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Question

Question: Prove the following: \(\dfrac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}=-\dfrac{\sin 2x}{\cos 10x}\)...

Prove the following:
cos9xcos5xsin17xsin3x=sin2xcos10x\dfrac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}=-\dfrac{\sin 2x}{\cos 10x}

Explanation

Solution

Hint: For solving this question, we will simplify the term on the left-hand side and prove that it is equal to the term on the right-hand side. And we will use trigonometric formulas like cosCcosD=2sin(C+D2)sin(CD2)\cos C-\cos D=-2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right) and sinCsinD=2cos(C+D2)sin(CD2)\sin C-\sin D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right) for simplifying the term on the left-hand side. After that, we will easily prove the desired result.

Complete step-by-step answer:

Given:
We have to prove the following equation:
cos9xcos5xsin17xsin3x=sin2xcos10x\dfrac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}=-\dfrac{\sin 2x}{\cos 10x}
Now, we will simplify the term on the left-hand side and prove that it is equal to the term on the right-hand side.
Now, before we proceed we should know the following formulas:
cosCcosD=2sin(C+D2)sin(CD2)...................(1) sinCsinD=2cos(C+D2)sin(CD2)......................(2) \begin{aligned} & \cos C-\cos D=-2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)...................\left( 1 \right) \\\ & \sin C-\sin D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)......................\left( 2 \right) \\\ \end{aligned}
Now, we will use the above two formulas to simplify the term on the left-hand side.
On the left-hand side, we have cos9xcos5xsin17xsin3x\dfrac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x} .
Now, we will use the formula from the equation (1) to write cos9xcos5x=2sin7xsin2x\cos 9x-\cos 5x=-2\sin 7x\sin 2x in the term on the left-hand side. Then,
cos9xcos5xsin17xsin3x 2sin(9x+5x2)sin(9x5x2)sin17xsin3x 2sin7xsin2xsin17xsin3x \begin{aligned} & \dfrac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x} \\\ & \Rightarrow \dfrac{-2\sin \left( \dfrac{9x+5x}{2} \right)\sin \left( \dfrac{9x-5x}{2} \right)}{\sin 17x-\sin 3x} \\\ & \Rightarrow \dfrac{-2\sin 7x\sin 2x}{\sin 17x-\sin 3x} \\\ \end{aligned}
Now, we will use the formula from the equation (2) to write sin17xsin3x=2cos10xsin7x\sin 17x-\sin 3x=2\cos 10x\sin 7x in the above expression. Then,
2sin7xsin2xsin17xsin3x 2sin7xsin2x2cos(17x+3x2)sin(17x3x2) 2sin7xsin2x2cos10xsin7x sin7xsin2xsin7xcos10x sin2xcos10x \begin{aligned} & \dfrac{-2\sin 7x\sin 2x}{\sin 17x-\sin 3x} \\\ & \Rightarrow \dfrac{-2\sin 7x\sin 2x}{2\cos \left( \dfrac{17x+3x}{2} \right)\sin \left( \dfrac{17x-3x}{2} \right)} \\\ & \Rightarrow \dfrac{-2\sin 7x\sin 2x}{2\cos 10x\sin 7x} \\\ & \Rightarrow \dfrac{-\sin 7x\sin 2x}{\sin 7x\cos 10x} \\\ & \Rightarrow -\dfrac{\sin 2x}{\cos 10x} \\\ \end{aligned}
Now, from the above result, we conclude that the value of the expression cos9xcos5xsin17xsin3x\dfrac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x} will be equal to the value of the expression sin2xcos10x-\dfrac{\sin 2x}{\cos 10x} . Then,
cos9xcos5xsin17xsin3x=sin2xcos10x\dfrac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}=-\dfrac{\sin 2x}{\cos 10x}
Now, from the above result, we conclude that the term on the left-hand side is equal to the term on the right-hand side.
Thus, cos9xcos5xsin17xsin3x=sin2xcos10x\dfrac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}=-\dfrac{\sin 2x}{\cos 10x} .
Hence, proved.

Note: Here, the student should first understand what we have to prove in the question. After that, we should proceed in a stepwise manner and apply trigonometric formulas like cosCcosD=2sin(C+D2)sin(CD2)\cos C-\cos D=-2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right) and sinCsinD=2cos(C+D2)sin(CD2)\sin C-\sin D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right) correctly. Moreover, while simplifying we should be aware of the result and avoid calculation mistakes while solving.