What is $cos (2 {sin}^{- 1} x)$ $cos ( 2 Sin^-1 x)$ ?

Question

30696 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-07T09:32:48

$\Rightarrow cos (2 {sin}^{- 1} x) = 1 - 2 x^{2}$ $=>cos(2sin^-1x)=1-2x^2$

We have to find , $cos (2 {sin}^{- 1} x)$ $cos(2sin^-1x)$

We know that,

$cos 2 θ = 1 - 2 {sin}^{2} θ$ $cos2theta=1-2sin^2theta$

$cos 2 θ = 1 - 2 {[sin θ]}^{2}$ $cos2theta=1-2[sintheta]^2$

We take, $θ = {sin}^{- 1} x$ $theta=sin^-1x$

$\Rightarrow cos (2 {sin}^{- 1} x) = 1 - 2 {[sin ({sin}^{- 1} x)]}^{2}$ $=>cos(2sin^-1x)=1-2[sin(sin^-1x)]^2$

$=>cos(2sin^-1x)=1-2(x)^2...to[becausesin(sin^-1theta)=theta]$

$=>cos(2sin^-1x)=1-2x^2$

What is cos ( 2 Sin^-1 x)cos(2sin−1x)cos ( 2 Sin^-1 x)?