Solveeit Logo

Question

Question: How do you prove that \[\dfrac{{\cos 7x + \cos 5x}}{{\sin 7x - \sin 5x}} = \cot x\] ?...

How do you prove that cos7x+cos5xsin7xsin5x=cotx\dfrac{{\cos 7x + \cos 5x}}{{\sin 7x - \sin 5x}} = \cot x ?

Explanation

Solution

This question though looks tough but is super easy. What we will do is simply use the sum and difference formulas of trigonometry to solve the LHS. That is the best way to solve the problem above. We will first generalize the above equation and then put the values of the angles from the original equation.

Complete step by step answer:
Given that, LHS=cos7x+cos5xsin7xsin5xLHS = \dfrac{{\cos 7x + \cos 5x}}{{\sin 7x - \sin 5x}}
This is of the form, cosx+cosysinx+siny\dfrac{{\cos x + \cos y}}{{\sin x + \sin y}}
So we will solve this with the help of the formulas below,
cosx+cosy=2cos(x+y2).cos(xy2)\cos x + \cos y = 2\cos \left( {\dfrac{{x + y}}{2}} \right).\cos \left( {\dfrac{{x - y}}{2}} \right)
sinx+siny=2cos(x+y2).sin(xy2)\Rightarrow \sin x + \sin y = 2\cos \left( {\dfrac{{x + y}}{2}} \right).\sin \left( {\dfrac{{x - y}}{2}} \right)

Now taking the ratio we get,
cosx+cosysinx+siny=2cos(x+y2).cos(xy2)2cos(x+y2).sin(xy2)\dfrac{{\cos x + \cos y}}{{\sin x + \sin y}} = \dfrac{{2\cos \left( {\dfrac{{x + y}}{2}} \right).\cos \left( {\dfrac{{x - y}}{2}} \right)}}{{2\cos \left( {\dfrac{{x + y}}{2}} \right).\sin \left( {\dfrac{{x - y}}{2}} \right)}}
Since the initial cos functions are same we can cancel them,
cosx+cosysinx+siny=cos(xy2)sin(xy2)\dfrac{{\cos x + \cos y}}{{\sin x + \sin y}} = \dfrac{{\cos \left( {\dfrac{{x - y}}{2}} \right)}}{{\sin \left( {\dfrac{{x - y}}{2}} \right)}}
Remaining functions have same angle nad as we know ratio of cos to sin is cot function,
cosx+cosysinx+siny=cot(xy2)\dfrac{{\cos x + \cos y}}{{\sin x + \sin y}} = \cot \left( {\dfrac{{x - y}}{2}} \right)

Now replace x by 7x and y by 5x,
cos7x+cos5xsin7xsin5x=cot(7x5x2)\dfrac{{\cos 7x + \cos 5x}}{{\sin 7x - \sin 5x}} = \cot \left( {\dfrac{{7x - 5x}}{2}} \right)
On solving we get,
cos7x+cos5xsin7xsin5x=cot(2x2)\dfrac{{\cos 7x + \cos 5x}}{{\sin 7x - \sin 5x}} = \cot \left( {\dfrac{{2x}}{2}} \right)
cos7x+cos5xsin7xsin5x=cotx\Rightarrow \dfrac{{\cos 7x + \cos 5x}}{{\sin 7x - \sin 5x}} = \cot x
This equals LHS=RHSLHS = RHS
Hence proved.

Note: These problems are totally dependent on formulas, but even a small mistake while writing a formula can cause trouble throughout the problem. The cancellation was possible because the functions were the same. If the functions are same but angles are different then in that case this is not possible. We were able to take the ratio of cos and sin function only because their angles were the same. So do carefully check the functions and the angles.In general if a question is heading “ prove that” , writing “hence proved “ is the proper way of ending the solution or proof.