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प्रश्न
Evaluate:
`sin[cos^-1 (3/5)]`
उत्तर
Let x = `cos^-1 (3/5)` .......(i)
∴ cos x = `3/5`
∴ sin x = `sqrt(1 - cos^2x)`
= `sqrt(1 - 9/25)`
= `sqrt(16/25)`
= `4/5`
⇒ x = `sin^-1 (4/5)` .......(ii)
From (i) and (ii), we get
`sin [cos^-1 (3/5)] = sin[sin^-1 (4/5)]`
= `4/5`
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