sinx-sin5x+sin9x-sin13x/cosx-cos5x -cos9x+cos13x= cot 4x
Answers
Answer:the method is correct once check it
Answer:
sinx-sin5x+sin9x-sin13x/cosx-cos5x -cos9x+cos13x= cot 4x
Step-by-step explanation:
sinx-sin5x+sin9x-sin13x/cosx-cos5x -cos9x+cos13x= cot 4x
LHS Numerator
= Sinx - Sin5x + Sin9x - Sin13x
= Sinx - Sin13x + Sin9x - Sin5x
Using SinA - SinB = 2Cos(A + B)/2 . Sin(A-B)/2
= 2Cos7xSin(-6x) + 2Co7xSin2x
= 2Cos7x(Sin2x - Sin6x)
= 2Cos7x ( 2Cos4xSin(-2x))
= - 4Cos7x Cos4xSin2x
LHS Denominator
Cosx - Cos5x - cos9x + cos13x
= Cosx + Cos13x - (Cos5x + Cos9x)
Using CosA + CosB = 2Cos(a+b)/2 Cos(a-b)/2
= 2Cos7xCos(6x) - 2Cos7xCos2x
= 2Cos7x(Cos(6x) - Cos2x)
Using CosA - CosB = -2Sin(a+b)/2 Sin(a-b)/2
= 2Cos7x(-2Sin4xSin2x)
= - 4Cos7xSin4xSin2x
LHS = - 4Cos7x Cos4xSin2x / (- 4Cos7xSin4xSin2x)
= Cos4x/Sin4x
= Cot4x
= RHS
QED
Proved
sinx-sin5x+sin9x-sin13x/cosx-cos5x -cos9x+cos13x= cot 4x