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
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