Question -
Answer -
Consider
L.H.S. = cos 6x
It can be written as
= cos 3(2x)
Using the formula cos3A┬а= 4 cos3┬аA┬атАУ 3 cos┬аA
= 4 cos3┬а2x┬атАУ3 cos┬а2x
Again by using formulacos 2x┬а= 2 cos2┬аx┬атАУ 1
= 4 [(2 cos2┬аx┬атАУ1)3┬атАУ 3 (2 cos2┬аx┬атАУ 1)
By furthersimplification
= 4 [(2 cos2┬аx)┬а3┬атАУ(1)3┬атАУ 3 (2 cos2┬аx)┬а2┬а+3 (2 cos2┬аx)] тАУ 6cos2┬аx┬а+ 3
We get
= 4 [8cos6x┬атАУ1 тАУ 12 cos4x┬а+ 6 cos2x] тАУ 6 cos2x┬а+3
By multiplication
= 32 cos6x┬атАУ4 тАУ 48 cos4x┬а+ 24 cos2┬аx┬атАУ 6cos2x┬а+ 3
On further calculation
= 32 cos6x┬атАУ48 cos4x┬а+ 18 cos2x┬атАУ 1
= R.H.S.