of this expression is driving me crazy...

sin(2sin^(-1)(-2/5))

anyone willing to enlighten me?

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- Jun 10th 2008, 07:51 PMgabrie30Finding the exact value
of this expression is driving me crazy...

sin(2sin^(-1)(-2/5))

anyone willing to enlighten me? - Jun 10th 2008, 08:31 PMmr fantastic
Let $\displaystyle \alpha = \sin^{-1} \left( - \frac{2}{5} \right) \Rightarrow \sin \alpha = - \frac{2}{5}$.

You want the value of $\displaystyle \sin (2 \alpha) = 2 \sin \alpha \cos \alpha$.

I will assume you can find $\displaystyle \cos \alpha$ given $\displaystyle \sin \alpha = - \frac{2}{5}$ ......