If sin theta = frac{1}{2} (x + frac{1}{x}),th | vedclass

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(A) Given $\sin 	heta = \frac{1}{2} (x + \frac{1}{x})$.Since the range of $\sin 	heta$ is $[-1, 1]$,and for any real $x 
eq 0$,$|x + \frac{1}{x}| \

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(A) Given $\sin 	heta = \frac{1}{2} (x + \frac{1}{x})$.Since the range of $\sin 	heta$ is $[-1, 1]$,and for any real $x 
eq 0$,$|x + \frac{1}{x}| \geq 2$,we have $|\sin 	heta| = \frac{1}{2} |x + \frac{1}{x}| \geq 1$.Thus,$\sin 	heta$ can only be $1$ or $-1$.If $\sin 	heta = 1$,then $x + \frac{1}{x} = 2$,which implies $x = 1$.Then $\sin 3 	heta = \sin(3 	imes 90^\circ) = \sin 270^\circ = -1$.Also,$\frac{1}{2} (x^3 + \frac{1}{x^3}) = \frac{1}{2} (1^3 + \frac{1}{1^3}) = \frac{1}{2} (2) = 1$.Therefore,$\sin 3 	heta + \frac{1}{2} (x^3 + \frac{1}{x^3}) = -1 + 1 = 0$.If $\sin 	heta = -1$,then $x + \frac{1}{x} = -2$,which implies $x = -1$.Then $\sin 3 	heta = \sin(3 	imes 270^\circ) = \sin 810^\circ = \sin 90^\circ = 1$.Also,$\frac{1}{2} (x^3 + \frac{1}{x^3}) = \frac{1}{2} ((-1)^3 + \frac{1}{(-1)^3}) = \frac{1}{2} (-2) = -1$.Therefore,$\sin 3 	heta + \frac{1}{2} (x^3 + \frac{1}{x^3}) = 1 - 1 = 0$.

If $\sin \theta = \frac{1}{2} (x + \frac{1}{x})$,then $\sin 3 \theta + \frac{1}{2} (x^3 + \frac{1}{x^3}) = $

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