91Ó°ÊÓ

Break down the given angle \( \frac{17 \pi}{12} \) as \( \frac{9 \pi}{4}-\frac{5 \pi}{6} \). This manipulation allows us to deal with simpler fractions of pi.

Notice that the angle \( \frac{9 \pi}{4} \) is equivalent to \( 2\pi + \frac{\pi}{4} \), which corresponds to \( \frac{\pi}{4} \) because \( 2\pi \) stands for a full circle rotation. Thus, the cosine, sine and tangent of this angle equals to cosine, sine and tangent of \( \frac{\pi}{4} \) respectively. By comparison, \( \frac{5 \pi}{6} \) is already at its simplest form.

Using subtraction formulas we have \(\sin(\alpha - \beta) = \sin\alpha \cos\beta - \cos\alpha \sin\beta\),\(\cos(\alpha - \beta) = \cos\alpha \cos\beta + \sin\alpha \sin\beta\),and \(\tan(\alpha - \beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha \tan\beta}\).Plug in \( \frac{\pi}{4} \) as \( \alpha \) and \( \frac{5 \pi}{6} \) as \( \beta \). For \( \frac{\pi}{4} \), \( \sin(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \) and \( \tan(\frac{\pi}{4}) = 1 \). For \( \frac{5 \pi}{6} \), \( \sin(\frac{5 \pi}{6}) = \frac{1}{2} \), \( \cos(\frac{5 \pi}{6}) = -\frac{\sqrt{3}}{2} \) and \( \tan(\frac{5 \pi}{6}) = -\sqrt{3} \). Now we can calculate the sine, cosine and tangent of the given angle.

Substitute the values obtained in Step 3 to the subtraction formulas.For sine, we obtain \( \sin(\frac{17 \pi}{12}) = \sin(\frac{\pi}{4}) \cos(\frac{5 \pi}{6}) - \cos(\frac{\pi}{4}) \sin(\frac{5 \pi}{6}) = \frac{\sqrt{2}}{2} \cdot -\frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = -\frac{\sqrt{6}+\sqrt{2}}{4} \).For cosine, we have \( \cos(\frac{17 \pi}{12}) = \cos(\frac{\pi}{4}) \cos(\frac{5 \pi}{6}) + \sin(\frac{\pi}{4}) \sin(\frac{5 \pi}{6}) = \frac{\sqrt{2}}{2} \cdot -\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = -\frac{\sqrt{6}-\sqrt{2}}{4} \).For tangent, we have \( \tan(\frac{17 \pi}{12}) = \frac{\tan(\frac{\pi}{4}) - \tan(\frac{5 \pi}{6})}{1 + \tan(\frac{\pi}{4}) \tan(\frac{5 \pi}{6})} = \frac{1 - -\sqrt{3}}{1 + 1 \cdot -\sqrt{3}} = \frac{2+\sqrt{3}}{1-\sqrt{3}} \cdot \frac{1+\sqrt{3}}{1+\sqrt{3}} = 2\sqrt{3}-1 \).