Question

sin⁻¹(2x√1-x²)= 2sin⁻¹x, | x | < 1/√2,prove it

Answer 1

we have to prove that,

sin⁻¹(2x√1-x²)= 2sin⁻¹x , where |x| < 1/√2

2sin^-1x = A ....(1)

⇒sin^-1x = A/2

⇒sin(A/2) = x/1 = p/h

from Pythagoras theorem, b = √(h² - p²)

= √(1² - x²) = √(1 - x²)

then, cos(A/2) = b/h = √(1 - x²)/1 = √(1 - x²)

we know, sin2θ = 2sinθ.cosθ

so, sinA = 2sin(A/2).cos(A/2)

= 2 × x × √(1 - x²)

= 2x√(1 - x²)

hence, sinA = 2x√(1 - x²)

⇒ A = sin^-1{2x√(1 - x²)} ......(2)

from equations (1) and (2),

sin^-1{2x√(1 - x²)} = 2sin^-1x

Answer 2

Step-by-step explanation:

To prove :

sin⁻¹(2x√1-x²)= 2sin⁻¹x, | x | < 1/√2

Let

$x = sin \: a \\ \\or\\\\a={sin}^{ - 1}x$

put this in both side of equation

${sin}^{ - 1} (2 \: sin \: a \sqrt{1 - {sin}^{2}a } ) = 2 {sin}^{ - 1} (sin \: a) \\ \\ {sin}^{ - 1} (2 \: sin \: a \: \sqrt{ {cos}^{2}a } ) = 2 {sin}^{ - 1} (sin \: a) \\ \\ {sin}^{ - 1} (2 \: sin \: a \:cos \: a) = 2 {sin}^{ - 1} (sin \: a) \\ \\ {sin}^{ - 1} (sin \: 2a) = 2 {sin}^{ - 1} (sin \: a) \\ \\ because \: {sin}^{ - 1}(sin \: x) = x \: if \: x \: \epsilon \: [- 1, \: 1] \\ \\ 2a = 2a \\ \\2{sin}^{ - 1}x =2{sin}^{ - 1}x\\\\ hence \: proved$

Hope it helps you