Question

Show that `sin^-1 (- 3/5) - sin^-1 (- 8/17) = cos^-1 (84/85)`

Answer 1

`sin^-1 (- 3/5) - sin^-1 (- 8/17)`

= `- sin^-1 (3/5) + sin^-1 (8/17)`

= `sin^-1 (8/17) - sin^-1 (3/5)`

AB = `sqrt(17^2 - 8^2) = sqrt225` = 15

Let `sin^-1 (8/17)` = A

`8/17` = sin A

sin A = `8/17`

∴ cos A = `15/17`

Let `sin^-1 (3/5)` = B

sin B = `3/5`

∴ cos B = `4/5`

Consider cos(A – B) = cos A cos B + sin A sin B

`= 15/17 xx 4/5 + 8/17 xx 3/5`

`= 60/85 + 24/85`

cos (A – B) = `84/85`

∴ A – B = `cos^-1 (84/85)`

i.e., `sin^-1 (8/17) - sin^-1 (3/5) = cos^-1 (84/85)`

i.e., `sin^-1 (-3/5) - sin^-1 (-8/17) = cos^-1 (84/85)`