Problems from Trigonometric ratios

Find the value of the trigonometric functions (sine, cosine and tangent) for $$x=\dfrac{5\pi}{6}$$ rad.

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Development:

The angle $$\dfrac{5\pi}{6}$$ is an angle in the second quadrant, that is $$\dfrac{\pi}{2} < \dfrac{5\pi}{6} < \pi$$, therefore we have: $$$\sin(x)=\sin(\dfrac{5\pi}{6})=\sin(\pi-\dfrac{5\pi}{6})=\sin(\dfrac{\pi}{6})=\dfrac{1}{2}$$$

$$$\cos(x)=\cos(\dfrac{5\pi}{6})=-\cos(\pi-\dfrac{5\pi}{6})=-\cos(\dfrac{\pi}{6})=-\dfrac{\sqrt{3}}{2}$$$

$$$\tan(x)=\tan(\dfrac{5\pi}{6})=-\tan(\pi-\dfrac{5\pi}{6})=-\tan(\dfrac{\pi}{6})=-\dfrac{\sqrt{3}}{3}$$$

Solution:

$$\sin(x)=\dfrac{1}{2}$$

$$\cos(x)=-\dfrac{\sqrt{3}}{2}$$

$$\tan(x)=-\dfrac{\sqrt{3}}{3}$$

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