Statement (S1): sin 55^{circ} + sin 53^{circ} | vedclass

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(D) $I$. Evaluate $(S1): \sin 55^{\circ} + \sin 53^{\circ} - \sin 19^{\circ} - \sin 17^{\circ}$Using $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \fra

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$(S1)$ is false,$(S2)$ is true

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(D) $I$. Evaluate $(S1): \sin 55^{\circ} + \sin 53^{\circ} - \sin 19^{\circ} - \sin 17^{\circ}$Using $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$:$= (\sin 55^{\circ} - \sin 17^{\circ}) + (\sin 53^{\circ} - \sin 19^{\circ})$$= 2 \cos 36^{\circ} \sin 19^{\circ} + 2 \cos 36^{\circ} \sin 17^{\circ}$$= 2 \cos 36^{\circ} (\sin 19^{\circ} + \sin 17^{\circ})$$= 2 \cos 36^{\circ} (2 \sin 18^{\circ} \cos 1^{\circ})$Since $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$ and $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$:$= 2 \left(\frac{\sqrt{5}+1}{4}ight) \cdot 2 \left(\frac{\sqrt{5}-1}{4}ight) \cos 1^{\circ}$$= 4 \left(\frac{5-1}{16}ight) \cos 1^{\circ} = \cos 1^{\circ}$.Thus,$(S1)$ is false as $\cos 1^{\circ} 
eq \cos 2^{\circ}$.$II$. Evaluate $(S2): f(x) = \frac{1}{3 - \cos 2x}$Since $-1 \leq \cos 2x \leq 1$,we have $-1 \leq -\cos 2x \leq 1$.Adding $3$: $2 \leq 3 - \cos 2x \leq 4$.Taking reciprocals: $\frac{1}{4} \leq \frac{1}{3 - \cos 2x} \leq \frac{1}{2}$.Thus,$(S2)$ is true.

Statement $(S1): \sin 55^{\circ} + \sin 53^{\circ} - \sin 19^{\circ} - \sin 17^{\circ} = \cos 2^{\circ}$.
Statement $(S2):$ The range of $\frac{1}{3 - \cos 2x}$ is $\left[\frac{1}{4}, \frac{1}{2}\right]$.
Which one of the following is correct?

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Statement $(S1): \sin 55^{\circ} + \sin 53^{\circ} - \sin 19^{\circ} - \sin 17^{\circ} = \cos 2^{\circ}$.Statement $(S2):$ The range of $\frac{1}{3 - \cos 2x}$ is $\left[\frac{1}{4}, \frac{1}{2}\right]$.Which one of the following is correct?