Question

$3 {sin}^{ - 1}x = {sin}^{ - 1} (3x - 4 {x}^{3} )$
Prove the following.
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Answer 1

Step-by-step explanation:

$3 {sin}^{ - 1}x = {sin}^{ - 1} (3x - 4 {x}^{3}$

Answer 2

Step-by-step explanation:

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$3 {sin}^{ - 1}x = {sin}^{ - 1} (3x - 4 {x}^{3} )$

Put x= sin Q

Q = sin inverse x - - - - - - - - 1

The equation becomes,

Sin inverse ( 3sinQ - 4 sin cube Q)

As 3sinQ - 4SincubeQ = Sin3Q then,

Sin inverse(sin3Q) = 3Q = 3sin inverse

x(from 1)

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