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Question Number 130580 by bramlexs22 last updated on 27/Jan/21

$$\mathrm{If}\mathrm{x}ϵ[-1,0)\mathrm{then}{\cos }^{-1}({2\mathrm{x}}^2-1)-{2\sin }^{-1}\left(\mathrm{x}\right)=?$$

Commented by EDWIN88 last updated on 27/Jan/21

$$\pi$$

Answered by mindispower last updated on 27/Jan/21

$$x=cos\left(\frac{\theta }{2}\right),\theta \in [\pi ,2\pi [$$$$\Rightarrow {cos}^{-}(2{cos}^2\left(\frac{\theta }{2}\right)-1)-2{sin}^{-}\left(cos\left(\frac{\theta }{2}\right)\right)$$$$={cos}^{-}\left(cos\left(\theta \right)\right)-2{sin}^{-}\left(sin\left(\frac{\pi -\theta }{2}\right)\right)$$$${sin}^{-}\left(sin\left(\frac{\pi -\theta }{2}\right)\right)=\frac{\pi -\theta }{2},\frac{\pi -\theta }{2}\in ]-\frac{\pi }{2},0[$$$${cos}^{-}\left(cos\left(\theta \right)\right)={cos}^{-}\left(cos(2\pi -\theta )\right)=2\pi -\theta$$$$2\pi -\theta -2\left(\frac{\pi -\theta }{2}\right)=\pi$$$$$$

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