91Ó°ÊÓ

We are given \(\frac{17 \pi}{12} = \frac{9 \pi}{4} - \frac{5 \pi}{6}\). This is the same as saying the angle \(\Theta = \frac{9 \pi}{4} - \frac{5 \pi}{6}\).

The sine, cosine and tangent of this angle can be calculated by applying the formulae for the sine, cosine and tangent of the difference of two angles. These are\[\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta\]\[\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta\]\[\tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}\]where \(\alpha = \frac{9 \pi}{4}\) and \(\beta = \frac{5 \pi}{6}\).

As both \(\alpha\) and \(\beta\) are parts of the quadrants of the unit circle, we can simply recall their sine, cosine and tangent. For sake of brevity, we can find the sine and cosine and then calculate the tangent by dividing sine by cosine.\[\sin(\Theta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta = \sin \frac{9 \pi}{4} \cos \frac{5 \pi}{6} - \cos \frac{9 \pi}{4} \sin \frac{5 \pi}{6} = 1 \cdot \frac{1}{2} - 0 \cdot \frac{\sqrt{3}}{2} = \frac{1}{2}\]\[\cos(\Theta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta = \cos \frac{9 \pi}{4} \cos \frac{5 \pi}{6} + \sin \frac{9 \pi}{4} \sin \frac{5 \pi}{6} = 0 \cdot \frac{1}{2} + 1 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}\]Then\[\tan(\Theta) = \frac{\sin(\Theta)}{\cos(\Theta)} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}\]