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

RHS : Let y = sin^ 1 [3x 4x^3] Put x = sin θ⇒θ = sin^ 1x ...(i) ∴ y = sin^ 1[3sinθ 4sin^3 θ] = sin^ 1(sin3θ) As x ϵ [ 1/2 , 1/2 ] ⇒ 1/2≤ x ≤1/2 ⇒ 1 ...

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Byju's Answer

Standard XII

Mathematics

Solving Simultaneous Trigonometric Equations

Prove that : ...

Question

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

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Solution

RHS : Let $y = s i n^{- 1} [3 x - 4 x^{3}] Put x = s i n θ \Rightarrow θ = s i n^{- 1} x . . . (i) ∴ y = s i n^{- 1} [3 s i n θ - 4 s i n^{3} θ] = s i n^{- 1} (s i n 3 θ) As x ϵ [- \frac{1}{2}, \frac{1}{2}] \Rightarrow - \frac{1}{2} \leq x \leq \frac{1}{2} \Rightarrow - \frac{1}{2} \leq s i n θ \leq \frac{1}{2} \Rightarrow - \frac{π}{6} \leq θ \leq \frac{π}{6} - \frac{π}{2} \leq 3 θ \leq \frac{π}{2} - 1 \leq s i n 3 θ \leq 1 ∴ y = 3 θ = 3 s i n^{- 1} x = L H S By using (i)$

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Similar questions

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

Q. Prove that

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

ϵ [- \frac{1}{2}, \frac{1}{2}]

Prove the following,

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

Q. 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}]

Q. Prove that 2

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

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}]

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Prove that :3sin−1x=sin−1(3x−4x3),xϵ[−12,12].

RHS : Let y=sin−1[3x−4x3]Put x=sinθ⇒θ=sin−1x...(i)∴y=sin−1[3sinθ−4sin3θ]=sin−1(sin3θ)As x ϵ[−12,12] ⇒−12≤x≤12 ⇒−12≤sinθ≤12⇒−π6≤θ≤π6 −π2≤3θ≤π2 −1≤sin3θ≤1∴y=3θ=3sin−1x=LHS By using (i)

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