Question

cos(2 sin-1 x) = 1/9, x>0
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Answer 1

Note that arcsin = sin inverse = sin^(-1)

Answer:

2/3

Step-by-step explanation:

Let 2arcsin x = A

arcsin x = A/2

sin(A/2) = x

Therefore,

=> cos²(A/2) + sin²(A/2) = 1

=> cos²(A/2) = 1 - x²

thus, cosA = 2cos²(A/2) - 1 {formula}

=> cosA = 2(1 - x²) - 1

=> cosA = 1 - 2x²

Substitute this value in the question:

=> cos(A) = 1/9

=> 1 - 2x² = 1/9

=> 8/9 = 2x²

=> ± 2/3 = x

As x > 0, x = 2/3