Asked by Anonymous
Prove:
sin^2x - sin^4x = cos^2x - cos^4x
What I have,
LS
= (sinx - sin^2x) (sinx + sin^2x)
= (sinx - 1 -cos^2x) (sinx + 1 - cos^2x)
= sin^2x + sinx - sinx - cos^2xsinx - cos^2xsinx - 1 - 1 + cos^4x
= sin^2x - 2cos^2xsinx - 2 + cos^4x
Where did I go wrong? Can anyone please help me prove this identity?
sin^2x - sin^4x = cos^2x - cos^4x
What I have,
LS
= (sinx - sin^2x) (sinx + sin^2x)
= (sinx - 1 -cos^2x) (sinx + 1 - cos^2x)
= sin^2x + sinx - sinx - cos^2xsinx - cos^2xsinx - 1 - 1 + cos^4x
= sin^2x - 2cos^2xsinx - 2 + cos^4x
Where did I go wrong? Can anyone please help me prove this identity?
Answers
Answered by
Reiny
(sinx - 1 -cos^2x) (sinx + 1 - cos^2x)
should have been
(sinx - 1 <b>+</b> cos^2x) (sinx + 1 - cos^2x) and then the next line should be
sin^2x + sinx - cos^2xsinx - sinx - 1 + cos^2 x + cos^2xsinx - cos^4x
= cos^2 x - cos^4 x
= RS
an easier way would have been
LS = sin^2x(1-sin^2x) by common factor
=sin^2x(cos^2x)
RS = cos^2x(1-cos^2x) also by common factor
=cos^2x(sin^2x)
= LS
should have been
(sinx - 1 <b>+</b> cos^2x) (sinx + 1 - cos^2x) and then the next line should be
sin^2x + sinx - cos^2xsinx - sinx - 1 + cos^2 x + cos^2xsinx - cos^4x
= cos^2 x - cos^4 x
= RS
an easier way would have been
LS = sin^2x(1-sin^2x) by common factor
=sin^2x(cos^2x)
RS = cos^2x(1-cos^2x) also by common factor
=cos^2x(sin^2x)
= LS
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