If sin theta_1 + sin theta_2 + sin theta_3 = | vedclass

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(D) Given that $\sin 	heta_1 + \sin 	heta_2 + \sin 	heta_3 = 3$.Since the maximum value of the $\sin$ function is $1$,the only way for the sum of t

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(D) Given that $\sin 	heta_1 + \sin 	heta_2 + \sin 	heta_3 = 3$.Since the maximum value of the $\sin$ function is $1$,the only way for the sum of three $\sin$ values to be $3$ is if each individual term is equal to $1$.Therefore,$\sin 	heta_1 = 1, \sin 	heta_2 = 1,$ and $\sin 	heta_3 = 1$.This implies that $	heta_1 = 	heta_2 = 	heta_3 = \frac{\pi}{2}$.Now,we calculate the sum of the cosines: $\cos 	heta_1 + \cos 	heta_2 + \cos 	heta_3$.Since $\cos(\frac{\pi}{2}) = 0$,we have $\cos 	heta_1 = 0, \cos 	heta_2 = 0,$ and $\cos 	heta_3 = 0$.Thus,$0 + 0 + 0 = 0$.

If $\sin \theta_1 + \sin \theta_2 + \sin \theta_3 = 3,$ then $\cos \theta_1 + \cos \theta_2 + \cos \theta_3 =$

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