Sum

Show that `sin^-1(3/5) + sin^-1(8/17) = cos^-1(36/85)`

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#### Solution

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

∴ sin x = `3/5` and 0 < x < `pi/2`

∴ cos x > 0

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

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

= `sqrt(1 - 9/25)`

= `4/5`

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

∴ sin y = `8/17` and 0 < y < `pi/2`

∴ cos y > 0

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

= `sqrt(1 - (8/17)^2`

= `sqrt(1 - 64/289)`

= `15/17`

But cos(x + y) = cosx cosy – sinx siny

= `4/5(15/17) - 3/15(8/17)`

= `(60 - 24)/85`

= `36/85`

∴ x + y = `cos^-1(36/85)`

∴ `sin^-1(3/5) + sin^-1(8/17) = cos^-1(36/85)`

Concept: Inverse Trigonometric Functions

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