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Question
If y = `sin^-1[("a"cosx - "b"sinx)/sqrt("a"^2 + "b"^2)]`, then find `("d"y)/("d"x)`
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Solution
y = `sin^-1[("a"cosx - "b"sinx)/sqrt("a"^2 + "b"^2)]`
= `sin^-1["a"/sqrt("a"^2 + "b"^2) cosx - "b"/sqrt("a"^2 + "b"^2) sinx]`
Put `"a"/sqrt("a"^2 + "b"^2)` = sin t and `"b"/sqrt("a"^2 + "b"^2)` = cos t
Also, sin2t + cos2t = `("a"^2)/("a"^2 + "b"^2) + ("b"^2)/("a"^2 + "b"^2)` = 1
and tan t = `"a"/"b"`
∴ t = `tan^-1("a"/"b")`
∴ y = sin–1(sin t cos x – cos t sin x)
= sin–1[sin(t – x)]
= t – x
= `tan^-1("a"/"b") - x`
Differentiating w.r.t. x, we get
`("d"y)/("d"x) = "d"/("d"x)[tan^-1("a"/"b") - x]`
= 0 – 1
= –1
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