Consider three points P = (-sin(beta - alpha) | vedclass

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(C) Let $P = (x_1, y_1) = (-\sin(\beta - \alpha), -\cos \beta)$ and $Q = (x_2, y_2) = (\cos(\beta - \alpha), \sin \beta)$.We observe that $R = (\cos(\

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$R$ lies on the line segment $QP$

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(C) Let $P = (x_1, y_1) = (-\sin(\beta - \alpha), -\cos \beta)$ and $Q = (x_2, y_2) = (\cos(\beta - \alpha), \sin \beta)$.We observe that $R = (\cos(\beta - \alpha + 	heta), \sin(\beta - 	heta))$.Using trigonometric identities,$\cos(\beta - \alpha + 	heta) = \cos(\beta - \alpha)\cos 	heta - \sin(\beta - \alpha)\sin 	heta = x_2 \cos 	heta + x_1 \sin 	heta$.Similarly,$\sin(\beta - 	heta) = \sin \beta \cos 	heta - \cos \beta \sin 	heta = y_2 \cos 	heta + y_1 \sin 	heta$.Thus,$R = (x_2 \cos 	heta + x_1 \sin 	heta, y_2 \cos 	heta + y_1 \sin 	heta)$.This shows that $R$ divides the line segment $PQ$ internally in the ratio $\sin 	heta : \cos 	heta$.Since $0 < 	heta < \frac{\pi}{4}$,both $\sin 	heta$ and $\cos 	heta$ are positive,so $R$ lies on the line segment $PQ$ (or $QP$).

Consider three points $P = (-\sin(\beta - \alpha), -\cos \beta)$,$Q = (\cos(\beta - \alpha), \sin \beta)$,and $R = (\cos(\beta - \alpha + \theta), \sin(\beta - \theta))$,where $0 < \alpha, \beta, \theta < \frac{\pi}{4}$. Then:

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