How do you find the value for $cos [2 {sin}^{- 1} (\frac{1}{2})]$ $cos[2 sin^ -1 (1/2)]$ ?

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Answer 1 · 2018-03-27T12:01:13

$cos [2 {sin}^{- 1} (\frac{1}{2})] = \frac{1}{2}$ $cos [2 sin^-1(1/2)]=1/2$

$cos [2 {sin}^{- 1} (\frac{1}{2})]$ $cos [2 sin^-1(1/2)]$ . Let ${sin}^{- 1} (\frac{1}{2}) = θ$ $sin^-1(1/2)=theta$

$:. sin theta=1/2 ; sin 30= 1/2 :. theta=30^0$

$:.cos [2 sin^-1(1/2)]= cos [2*theta]= cos (2*30)= cos 60=1/2$

$:.cos [2 sin^-1(1/2)]=1/2$ [Ans]

How do you find the value for cos[2 sin^ -1 (1/2)]cos[2sin−1(12)]cos[2 sin^ -1 (1/2)]?