Question

Find the exact value, without a

Answer 1

Answer:

$-2 -\sqrt{3}$

Step-by-step explanation:

First consider numerator

$sin \frac{\frac{7\pi}{6}}{2} = sin \frac{7\pi}{12}= sin (\frac{\pi}{4} + \frac{\pi}{3})$

Using the formula : sin (A + B) = sin A cos B + cos A sin B

$sin \frac{\pi}{4} = \frac{\sqrt{2} }{2}, \ cos \frac{\pi}{4} = \frac{\sqrt{2} }{2}\\\\sin \frac{\pi}{3} = \frac{\sqrt{2} }{2}, \ cos \frac{\pi}{3} = \frac{1}{2}$

$sin(\frac{\pi}{4} + \frac{\pi}{3}) = sin \frac{\pi}{4} \cdot cos \frac{\pi}{3} + cos \frac{\pi}{4} \cdot sin \frac{\pi}{3}$

                $=\frac{\sqrt{2} }{2} \cdot \frac{1}{2} + \frac{\sqrt{2} }{2} \cdot \frac{\sqrt{3} }{2} \\\\= \frac{\sqrt{2} }{4} + \frac{\sqrt{6} }{4}\\\\=\frac{\sqrt{2} +\sqrt{6} }{4}$

Second consider denominator

$cos \frac{\frac{7\pi}{6}}{2} = cos \frac{7\pi}{12}= cos (\frac{\pi}{4} + \frac{\pi}{3})$

Using the formula : cos (A + B) = cos A cos B - sin A sin B

$sin \frac{\pi}{4} = \frac{\sqrt{2} }{2}, \ cos \frac{\pi}{4} = \frac{\sqrt{2} }{2}\\\\sin \frac{\pi}{3} = \frac{\sqrt{2} }{2}, \ cos \frac{\pi}{3} = \frac{1}{2}$

$cos(\frac{\pi}{4} + \frac{\pi}{3}) = cos \frac{\pi}{4} \cdot cos \frac{\pi}{3} -sin \frac{\pi}{4} \cdot sin \frac{\pi}{3}$

$=\frac{\sqrt{2} }{2} \cdot \frac{1}{2} - \frac{\sqrt{2} }{2} \cdot \frac{\sqrt{3} }{2}\\\\=\frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4}\\\\= \frac{\sqrt{2} -\sqrt{6} }{4}$

Therefore,

           $tan \frac{7\pi}{12} = \frac{sin \frac{7\pi}{12}}{cos\frac{7\pi}{12}}$

                     $= \frac{\frac{\sqrt{2} +\sqrt{6} }{4}} {\frac{\sqrt{2} -\sqrt{6} }{4} }\\\\=\frac{\sqrt{2} +\sqrt{6} }{4} \times \frac{4 }{\sqrt{2} -\sqrt{6}}\\\\=\frac{\sqrt{2} +\sqrt{6} }{\sqrt{2} -\sqrt{6}}$

Either we can stop here or Rationalize the denominator:

$\frac{\sqrt{2} +\sqrt{6} }{\sqrt{2} -\sqrt{6}} \times \frac{\sqrt{2} +\sqrt{6} }{\sqrt{2} +\sqrt{6}} = \frac{(\sqrt{2} +\sqrt{6})^{2} }{(\sqrt{2})^2 -(\sqrt{6})^2} = \frac{2 + 6 +2\sqrt{12} }{2-6} = \frac{8+2\sqrt{12} }{-4} = \frac{8+ 4\sqrt{3} }{-4} = -2-\sqrt{3}$