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