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

Prove the following expression cos⁡9x−cos⁡5xsin⁡17x−sin⁡3x=−sin⁡2xcos⁡10x\dfrac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}=\dfrac{-\sin 2x}{\cos 10x}

Explanation

Solution

Hint: Using trigonometric fórmula cos⁡x−cos⁡y=−2sin⁡x+y2sin⁡x−y2\cos x-\cos y=-2\sin \dfrac{x+y}{2}\sin \dfrac{x-y}{2} and sin⁡x−sin⁡y=2cos⁡x+y2sin⁡x−y2\sin x-\sin y=2\cos \dfrac{x+y}{2}\sin \dfrac{x-y}{2}. Take the LHS and Then we just need to simplify the expression. On simplifying we will get the RHS.

Complete step-by-step answer:
Mark the expression as (i) so we have cos⁡9x−cos⁡5xsin⁡17x−sin⁡3x=−sin⁡2xcos⁡10x...(i)\dfrac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}=\dfrac{-\sin 2x}{\cos 10x}...(i)
Trigonometry is a branch which gives access to distinct degrees and values of something. With the help of trigonometry, any angle between lines or angle of a triangle can be found out. There are direct trigonometric values such as sin 0=0, sin 30∘30^\circ = 0.5 and so on. Relevant is the case with cosine, tangent, secant, cosecant and cotangent.
With the help of trigonometry, relation between angles and sides of a triangle can be found out. Consider any side of the trigonometric expression according to ease.
Here consider left hand side of the expression (i)(i) which is cos⁡9x−cos⁡5xsin⁡17x−sin⁡3x\dfrac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}
Now apply the formula cos⁡x−cos⁡y=−2sin⁡x+y2sin⁡x−y2\cos x-\cos y=-2\sin \dfrac{x+y}{2}\sin \dfrac{x-y}{2} as a substitution in the numerator and sin⁡x−sin⁡y=2cos⁡x+y2sin⁡x−y2\sin x-\sin y=2\cos \dfrac{x+y}{2}\sin \dfrac{x-y}{2} as a substitution in the denominator.
This implies that the left hand side of expression (i) changes into,
cos⁡9x−cos⁡5xsin⁡17x−sin⁡3x=−2sin⁡(9x+5x2)sin⁡(9x−5x2)2cos⁡(17x+3x2)sin⁡(17x−3x2)\dfrac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}=\dfrac{-2\sin \left( \dfrac{9x+5x}{2} \right)\sin \left( \dfrac{9x-5x}{2} \right)}{2\cos \left( \dfrac{17x+3x}{2} \right)\sin \left( \dfrac{17x-3x}{2} \right)}
Now we further solve it to get the simplest terms. This implies that −2sin⁡(14x2)sin⁡(4x2)2cos⁡(20x2)sin⁡(14x2)=−2sin⁡7xsin⁡2x2cos⁡10xsin⁡7x\dfrac{-2\sin \left( \dfrac{14x}{2} \right)\sin \left( \dfrac{4x}{2} \right)}{2\cos \left( \dfrac{20x}{2} \right)\sin \left( \dfrac{14x}{2} \right)}=\dfrac{-2\sin 7x\sin 2x}{2\cos 10x\sin 7x}
So, till here it implies that the value of the expression is changed to cos⁡9x−cos⁡5xsin⁡17x−sin⁡3x=−2sin⁡7xsin⁡2x2cos⁡10xsin⁡7x\dfrac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}=\dfrac{-2\sin 7x\sin 2x}{2\cos 10x\sin 7x}
Now we cancel the numerator and denominator. This implies −2sin⁡7xsin⁡2x2cos⁡10xsin⁡7x=−sin⁡2xcos⁡10x\dfrac{-2\sin 7x\sin 2x}{2\cos 10x\sin 7x}=\dfrac{-\sin 2x}{\cos 10x}
This is equal to the right hand side of expression (i)(i)
Hence, cos⁡9x−cos⁡5xsin⁡17x−sin⁡3x=−sin⁡2xcos⁡10x\dfrac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}=\dfrac{-\sin 2x}{\cos 10x} is proved.

Note: Use of trigonometric formulas like cos⁡x−cos⁡y=−2sin⁡x+y2sin⁡x−y2\cos x-\cos y=-2\sin \dfrac{x+y}{2}\sin \dfrac{x-y}{2}and sin⁡x−sin⁡y=2cos⁡x+y2sin⁡x−y2\sin x-\sin y=2\cos \dfrac{x+y}{2}\sin \dfrac{x-y}{2}needs focus while applying. This is because if focus is lost in between solving the negative signs will not solve properly. It will imply a wrong answer. The best trick is to solve it step by step so that solution while solving is understood clearly.