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11TH · MATH

Question 6

menu_bookSolutionmenu_bookTheorymenu_book•(i)sin⁡2(α+β2)−sin⁡2(α−β2)=sin⁡2αsin⁡β\sin^2\left(\alpha + \dfrac{\beta}{2}\right) - \sin^2\left(\alpha - \dfrac{\beta}{2}\right) = \sin 2\alpha \sin\betasin2(α+2β​)−sin2(α−2β​)=sin2αsinβmenu_book•(ii)sin⁡2α+sin⁡2β+cos⁡2(α+β)+2sin⁡αsin⁡βcos⁡(α+β)=1\sin^2\alpha + \sin^2\beta + \cos^2(\alpha + \beta) + 2\sin\alpha\sin\beta\cos(\alpha + \beta) = 1sin2α+sin2β+cos2(α+β)+2sinαsinβcos(α+β)=1
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11TH · MATH

Question 6

menu_bookSolutionmenu_bookTheorymenu_book•(i)sin⁡2(α+β2)−sin⁡2(α−β2)=sin⁡2αsin⁡β\sin^2\left(\alpha + \dfrac{\beta}{2}\right) - \sin^2\left(\alpha - \dfrac{\beta}{2}\right) = \sin 2\alpha \sin\betasin2(α+2β​)−sin2(α−2β​)=sin2αsinβmenu_book•(ii)sin⁡2α+sin⁡2β+cos⁡2(α+β)+2sin⁡αsin⁡βcos⁡(α+β)=1\sin^2\alpha + \sin^2\beta + \cos^2(\alpha + \beta) + 2\sin\alpha\sin\beta\cos(\alpha + \beta) = 1sin2α+sin2β+cos2(α+β)+2sinαsinβcos(α+β)=1
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