# Solve the following simultaneous equations: sin-1x+sin -1y=2pi/3, cos-1x+cos-1y=pi/2

1
Log in to add a comment

Log in to add a comment

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

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