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Prove that: ^{-1} x = 2 ^{-1} {{1 -x}{2}} = 2 ^{-1} {{1 + x}{3

Question

Prove that:
$cos^{- 1} x = 2 sin^{- 1} \frac{1 - x}{2} = 2 cos^{- 1} \frac{1 + x}{3}$

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Prove that:

cos^{- 1} x = 2 sin^{- 1} \frac{1 - x}{2} = 2 cos^{- 1} \frac{1 + x}{3}

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Text solutionVerified

Let

sin^{- 1} \frac{1 - x}{2} = θ ⟹ sin θ = \frac{1 - x}{2}

As we know that

cos 2 A = 1 - 2 sin^{2} A

So

cos 2 θ = 1 - 2 sin^{2} θ = 1 - 2 \times (\frac{1 - x}{2})^{2} = 1 - 2 \times \frac{1 - x}{2} = 1 - (1 - x) = x

⟹ 2 θ = cos^{- 1} x

⟹ 2 sin^{- 1} \frac{1 - x}{2} = cos^{- 1} x \dots (1)

Let

cos^{- 1} \frac{1 + x}{2} = ϕ ⟹ cos ϕ = \frac{1 + x}{2}

As we know that

cos 2 A = 2 cos^{2} A - 1

So

cos 2 ϕ = 2 cos^{2} ϕ - 1 = 2 \times (\frac{1 + x}{2})^{2} - 1 = 2 \times \frac{1 + x}{2} - 1 = 1 + x - 1 = x

⟹ 2 ϕ = cos^{- 1} x

⟹ 2 cos^{- 1} \frac{1 + x}{2} = cos^{- 1} x \dots (2)

From

(1) & (2)

cos^{- 1} x = 2 sin^{- 1} \frac{1 - x}{2} = 2 cos^{- 1} \frac{1 + x}{2}

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Question Text	Prove that: $cos^{- 1} x = 2 sin^{- 1} \frac{1 - x}{2} = 2 cos^{- 1} \frac{1 + x}{3}$
Answer Type	Text solution:1
Upvotes	150