solve the following simultaneous equations: sin-1x+sin -1y=2pi/3, cos-1x+cos-1y=pi/2
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Let Sin⁻¹ x = A, Sin⁻¹ y = B
Sin A = x Sin B = y
Cos A = √(1-x²) Cos B = √(1-y²)
Cos⁻¹ x = C
=> Cos C = x = Sin A = Cos (90-A)
=> C = 90 - A or 90 + A
Cos⁻¹ y = D
=> Cos D = y = SIn B
=> D = 90 + B or 90 - B
Now given equations are,
A + B = 2π/3 -- equation 1
C+D = π/2 --- equation 2
1) C+D= (90 + A) + (90 + B) = π/2
A+B = -π/2 not possible
2) C+D = (90+A)+(90-B) = π/2
A-B = -π/2 -- equation 3
Hence, from equation 1 and 3, A = 15 deg. B = 105 deg
3) C+D = (90-A) + (90+B) = π/2
B-A = -π/2 , then B =15 deg, A = 105 deg
4) C+D = 90-A+90-B = π/2
A+B = π/2 this is not possible
x = Sin A = Sin 15 deg and y = Sin 105 deg or Cos 15 deg
or, x = Cos 15 and y = Sin 15
Sin A = x Sin B = y
Cos A = √(1-x²) Cos B = √(1-y²)
Cos⁻¹ x = C
=> Cos C = x = Sin A = Cos (90-A)
=> C = 90 - A or 90 + A
Cos⁻¹ y = D
=> Cos D = y = SIn B
=> D = 90 + B or 90 - B
Now given equations are,
A + B = 2π/3 -- equation 1
C+D = π/2 --- equation 2
1) C+D= (90 + A) + (90 + B) = π/2
A+B = -π/2 not possible
2) C+D = (90+A)+(90-B) = π/2
A-B = -π/2 -- equation 3
Hence, from equation 1 and 3, A = 15 deg. B = 105 deg
3) C+D = (90-A) + (90+B) = π/2
B-A = -π/2 , then B =15 deg, A = 105 deg
4) C+D = 90-A+90-B = π/2
A+B = π/2 this is not possible
x = Sin A = Sin 15 deg and y = Sin 105 deg or Cos 15 deg
or, x = Cos 15 and y = Sin 15
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