Question

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?

• A. Both (S1) and (S2) are true
• B. Both (S1) and (S2) are false
• C. (S1) is true, (S2) is false
• D. (S1) is false, (S2) is true

Hint:

For proving trigonometric identities, simplify both sides and compare. For range problems, use the minimum and maximum values of the trigonometric functions involved.

Answer 1

The Correct Option is D

We need to analyze two statements: \[ \text{(S1): } \sin 55^\circ + \sin 53^\circ - \sin 19^\circ - \sin 17^\circ = \cos 2^\circ \] \[ \text{(S2): } \text{The range of } \frac{1}{3 - \cos 2x} \text{ is } \left[\frac{1}{4}, \frac{1}{2} \right] \] 

Step 1: Verifying Statement (S1) 
We use the sine addition-subtraction identities: \[ \sin A + \sin B = 2 \sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right) \] Applying this identity, \[ \sin 55^\circ + \sin 53^\circ = 2 \sin \left(\frac{55^\circ + 53^\circ}{2} \right) \cos \left(\frac{55^\circ - 53^\circ}{2} \right) = 2 \sin 54^\circ \cos 1^\circ \] \[ \sin 19^\circ + \sin 17^\circ = 2 \sin \left(\frac{19^\circ + 17^\circ}{2} \right) \cos \left(\frac{19^\circ - 17^\circ}{2} \right) = 2 \sin 18^\circ \cos 1^\circ \] Now, \[ \sin 55^\circ + \sin 53^\circ - \sin 19^\circ - \sin 17^\circ = 2 \cos 1^\circ (\sin 54^\circ - \sin 18^\circ) \] Since \( \sin 54^\circ \approx 0.809 \) and \( \sin 18^\circ \approx 0.309 \), \[ \sin 54^\circ - \sin 18^\circ = 0.809 - 0.309 = 0.5 \] Thus, \[ \text{LHS} = 2 \cos 1^\circ \times 0.5 = \cos 1^\circ \approx 0.999 \] Since \( \cos 2^\circ \approx 0.999 \), the two sides are close but **not exactly equal**. 

Conclusion: (S1) is False. Step 2: Verifying Statement (S2) 
Given, \[ f(x) = \frac{1}{3 - \cos 2x} \] Since \( \cos 2x \in [-1, 1] \), - Maximum value of \( 3 - \cos 2x = 3 - (-1) = 4 \)
- Minimum value of \( 3 - \cos 2x = 3 - 1 = 2 \)
Thus, \[ f(x) = \frac{1}{3 - \cos 2x} \in \left[\frac{1}{4}, \frac{1}{2} \right] \] 

Conclusion: (S2) is True. Final Answer: (D) (S1) is false, (S2) is true.