which is the reference angle.