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प्रश्न
If 2cos α = `x + 1/x` and 2 cos β = `y + 1/x`, show that `x^"m"/y^"n" - y^"n"/x^"m"` = 2i sin(mα – nβ)
उत्तर
`x^"m"/y^"n" = (cps alpha + "i" sin alpha)^"m"/(cos beta + "i" sin beta)^"n"`
By de Moivre's theorem
= `(cos "m"alpha + "i" sin "m"alpha)/(cos "n"beta + "i" sin "n"beta)`
= `cos("m"alpha - "n"beta) - "i" sin("m"alpha - "n"beta)`
`y^"n"/x^"m" = cos("m"alpha - "n"beta) - "i"sin("m"alpha - "n"beta)`
∴ `x^"m"/y^"n" - y^"n"/x^"m" = cos("m"alpha - "n"beta) + "i" sin("m"alpha - "n"beta)`
`- cos("m"alpha - "n"beta) + "i" sin("m"alpha - "n"beta) = 2"i" sin("m"alpha - "n"beta)`
Hence proved.
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