Vertify that ((sin x)(sin x)-(cos x)(cos x))=1-2sinxcosx is an identity?

Question

Any help? Thanks.

Answer 1

LHS=sin^2x-cos^2x=(sinx+cosx)(sinx-cosx)
RHS=1-2sinxcosx
=(sin^2x+cos^2x)-2sinxcosx as sin^2x+cos^2x=1
=(sinx-cosx)^2
LHS and RHS can be equal only ifsinx+cosx=sinx-cosx
so it is not an identity
an identity must hole for all values

Answer 2

Gopal is right, but I suspect the original question was
(sin x - cos x)^2 = etc, and you made the mistake of thinking that you square each term separately. In fact, it should be

(sin x - cos x)(sin x - cos x) and when you expand you get four terms: sin x * each of the two terms in the second parentheses
-cos x * each of the two terms in the second parentheses.

Two of these four terms are
(sin x)^2 + (cos x)^2 which is identically equal to 1, and the other two give -2 sin x cos x. I leave you to finish it.

Answer 3

( (sinx)(sinx) - ( cosx)(cosx) )
=( sinx + cosx )^2 -4sinxcosx [ (a^2-b^2) = (a+b)^2 - 4ab]
=sin^2x + cos^2x +2sinxcox - 4sinxcosx
= 1-2sinxcosx
Therefore it is an Identity

Answer 4

if by this you mean

(sin(x) * sin(x)) - (cos(x) * cos(x))
(sin(x))^2 - (cos(x))^2
(1 - cos(x)^2) - cos(x)^2
1 - cos(x)^2 - cos(x)^2
1 - 2cos(x)^2

1 - 2cos(x)^2 = cos(2x)

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1 - 2sinxcosx = 1 - sin(2x)

Answer 5

loser. i don't care. go get a job. and if youre under 16 then you shouldn't be on this site, kiddo.