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

Answer 1

Subject: Re: Proving trig identiies...small error..thisis the right one
From:
Date: Monday June 14, 2004 at 08:35:54

RHS= 2csc^2x=2(cot^2x+1)

note :-(csc^2x=cot^2x+1)

=2cot^2x+2

=2cos^2x/sin^2x +2

note :-((cot^2x=cos^2x/sin^2x)

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

note :-((sin^2x=1-cos^2x)

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

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

=2/(1-cos^2x)

=2/(1-cosx)(1+cosx)

note :-((1-cos^2x=(1-cosx)(1+cosx))


THEREFORE lHS =2/(1-cosx)(1+cosx) (1-cos^2x=(1-cosx)(1+cosx))
now for the RHS we get

RHS= 1/1+COSX + 1/1-COSX

=(1-cosx) + ( 1 + cosx)/(1-cosx)(1-cosx)

note :-(( through expansion)

=2/(1-cosx)(1+cosx)

note :-(cosx on the numerator cancel out)

therefore RHS= LHS

THEREFORE 2csc^2x= 1/1+cosx + 1/1-cosx

as required



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Answer 2

Always true from the pythagorean theorem
sin=o/h
cos=a/h
we know that o^2+a^2=h^2 divide all terms by h^2
(o/h)^2+(a/h)^2=(h/h)^2
sin^2 +cos^2=1

the -x doesn't change anything since (-z)^2=z^2

Answer 3

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

Info found at www.math.com/tables/trig/identities.htm

Answer 4

What's sin, cos and x? Then you'll have your answer.