Question

prove the following identity

sin^4(x)-cox^4(x)+cos2x=0

kladfalkjflk i just cant get zero its so irritating

Answer 1

hint : a^2 - b^2 = (a+b)(a-b)

Answer 2

hint2 : cos^2x - sin^2x = cos(2x)

Answer 3

i did that alkdjfaldkfj but then i had an extra square on top

Answer 4

hint3 : cos^2x + sin^2x = 1

Answer 5

where is the
cos^2x + sin^2x = 1? o-o

Answer 6

\[\large \sin^4(x)-\cos^4(x) = (\sin^2x + \cos^2x)(\sin^2x - \cos^2x)\]

Answer 7

isn't there a square outside everything?

Answer 8

a^4-b^4 = (a^2-b^2)^2 doesn't it?

Answer 9

no, (a+b)^2= a^2+2ab+b^2

Answer 10

maybe define sin^2x = a and cos^2x = b

Answer 11

okay

so i'll get

(a^2-b^2)
=(a-b)(a+b)

then sub back in
(sin^2x-cos^2x)(sin^2x+cos^2x)?

Answer 12

that looks good

Answer 13

so (sin^2x-cos^2x) = -cos2x

(sin^2x+cos^2x)=1

Answer 14

therefore answer would be

-cos2x+cos2x=0


finally good grief
thankyu

Answer 15

:)