Question

Prove that cos 4x+cos 3x+cos 2x/sin 4x+sin 3x+sin 2x=cot 3x

Answer 1

Answer:

Step-by-step explanation:

Taking L.H.S and solving numerator and denominator separately:

Solving Numerator,

We know that

Cosx + Cosy = 2 Cos(x+y/2).Cos(x-y/2)

Cos4x + Cos2x = 2 Cos 6x/2.Cos2x/2 = 2Cos3xCosx.

Now Cos 4x+ Cos 3x + Cos 2x

         =  2Cos3xCosx + Cos3x

         = 2Cos3x(Cosx + 1)

Solving Denominator,

We know that

Sinx + Siny = 2 Sin(x+y/2).Cos(x-y/2)

Sin4x + Sin2x = 2 Sin6x/2.Cos2x/2 = 2Sin3xCosx.

Now Sin 4x+ Sin 3x + Sin 2x

         =  2Sin3xCosx + Sin3x

         = 2Sin3x(Cosx + 1)

Now substitute in Numerator/Denominator

cos 4x+cos 3x + cos 2x / sin 4x+sin 3x+sin 2x

= 2Cos3x (Cosx + 1) / 2Sin3x (Cosx + 1)

= Cot3x

= R.H.S

Hence Proved.

Answer 2

AnswEr:

$\\ \sf \: LHS = \frac{cos4x \: + \: cos \: 3x \: + \: cos \: 2x}{sin \: 4x \: + sin \: 3x \: + sin \: 2x} \\ \\ \\ \implies \sf \: LHS = \frac{(cos \: 4x \: + \: cos \: 2x) + cos \: 3x}{(sin \: 4x + sin \: 2x) + sin \: 3x} \\ \\ \\ \implies \sf \: LHS = \frac{2 \: cos \: ( \frac{4x + 2x}{2}) \: cos \: ( \frac{4x - 2x}{2} ) + cos \: 3x}{2 \: sin \: ( \frac{4x + 2x}{2}) \: cos \: ( \frac{4x - 2x}{2} ) + sin \: 3x } \\ \\ \\ \implies \sf \: LHS = \frac{2 \: cos \: 3x \: cos \: x \: + cos \: 3x}{2 \: sin \: 3x \: cos \: x + \: sin \: 3x} \\ \\ \\ \implies \sf \: LHS = \frac{cos \: 3x(2 \: cos \: x + 1)}{sin \: 3x(2 \: cos \: x + 1)} \\ \\ \\ \implies \sf \: LHS = \frac{cos \: 3x}{sin \: 3x} \\ \\ \\ \implies \sf \blue {cot \: 3x \: = RHS}$