Question

2sin⁻¹ 3/5 + cos⁻¹ 24/25 =π/2,Prove it.

Answer 1

LHS = 2sin^-1(3/5) + cos^-1(24/25)

Let 2sin^-1(3/5) = A
sin^-1(3/5) = A/2
sinA/2 = 3/5 => cosA/2 = 4/5
cosA = 2cos²A/2 - 1 [ from formula]
cosA = 2 × (4/5)² -1
cosA = 2 × 16/25 - 1 = (32 - 25)/25 = 7/25
hence, cos^-1(7/25) = A

therefore we can write 2sin^-1(3/5) = cos^-1(7/25)

now, LHS = 2sin^-1(3/5) + cos^-1(24/25) = cos^-1(7/25) + cos^-1(24/25)

we know, cos^-1x + cos^-1y = cos^-1[xy - √(1 - x^2)(√1 - y^2)]

so, cos^-1(7/25) + cos^-1(24/25) = cos^-1[7/25 × 24/25 - √(1 - 7²/25²) × √(1 - 24²/25²)]

= cos^-1[168/625 - √{(25² - 7²)/25²} × √{(24² - 25²)/25²}

= cos^-1[168/625 - 24/25 × 7/25 ]

= cos^-1[168/625 - 168/625]

= cos^-1 (0) = π/2 = RHS