i think it is impossible
cos2x = cos^4 - sin ^4
please help
how to prove this
2007-09-04 21:52:35 · 7 answers · asked by master bert 2 in Science & Mathematics ➔ Mathematics
take the RHS
cos^4x - sin ^4x
by the formula a^2-b^2=(a-b)(a+b)
=(cos^2x-sin^2x)(cos^2x+sin^2x)
=(cos^2x-sin^2x) * 1
=cos^2x-sin^2x
=(1-sin^2x)-sin^2x
=1-2sin^2x
by indentity 1-2sin^2x=cos2x
=cos2x
so LHS=RHS
2007-09-04 22:02:31 · answer #1 · answered by shubham_nath 3 · 0⤊ 0⤋
Prove the identity. I assume you mean
cos 2x = (cos^4 x) - (sin^4 x)
Right Hand Side = (cos^4 x) - (sin^4 x)
= (cos²x - sin²x)(cos²x + sin²x) = (cos²x - sin²x)*1
= cos²x - sin²x = cos 2x = Left Hand Side
2007-09-05 06:22:52 · answer #2 · answered by Northstar 7 · 0⤊ 0⤋
cos^4(x) - sin^4(x) = [ cos^2 (x) - sin^2(x) ] [ cos^2 (x) + sin^2 (x) ]
= cos 2x [ because cos^2(x) + sin^2(x) = 1 ]
2007-09-05 05:00:05 · answer #3 · answered by Madhukar 7 · 0⤊ 0⤋
RHS
cos^4x - sin^4x
(cos ² x - sin ² x) (cos ² x + sin ² x)
(cos 2x ) ( 1 )
cos 2x
LHS
cos 2x
LHS = RHS
2007-09-05 08:24:11 · answer #4 · answered by Como 7 · 1⤊ 0⤋
RHS
= Cos^4x - Sin^4x
= (Cos^2x)^2 - (Sin^2x)^2
. [Using (a^4 - b^4) = a^2^2 - b^2^2]
= (Cos^2x + Sin^2x) ( Cos^2x - Sin^2x)
. [Using (a^2 - b^2) = (a+b)(a-b)]
= 1 ( Cos^2x - Sin^2x)
. [Since, Cos^2x + Sin^2x = 1]
= Cos^2x - Sin^2x
= Cos2x
=RHS
2007-09-05 05:17:26 · answer #5 · answered by bloodydraculaonline 1 · 0⤊ 0⤋
that is indeed impossible because from the identity...
cos(x+x) = cos(2x) = cosxcosx - sinxsinx
= cos^2x - sin^2x
it can never be equal to cos^4 - sin^4
2007-09-05 05:01:18 · answer #6 · answered by tootoot 3 · 0⤊ 1⤋
cos2x
= cos^2 (x) -- sin^2 (x)
= [ 1 ] [ cos^2 (x) -- sin^2 (x)]
= [cos^2x + sin^2x][cos^2x -- sin^2x]
= cos^4 (x) -- sin^4 (x)
2007-09-05 05:10:16 · answer #7 · answered by sv 7 · 0⤊ 0⤋