If sin theta - cos theta = frac{1}{sqrt{3}},t | vedclass

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(D) Given $\sin 	heta - \cos 	heta = \frac{1}{\sqrt{3}}$.Squaring both sides: $(\sin 	heta - \cos 	heta)^2 = (\frac{1}{\sqrt{3}})^2$.$1 - \sin(2

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$\frac{43}{27}$

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(D) Given $\sin 	heta - \cos 	heta = \frac{1}{\sqrt{3}}$.Squaring both sides: $(\sin 	heta - \cos 	heta)^2 = (\frac{1}{\sqrt{3}})^2$.$1 - \sin(2	heta) = \frac{1}{3} \Rightarrow \sin(2	heta) = 1 - \frac{1}{3} = \frac{2}{3}$.Now,$\cos(4	heta) = 1 - 2\sin^2(2	heta) = 1 - 2(\frac{2}{3})^2 = 1 - 2(\frac{4}{9}) = 1 - \frac{8}{9} = \frac{1}{9}$.Next,$\sin(6	heta) = 3\sin(2	heta) - 4\sin^3(2	heta) = 3(\frac{2}{3}) - 4(\frac{2}{3})^3 = 2 - 4(\frac{8}{27}) = 2 - \frac{32}{27} = \frac{54 - 32}{27} = \frac{22}{27}$.Finally,$\sin(2	heta) + \cos(4	heta) + \sin(6	heta) = \frac{2}{3} + \frac{1}{9} + \frac{22}{27} = \frac{18 + 3 + 22}{27} = \frac{43}{27}$.

If $\sin \theta - \cos \theta = \frac{1}{\sqrt{3}}$,then $\sin(2\theta) + \cos(4\theta) + \sin(6\theta) = $

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