सिद्ध कीजिए कि - `"cos 2x cos"(x)/(2)-"cos 3x cos"(9x)/(2)="sin 5x sin"(5x)/(2)`.

Explanation

`L.H.S. = "cos 2x cos"(x)/(2)-"cos 3x cos"(9x)/(2)`
`=(1)/(2)["2 cos 2x cos"(x)/(2)-"2 cos 3x cos"(9x)/(2)]`
`=(1)/(2)[cos(2x+(x)/(2))+cos(2x-(x)/(2)) - cos(3x+(9x)/(2))-cos(3x-(9x)/(2))]`
`[because 2cos A cos B = cos (A+B)+cos (A-B)]`
`=(1)/(2)["cos"(5x)/(2)+"cos"(3x)/(2)-"cos"(15x)/(2)-"cos"(3x)/(2)]`
`=(1)/(2)["cos"(5x)/(2)-"cos"(15x)/(2)]`
`=(1)/(2)[-2sin{((5x)/(2)+(15x)/(2))/(2)}sin{((5x)/(2)-(15x)/(2))/(2)}]`
`[because cos -cos D = - "2 sin"(C+D)/(2)"sin"(C-D)/(2)]`
`rArr L.H.S. =-sin 5x sin(-(5x)/(2))`
`= "sin 5x sin"(5x)/(2)= R.H.S. " "` इति सिद्धम्