Consider the left hand side of the given expression: $\text{LHS}=\cos\text{x}\cos\frac{\text{x}}{2}-\cos{3\text{x}}\cos\frac{9\text{x}}{2}$ We know that $2\cos\text{A}\cos\text{B}=\cos\text{A+B}+\cos(\text{A}-\text{B})$ Thus, $\text{LHS}=\frac{1}{2}\Big[\cos\Big(\text{x}+\frac{\text{x}}{2}\Big)+\cos\Big(\text{x}-\frac{\text{x}}{2}\Big)-\frac{1}{2}\Big[\cos\Big(3\text{x}+\frac{9\text{x}}{2}\Big)+\cos\Big(3\text{x}-\frac{9\text{x}}{2}\Big)\Big]$ $=\ \frac{1}{2}\Big[\cos\Big(\frac{3\text{x}}{2}\Big)+\cos\Big(\frac{\text{x}}{2}\Big)\Big]-\frac{1}{2}\Big[\cos\Big(\frac{15\text{x}}{2}\Big)+\cos\Big(-\frac{3\text{x}}{2}\Big)\Big]$ $=\ \frac{1}{2}\Big[\cos\Big(\frac{3\text{x}}{2}\Big)-\cos\Big(\frac{\text{x}}{2}\Big)\Big]-\frac{1}{2}\Big[\cos\Big(\frac{15\text{x}}{2}\Big)+\cos\Big(-\frac{3\text{x}}{2}\Big)\Big]$$[\because\ \cos(-\theta)=\cos\theta]$ $=\ \frac{1}{2}\Big[\cos\Big(\frac{3\text{x}}{2}\Big)+\cos\Big(\frac{\text{x}}{2}\Big)-\cos\Big(\frac{15\text{x}}{2}\Big)-\cos\Big(\frac{3\text{x}}{2}\Big)\Big]$ $=\ \frac{1}{2}\Big[\cos\Big(\frac{\text{x}}{2}\Big)-\cos\Big(\frac{15\text{x}}{2}\Big)\Big]$ Also we know that, $\cos\text{D}-\cos\text{C}=2\sin\frac{\text{C+D}}{2}\sin\frac{\text{C}-\text{D}}{2}$ Therefore, $\text{LHS}=\frac{1}{2}\times2\sin\frac{\frac{15\text{x}}{2}+\frac{\text{x}}{2}}{2}\sin\frac{\frac{15\text{x}}{2}-\frac{\text{x}}{2}}{2}$ $=\ \sin\frac{\frac{16\text{x}}{2}}{2}\sin\frac{\frac{14\text{x}}{2}}{2}$ $=\ \sin\frac{8\text{x}}{2}\sin\frac{7\text{x}}{2}$ $=\ \text{RHS}$ Note: Question given in the book is incorrect. RHS should be equal to $\sin\frac{8\text{x}}{2}\sin\frac{7\text{x}}{2}.$
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