Questions: Evaluate (if possible) the sine, cosine, and tangent of the real number. t = (7 π)/6

Solution Steps

To evaluate \( \sin\left(\frac{7\pi}{6}\right) \), we can use the sine addition formula: \[ \sin(a+b) = \sin(a)\cos(b) + \cos(a)\sin(b) \] We rewrite \( \frac{7\pi}{6} \) as \( \frac{\pi}{6} + \pi \): \[ \sin\left(\frac{7\pi}{6}\right) = \sin\left(\frac{\pi}{6} + \pi\right) = \sin\left(\frac{\pi}{6}\right)\cos(\pi) + \cos\left(\frac{\pi}{6}\right)\sin(\pi) \] Substituting the known values: \[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}, \quad \cos(\pi) = -1, \quad \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}, \quad \sin(\pi) = 0 \] Thus, \[ \sin\left(\frac{7\pi}{6}\right) = \frac{1}{2} \cdot (-1) + \frac{\sqrt{3}}{2} \cdot 0 = -\frac{1}{2} \]

To evaluate \( \cos\left(\frac{7\pi}{6}\right) \), we can use the cosine addition formula: \[ \cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b) \] We rewrite \( \frac{7\pi}{6} \) as \( \frac{\pi}{6} + \pi \): \[ \cos\left(\frac{7\pi}{6}\right) = \cos\left(\frac{\pi}{6} + \pi\right) = \cos\left(\frac{\pi}{6}\right)\cos(\pi) - \sin\left(\frac{\pi}{6}\right)\sin(\pi) \] Substituting the known values: \[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}, \quad \cos(\pi) = -1, \quad \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}, \quad \sin(\pi) = 0 \] Thus, \[ \cos\left(\frac{7\pi}{6}\right) = \frac{\sqrt{3}}{2} \cdot (-1) - \frac{1}{2} \cdot 0 = -\frac{\sqrt{3}}{2} \]

To evaluate \( \tan\left(\frac{7\pi}{6}\right) \), we use the definition of tangent: \[ \tan(x) = \frac{\sin(x)}{\cos(x)} \] Substituting the values we found: \[ \tan\left(\frac{7\pi}{6}\right) = \frac{\sin\left(\frac{7\pi}{6}\right)}{\cos\left(\frac{7\pi}{6}\right)} = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \]

Final Answer