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sqrt(2)cosA=cosB+cos^3B, sqrt(2)sinA=sin...

`sqrt(2)cosA=cosB+cos^3B`, `sqrt(2)sinA=sinB-sin^3B` , find `sin(A-B)`

Text Solution

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`sqrt2cos A = cosB + cos^3B`
`=>sqrt2cos A = cosB(1 + cos^2B)->(1)`
`sqrt2sinA = sinB - sin^3B`
`=>sqrt2sinA = sinB(1 - sin^2B)`
`=>sqrt2sinA = sinBcos^2B`
`=>cos^2B = sqrt2sinA /sinB`
Putting value of `cos^2B` in (1),
`=>sqrt2cos A = cosB(1 + (sqrt2sinA) /sinB) `
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