Question

Sin3x+Sin5x+Sin7x+Sin9x
tan6x
Cos9x+Cos7x+Cos5x+cos3x​

Answer 1

Answer:

$identity \: used \\ sinx + siny = 2sin \frac{x + y}{2} cos \frac{x - y}{2} \\ cosx + cosy = 2cos \frac{x + y}{2} cos \frac{x - y}{2}$

$\frac{sin9x + sin3x + sin7x + sin5x}{cos9x + cos3x + cos7x + cos5x} \\ = \frac{2sin6xcos3x + 2sin6xcosx}{2cos6xcos3x + 2cos6xcosx} \\ = \frac{2sin6x(cos3x + cosx)}{2cos6x(cos3x + cosx)} \\ = tan6x$

Answer 2

Answer:

hi mathdude bro.......