Question

Prove the following expression $\dfrac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}=\dfrac{-\sin 2x}{\cos 10x}$

Answer 1

Hint: Using trigonometric fórmula $\cos x-\cos y=-2\sin \dfrac{x+y}{2}\sin \dfrac{x-y}{2}$ and $\sin x-\sin y=2\cos \dfrac{x+y}{2}\sin \dfrac{x-y}{2}$. 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 $\dfrac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}=\dfrac{-\sin 2x}{\cos 10x}...(i)$
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 $30^\circ$ = 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 $(i)$ which is $\dfrac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}$
Now apply the formula $\cos x-\cos y=-2\sin \dfrac{x+y}{2}\sin \dfrac{x-y}{2}$ as a substitution in the numerator and $\sin x-\sin y=2\cos \dfrac{x+y}{2}\sin \dfrac{x-y}{2}$ as a substitution in the denominator.
This implies that the left hand side of expression (i) changes into,
\[\dfrac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}=\dfrac{-2\sin \left( \dfrac{9x+5x}{2} \right)\sin \left( \dfrac{9x-5x}{2} \right)}{2\cos \left( \dfrac{17x+3x}{2} \right)\sin \left( \dfrac{17x-3x}{2} \right)}\]
Now we further solve it to get the simplest terms. This implies that \[\dfrac{-2\sin \left( \dfrac{14x}{2} \right)\sin \left( \dfrac{4x}{2} \right)}{2\cos \left( \dfrac{20x}{2} \right)\sin \left( \dfrac{14x}{2} \right)}=\dfrac{-2\sin 7x\sin 2x}{2\cos 10x\sin 7x}\]
So, till here it implies that the value of the expression is changed to \[\dfrac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}=\dfrac{-2\sin 7x\sin 2x}{2\cos 10x\sin 7x}\]
Now we cancel the numerator and denominator. This implies \[\dfrac{-2\sin 7x\sin 2x}{2\cos 10x\sin 7x}=\dfrac{-\sin 2x}{\cos 10x}\]
This is equal to the right hand side of expression $(i)$
Hence, $\dfrac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}=\dfrac{-\sin 2x}{\cos 10x}$ is proved.

Note: Use of trigonometric formulas like $\cos x-\cos y=-2\sin \dfrac{x+y}{2}\sin \dfrac{x-y}{2}$and $\sin x-\sin y=2\cos \dfrac{x+y}{2}\sin \dfrac{x-y}{2}$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.