જો sin theta - cos theta = frac{1}{sqrt{3}} હ | vedclass

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Question

(D) આપેલ છે કે $\sin 	heta - \cos 	heta = \frac{1}{\sqrt{3}}$.બંને બાજુ વર્ગ કરતા: $(\sin 	heta - \cos 	heta)^2 = (\frac{1}{\sqrt{3}})^2$.$1 - \si

Vedclass · Accepted Answer

$\frac{43}{27}$

Vedclass · Answer

(D) આપેલ છે કે $\sin 	heta - \cos 	heta = \frac{1}{\sqrt{3}}$.બંને બાજુ વર્ગ કરતા: $(\sin 	heta - \cos 	heta)^2 = (\frac{1}{\sqrt{3}})^2$.$1 - \sin(2	heta) = \frac{1}{3} \Rightarrow \sin(2	heta) = 1 - \frac{1}{3} = \frac{2}{3}$.હવે,$\cos(4	heta) = 1 - 2\sin^2(2	heta) = 1 - 2(\frac{2}{3})^2 = 1 - 2(\frac{4}{9}) = 1 - \frac{8}{9} = \frac{1}{9}$.આગળ,$\sin(6	heta) = 3\sin(2	heta) - 4\sin^3(2	heta) = 3(\frac{2}{3}) - 4(\frac{2}{3})^3 = 2 - 4(\frac{8}{27}) = 2 - \frac{32}{27} = \frac{54 - 32}{27} = \frac{22}{27}$.અંતે,$\sin(2	heta) + \cos(4	heta) + \sin(6	heta) = \frac{2}{3} + \frac{1}{9} + \frac{22}{27} = \frac{18 + 3 + 22}{27} = \frac{43}{27}$.

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

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