In the unit circle, calculate the exact values of sine, cosine, and tangent for the angle ( heta = frac{5pi}{6} ).

To find the trigonometric functions for \( \theta = \frac{5\pi}{6} \), we first recognize that this angle is in the second quadrant.

In the second quadrant, sine is positive, and cosine is negative.

Start with sine:

$ \sin \left(\frac{5\pi}{6}\right) = \sin \left(\pi – \frac{\pi}{6}\right) = \sin \left(\frac{\pi}{6}\right) = \frac{1}{2} $

Next, cosine:

$ \cos \left(\frac{5\pi}{6}\right) = \cos \left(\pi – \frac{\pi}{6}\right) = -\cos \left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2} $

Finally, tangent:

$ \tan \left(\frac{5\pi}{6}\right) = \frac{\sin \left(\frac{5\pi}{6}\right)}{\cos \left(\frac{5\pi}{6}\right)} = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} $

Therefore,

$ \sin \left(\frac{5\pi}{6}\right) = \frac{1}{2}, \cos \left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}, \tan \left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{3} $

For ( heta = frac{5pi}{6} ), a second quadrant angle, we have:

$ sin left(frac{5pi}{6}
ight) = frac{1}{2} $

$ cos left(frac{5pi}{6}
ight) = -frac{sqrt{3}}{2} $

$ an left(frac{5pi}{6}
ight) = -frac{sqrt{3}}{3} $

Using angle subtraction identity:

$ sin left(pi – heta
ight) = sin heta $

$ cos left(pi – heta
ight) = -cos heta $

$ an left(pi – heta
ight) = – an heta $

For ( heta = frac{pi}{6} ):

$ sin left(frac{pi}{6}
ight) = frac{1}{2} $

$ cos left(frac{pi}{6}
ight) = frac{sqrt{3}}{2} $

$ an left(frac{pi}{6}
ight) = frac{1}{sqrt{3}} = frac{sqrt{3}}{3} $

Therefore,

$ sin left(frac{5pi}{6}
ight) = frac{1}{2}, cos left(frac{5pi}{6}
ight) = -frac{sqrt{3}}{2}, an left(frac{5pi}{6}
ight) = -frac{sqrt{3}}{3} $

For the angle $ \theta = \frac{5\pi}{6} $ in the unit circle, we can find the sine, cosine, and tangent as follows:

Sine value:

$$ \sin\left(\frac{5\pi}{6}\right) = \frac{1}{2} $$

Cosine value:

$$ \cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} $$

Tangent value is calculated as sine divided by cosine:

$$ \tan\left(\frac{5\pi}{6}\right) = \frac{\sin\left(\frac{5\pi}{6}\right)}{\cos\left(\frac{5\pi}{6}\right)} = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} $$

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