SIN INVERSE OF X PLUS SIN INVERSE Y FORMULA

Sum and difference of inverse of sin cos and tan functions :

Problem 1 :

sin-1 35 + sin-1 817 = sin-1 7785

Solution :

sin-1 35 + sin-1 817 = sin-1 7785sin-1 x + sin-1 y = sin-1 x1 - y2 + y1 - x2sin-1 35 + sin-1 817 = sin-1 351 - 8172 + 8171 - 352= sin-1 351 - 82172 + 8171 - 3252= sin-1 351 - 64289 + 8171 - 925 = sin-1 35289 - 64289 + 817 25 - 925 = sin-1 35225289 + 817 1625 = sin-1 35 ⋅225289 + 817 ⋅ 1625 = sin-1 35 ⋅1517 + 817 ⋅ 45 = sin-1 917 + 3285 = sin-1 7785

Hence proved.

Problem 2 :

sin-1 45 + sin-1 513 + sin-1 1665 = 𝜋2

Solution :

sin-1 45 + sin-1 513 + sin-1 1665 = 𝜋2sin-1 45 + sin-1 513sin-1 x + sin-1 y = sin-1 x1 - y2 + y1 - x2= sin-1 451 - 5132 + 5131 - 452= sin-1 451 - 52132 + 5131 - 4252 = sin-1 451 - 25169 + 5131 - 1625 = sin-1 45169 - 25169 + 513 25 - 1625 = sin-1 45144169 + 513 925= sin-1 45 ⋅144169 + 513 ⋅ 925 = sin-1 45 ⋅1213 + 513 ⋅ 35= sin-1 4865 + 1565 = sin-1 6365= sin-1 6365 + sin-1 1665sin-1 x + sin-1 y = sin-1 x1 - y2 + y1 - x2= sin-1 63651 - 16652 + 16651 - 63652 = sin-1 63651 - 162652 + 16651 - 632652 = sin-1 63651 - 2564225 + 16651 - 39694225 = sin-1 63654225 - 2564225 + 1665 4225 - 39694225 = sin-1 636539694225 + 1665 2564225 = sin-1 6365 ⋅39694225 + 1665 ⋅ 2564225 = sin-1 6365 ⋅6365 + 1665 ⋅ 1665= sin-1 39694225 + 2564225 = sin-1 42254225= sin-1 [1]= 𝜋2

Hence proved.

Solve :

Problem 3 :

sin-1 3x5 + sin-1 4x5 = sin-1 x

Solution :

sin-1 3x5 + sin-1 4x5 = sin-1 xsin-1 x + sin-1 y = sin-1 x1 - y2 + y1 - x2 sin-1 3x51 - 4x52 + 4x51 - 3x52 = sin-1 x sin-1 3x51 - 42x252 + 4x51 - 32x252 = sin-1 x sin-1 3x51 - 16x225 + 4x51 - 9x225 = sin-1 x sin-1 3x525 - 16x225 + 4x5 25 - 9x225 = sin-1 x sin-1 3x5⋅ 25 - 16x25 + 4x5 ⋅ 25 - 9x25 = sin-1 x3x5⋅ 25 - 16x25 + 4x5 ⋅ 25 - 9x25 = xMultiplying 25 on both sides.3x ⋅25 - 16x2 + 4x ⋅25 - 9x2 = 25x 3x ⋅25 - 16x2 + 4x ⋅25 - 9x2 - 25x = 0 x325 - 16x2 + 425 - 9x2 - 25 =0 When x = 0325 - 16x2 + 425 - 9x2 - 25 = 0425 - 9x2 = 25 - 325 - 16x2Squaring on each sides.425 - 9x22 = 25 - 325 - 16x221625 - 9x2 = (25)2 + 3225 - 16x2 - 6(25)25 - 16x21625 - 9x2 = 625 + 925 - 16x2 - 15025 - 16x2 1625 - 9x2 - 625 - 925 - 16x2 + 15025 - 16x2 = 0400 - 144x2 - 625 - 225 + 144x2 + 15025 - 16x2 =0 400 - 625 - 225 + 15025 - 16x2 = 0-450 + 15025 - 16x2 = 0-450 = -15025 - 16x2-450-150 = 25 - 16x23 = 25 - 16x2Squaring on each sides.9 = 25 - 16x29 - 25 = -16x2-16 = -16x2x2 = 1x = ±1

Hence proved.

Problem 4 :

sin-1 (1 - x) - 2sin-1x = 𝜋2

Solution :

sin-1 (1 - x) - 2sin-1x = 𝜋2sin-1 (1 - x) = 𝜋2 + 2 sin-1x1 - x = sin𝜋2 + 2 sin-1x ∴ sin 𝜋2 + 𝜃 = cos 𝜃1 - x= cos 2 sin-1xLet sin-1x = 𝜃x = sin 𝜃1 - x = cos 2𝜃 1 - x = 1 - 2sin2𝜃1 - x = 1 - 2x22x2 - x = 0x(2x - 1) = 0x = 0 or 2x - 1 = 02x = 1x = 12

x = 0 or x = 1/2

Hence proved.

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