Question

Prove that...

Answer 1

@myininaya , @Hero

Answer 2

@amistre64, @ganeshie8

Answer 3

\(\large\tt \color{black}{1+cox+cos2x}\)

\(\large\tt \color{black}{=\dfrac{(2sin\frac{x}{2})(1+cox+cos2x)}{2sin\frac{x}{2}}}\)

\(\large\tt \color{black}{=\dfrac{2sin\frac{x}{2}+2sin\frac{x}{2}cox+2sin\frac{x}{2}cos2x}{2sin\frac{x}{2}}}\)

\(\large\tt \color{blue}{use~~identity~~2sinx~cosy=sin(x+y)-sin(y-x)}\)

\(\large\tt \color{black}{=\dfrac{2sin\frac{x}{2}+sin(\frac{x}{2}+x)-sin(x-\frac{x}{2})+sin(\frac{x}{2}+2x)-sin(2x-\frac{x}{2})}{2sin\frac{x}{2}}}\)

\(\large\tt \color{black}{=\dfrac{2sin\frac{x}{2}+sin(\frac{3x}{2})-sin(\frac{x}{2})+sin(\frac{5x}{2})-sin(\frac{3x}{2})}{2sin\frac{x}{2}}}\)

\(\large\tt \color{black}{=\dfrac{sin\frac{x}{2}+sin(\frac{5x}{2})}{2sin\frac{x}{2}}}\)

\(\large\tt \color{black}{\dfrac{1}{2}+\dfrac{sin(\frac{5x}{2})}{2sin\frac{x}{2}}}\)

Answer 4

@mathmath333 Thank you.

Answer 5

welcome