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(C) Given: $\sin 	heta_1+\sin 	heta_2=-\frac{21}{65}$ and $\cos 	heta_1+\cos 	heta_2=-\frac{27}{65}$.Squaring and adding both equations:$(\sin 	h

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$-\frac{3}{\sqrt{130}}$

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(C) Given: $\sin 	heta_1+\sin 	heta_2=-\frac{21}{65}$ and $\cos 	heta_1+\cos 	heta_2=-\frac{27}{65}$.Squaring and adding both equations:$(\sin 	heta_1+\sin 	heta_2)^2 + (\cos 	heta_1+\cos 	heta_2)^2 = \left(-\frac{21}{65}ight)^2 + \left(-\frac{27}{65}ight)^2$$(\sin^2 	heta_1 + \cos^2 	heta_1) + (\sin^2 	heta_2 + \cos^2 	heta_2) + 2(\cos 	heta_1 \cos 	heta_2 + \sin 	heta_1 \sin 	heta_2) = \frac{441+729}{4225}$$1 + 1 + 2 \cos(	heta_1 - 	heta_2) = \frac{1170}{4225}$$2(1 + \cos(	heta_1 - 	heta_2)) = \frac{1170}{4225} = \frac{18}{65}$$4 \cos^2 \left(\frac{	heta_1-	heta_2}{2}ight) = \frac{18}{65}$$\cos^2 \left(\frac{	heta_1-	heta_2}{2}ight) = \frac{18}{65 	imes 4} = \frac{9}{130}$$\cos \left(\frac{	heta_1-	heta_2}{2}ight) = \pm \frac{3}{\sqrt{130}}$.Since $(	heta_1-	heta_2)$ lies in the $3^{	ext{rd}}$ or $4^{	ext{th}}$ quadrant,the angle $\frac{	heta_1-	heta_2}{2}$ lies in the $2^{	ext{nd}}$ or $4^{	ext{th}}$ quadrant (or specifically,the range of $\frac{	heta_1-	heta_2}{2}$ is $(135^{\circ}, 180^{\circ})$ or $(270^{\circ}, 315^{\circ})$). However,given the sum of cosines is negative,the cosine of the half-angle must be negative.Thus,$\cos \left(\frac{	heta_1-	heta_2}{2}ight) = -\frac{3}{\sqrt{130}}$.

Suppose $\theta_1$ and $\theta_2$ are such that $(\theta_1-\theta_2)$ lies in the $3^{\text{rd}}$ or $4^{\text{th}}$ quadrant. If $\sin \theta_1+\sin \theta_2=-\frac{21}{65}$ and $\cos \theta_1+\cos \theta_2=-\frac{27}{65}$,then $\cos \left(\frac{\theta_1-\theta_2}{2}\right)=$

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