If cos theta - sin theta = sqrt{5} sin theta, | vedclass

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(A) Given: $\cos 	heta - \sin 	heta = \sqrt{5} \sin 	heta$ $\cos 	heta = (\sqrt{5} + 1) \sin 	heta$ $	an 	heta = \frac{1}{\sqrt{5} + 1} = \frac

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$\sqrt{5} \cos 	heta$

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(A) Given: $\cos 	heta - \sin 	heta = \sqrt{5} \sin 	heta$ $\cos 	heta = (\sqrt{5} + 1) \sin 	heta$ $	an 	heta = \frac{1}{\sqrt{5} + 1} = \frac{\sqrt{5} - 1}{4}$ Let $x = \cos 	heta + \sin 	heta$. $x^2 = \cos^2 	heta + \sin^2 	heta + 2 \sin 	heta \cos 	heta = 1 + 2 \sin 	heta \cos 	heta$ Since $\cos 	heta = (\sqrt{5} + 1) \sin 	heta$,we have $\sin 	heta = \frac{\cos 	heta}{\sqrt{5} + 1}$. Using $\cos^2 	heta + \sin^2 	heta = 1$: $\cos^2 	heta + \frac{\cos^2 	heta}{(\sqrt{5} + 1)^2} = 1$ $\cos^2 	heta [1 + \frac{1}{5 + 1 + 2\sqrt{5}}] = 1$ $\cos^2 	heta [\frac{6 + 2\sqrt{5} + 1}{6 + 2\sqrt{5}}] = 1$ $\cos^2 	heta = \frac{6 + 2\sqrt{5}}{7 + 2\sqrt{5}} = \frac{(6 + 2\sqrt{5})(7 - 2\sqrt{5})}{49 - 20} = \frac{42 - 12\sqrt{5} + 14\sqrt{5} - 20}{29} = \frac{22 + 2\sqrt{5}}{29}$ Alternatively,from $\cos 	heta - \sin 	heta = \sqrt{5} \sin 	heta$,we get $\cos 	heta = (\sqrt{5} + 1) \sin 	heta$. Then $\cos 	heta + \sin 	heta = (\sqrt{5} + 1) \sin 	heta + \sin 	heta = (\sqrt{5} + 2) \sin 	heta$. Given the options provided in the original prompt,the intended expression was likely $\cos 	heta + \sin 	heta = \sqrt{5} \cos 	heta$.

If $\cos \theta - \sin \theta = \sqrt{5} \sin \theta$,then $\cos \theta + \sin \theta = $

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