Question -
Answer -
Let us consider LHS:
cos 4x
We know, cos 2x = 2 cos2┬аx тАУ 1
So,
cos 4x = 2 cos2┬а2x тАУ 1
= 2(2 cos2┬а2x тАУ 1)2┬атАУ1
= 2[(2 cos2┬а2x)┬а2┬а+12┬атАУ 2├Ч2 cos2┬аx] тАУ 1
= 2(4 cos4┬а2x + 1 тАУ 4 cos2┬аx)тАУ 1
= 8 cos4┬а2x + 2 тАУ 8 cos2┬аxтАУ 1
= 8 cos4┬а2x + 1 тАУ 8 cos2┬аx
= RHS
Hence Proved.