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`sqrt(2)cosA=cosB+cos^3B`, `sqrt(2)sinA=sinB-sin^3B` , find `sin(A-B)` |
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Answer» `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) ` `=>sqrt2cosAsinB = sinBcosB+sqrt2sinAcosB` `=>sqrt2(sinAcosB-cosAsinB) = -sinBcosB` `=>sqrt2(sin(A-B)) = (-sin2B)/2` `=>sin(A-B) = (-sin2B)/(2sqrt2).` |
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