find the value of sin[ $\pi$ /6-1/sin(- $\sqrt{-3}$/2)]

asked by guest
on Dec 04, 2024 at 12:56 am



You asked:

Evaluate the expression: \(\sin\left( \frac{\pi}{6} - \frac{1}{\sin\left( \frac{-\sqrt{-3}}{2} \right)} \right)\)

MathBot Answer:

\[\sin\left( \frac{\pi}{6} - \frac{1}{\sin\left( \frac{-\sqrt{-3}}{2} \right)} \right) = \sin{\left(\frac{\pi}{6} - \frac{i}{\sinh{\left(\frac{\sqrt{3}}{2} \right)}} \right)}\]

\[\begin{aligned}\sin\left( \frac{\pi}{6} - \frac{1}{\sin\left( \frac{-\sqrt{-3}}{2} \right)} \right)& = \sin{\left(\frac{\pi}{6} - \frac{1}{\sin{\left(- \cdot \sqrt{-3} \cdot \frac{1}{2} \right)}} \right)}\\& = \sin{\left(\frac{\pi}{6} - \frac{1}{\sin{\left(- \cdot i \cdot \sqrt{3} \cdot \frac{1}{2} \right)}} \right)}\\& = \sin{\left(\frac{\pi}{6} - \frac{1}{\sin{\left(- \frac{1}{2} \cdot i \cdot \sqrt{3} \right)}} \right)}\\& = \sin{\left(\frac{\pi}{6} - \frac{1}{- \cdot i \cdot \sinh{\left(\frac{1}{2} \cdot \sqrt{3} \right)}} \right)}\\& = \sin{\left(\frac{\pi}{6} - i \cdot \frac{1}{\sinh{\left(\frac{1}{2} \cdot \sqrt{3} \right)}} \right)}\end{aligned}\]

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