Question

find the exact value of the following trigonometric function: \\(\cos 150^{\circ}\\)\
a \\(-\dfrac{\sqrt{3}}{2}\\)\
b \\(-\dfrac{\sqrt{3}}{3}\\)\
c \\(2\\)\
d undefined

Explanation:

Step1: Rewrite 150° as a difference

We know that \(150^{\circ}=180^{\circ} - 30^{\circ}\). So, we can use the cosine difference identity \(\cos(A - B)=\cos A\cos B+\sin A\sin B\), but here we have \(\cos(180^{\circ}-30^{\circ})\). The formula for \(\cos(180^{\circ}-\theta)\) is \(-\cos\theta\) (from the unit circle and trigonometric identities, since cosine is negative in the second quadrant and \(180^{\circ}-\theta\) is in the second quadrant when \(\theta\) is acute).

Step2: Apply the identity

Using \(\cos(180^{\circ}-\theta)=-\cos\theta\) with \(\theta = 30^{\circ}\), we get \(\cos(150^{\circ})=\cos(180^{\circ}-30^{\circ})=-\cos(30^{\circ})\).
We know that \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\), so substituting this value in, we have \(-\cos(30^{\circ})=-\frac{\sqrt{3}}{2}\).

Answer:

A. \(-\frac{\sqrt{3}}{2}\)