PROve

Asked by  | 17th Sep, 2008, 07:31: PM

Expert Answer:

let A=θ and B=α

cos4A - cos4B =  2cos22A - 1 - (2cos22B - 1 ) = 2( cos22A - cos22B )

= 2 (cos2A + cos2B) (cos2A - cos2B)

=2( 2cos2A - 1 + 1 - 2sin2B) (2cos2A - 1 -(2cos2B - 1 ) )

=8(cos2A - sin2B) (cos2A -cos2B)

=8(cosA+sinB) (cosA - sinB) (cosA+cosB) (cosA-cosB)

Answered by  | 18th Sep, 2008, 12:55: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.