Prove that - 3 sin-1x - sin-13x-4x3- x - -1-2-1-2- -

Let us assume that x = sin θ⇒θ = sin^ 1x R.H.S. = sin^ 1(3x 4x^3) = sin^ 1(3 sin θ 4sin^3 θ) nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; ...

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Sum of Trigonometric Ratios in Terms of Their Product

Prove that : ...

Question

Prove that : $3 {sin}^{- 1} x = {sin}^{- 1} (3 x - 4 x^{3}), x ϵ [\begin{matrix} - \frac{1}{2}, \frac{1}{2} \end{matrix}]$

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

ϵ [- \frac{1}{2}, \frac{1}{2}]

s i n^{- 1} = s i n^{- 1} (3 x - 4 x^{3}), x ϵ [\frac{- 1}{2}, \frac{1}{2}]

Prove that : $3 s i n^{- 1} x = s i n^{- 1} (3 x - 4 x^{3}), x ϵ [- \frac{1}{2}, \frac{1}{2}] .$

Prove: $3 {sin}^{- 1} x = {sin}^{- 1} (3 x - {4 x}^{3}), x \in [- \frac{1}{2}, \frac{1}{2}]$

t a n^{- 1} \frac{1}{2} - t a n^{- 1} \frac{1}{7} = \frac{π}{4}

s i n^{- 1} x = s i n^{- 1} (3 x - 4 x^{3}), x ϵ [\frac{- 1}{2}, \frac{1}{2}]

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