Question

Prove the following:

`sin^-1(3/5) + cos^-1(12/13) = sin^-1(56/65)`

Answer 1

Let `sin^-1(3/5) = x, cos^-1(12/13) = y and sin^-1(56/65)` = z.

Then `sin x = (3)/(5), "where"  0 < x < pi/(2)`

cos y = `(12)/(13), "where"  0 < y < pi/(2)`

and sin z = `(56)/(65), "where"  0 < z < pi/(2)`

∴ cos x > 0, sin y > 0

 

Finding sin x, cos x

Now, cos x = `sqrt(1 - sin^2 x)`

= `sqrt(1 - ((3)/(5))^2 `

= `sqrt(1 - (9)/(25)`  = `sqrt(16/25) = (4)/(5)`

 

Finding sin y, cos y

sin y = `sqrt(1 - cos^2y)`

= `sqrt(1 - ((12)/(13))^2`

= `sqrt(1 - (144)/(169)`  = `sqrt(25/169) = (5)/(13)`

 

We know that
`sin(x + y) = sin x  cos y + cos x  sin y`

= `(3/5)  "x"  (12/13) + (4/5)  "x"  (5/13)`

= `(36)/(65) + (20)/(65) = (56)/(65)`

 

`∴ sin(x + y) = (56)/(65)`

`∴ x + y = sin^-1 (56)/(65)`

Hence, `sin^-1(3/5) + cos^-1(12/13) = sin^-1(56/65)`.