what are the sine, cosine, and tangent of 11π/6 radians? a sin θ = √3/2 ; cos θ = 1/2 ; tan θ = √3/3 b sin θ…

Explanation:

Step1: Rewrite the angle

We can rewrite $\frac{11\pi}{6}$ as $2\pi-\frac{\pi}{6}$.

Step2: Use trig - function properties

Since $\sin(2\pi - \alpha)=-\sin\alpha$, $\cos(2\pi - \alpha)=\cos\alpha$ and $\tan(2\pi - \alpha)=-\tan\alpha$. For $\alpha = \frac{\pi}{6}$, we know that $\sin\frac{\pi}{6}=\frac{1}{2}$, $\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$ and $\tan\frac{\pi}{6}=\frac{\sqrt{3}}{3}$. So $\sin\frac{11\pi}{6}=-\sin\frac{\pi}{6}=-\frac{1}{2}$, $\cos\frac{11\pi}{6}=\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$, $\tan\frac{11\pi}{6}=-\tan\frac{\pi}{6}=-\frac{\sqrt{3}}{3}$.

Answer:

C. $\sin\theta =-\frac{1}{2};\cos\theta=\frac{\sqrt{3}}{2};\tan\theta =-\frac{\sqrt{3}}{3}$