2006-12-19 14:24:24 · 16 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
correction: (sinx+cosx)^2 + (sinx-cosx)^2 =?
2006-12-19 14:28:34 · update #1
(sinx + cos x)^2 + (sinx - cosx)^2
= (sin^ x + cos^ x + 2 sin x cos x) + (sin^ x + cos^ x - 2 sin x cos x )
Do you know the relationship : sin^ x + cos^ x = 1
= 1 + 2 sin x cos x + 1 - 2 sin x cos x
= 2
2006-12-19 14:30:40 · answer #1 · answered by Kinu Sharma 2 · 0⤊ 1⤋
sin^2 x + 2 * sinx * cosx + cos^2 x + sin^2x - 2 * sinx * cosx + cos^2 x = 2 * (sin^2 x + cos^2 x) = 2
2006-12-19 14:36:32 · answer #2 · answered by Rick 5 · 0⤊ 0⤋
1 AND 2 SINX COSX = 2X, IT'S FAIRLY EASY
(sinx + cos x)^2 + (sinx - cosx)^2
= (sin^ x + cos^ x + 2 sin x cos x) + (sin^ x + cos^ x - 2 sin x cos x )
Do you know the relationship : sin^ x + cos^ x = 1
= 1 + 2 sin x cos x + 1 - 2 sin x cos x
= 2
voila :-)
2006-12-19 14:33:38 · answer #3 · answered by Anonymous · 0⤊ 0⤋
1 and 2 sinx cosx = sin 2x... easy
2006-12-19 14:26:30 · answer #4 · answered by Anonymous · 0⤊ 1⤋
Remember, sin^2(x) + cos^2(x) = 1
(sinx+cosx)^2 = sin^2(x) + 2*sin(x)cos(x) + cos^2(x)
(sinx+cosx)^2 = 2*sin(x)cos(x) + 1
(sinx-cosx)^2 = sin^2(x) - 2*sin(x)cos(x) + cos^2(x)
(sinx-cosx)^2 = -2*sin(x)cos(x) + 1
(2*sin(x)cos(x) + 1) + (-2*sin(x)cos(x) + 1)
1 + 1
=2
Answer: 2
2006-12-19 14:30:22 · answer #5 · answered by sft2hrdtco 4 · 0⤊ 1⤋
(sinx+cos)^2 + (sinx-cosx)^2 = 2(sin^2x + cos^2x) = 2
2006-12-19 14:34:15 · answer #6 · answered by James Chan 4 · 0⤊ 0⤋
(1-cosx)/sinx + sinx/(1-cosx) = ((1-cosx)(1-cosx) + (sinx)(sinx))/sinx(1-cosx) = (1-2cosx+cos^2x+sin^2x)/sinx(1-cosx) [Note: cos^2x+sin^2x=1] = (2-2cosx)/sinx(1-cosx) = (2(1-cosx))/sinx(1-cosx) = 2/sinx
2016-05-22 22:52:52 · answer #7 · answered by ? 4 · 0⤊ 0⤋
(sinx+cosx)^2 + (sinx-cosx)^2
= (sin^2x + 2sinxcosx + cos^2x) + (sin^2x - 2sinxcosx + cos^2x)
cancel the terms which add up to zero
= sin^2x + cos^2x + sin^2x + cos^2x
which is
= (sin^2x + cos^2x) + (sin^2x + cos^2x)
and since sin^2x + cos^2x = 1
we get
= (1) + (1)
= 2
therefore,
(sinx+cosx)^2 + (sinx-cosx)^2 = 2
2006-12-19 16:48:29 · answer #8 · answered by dkrudge 2 · 0⤊ 0⤋
sin^2 x +2sinx cosx +cos^2 x +
sin^2 x-2sinx cosx +cos^2 x=2.
2006-12-19 14:31:29 · answer #9 · answered by mathematician 7 · 1⤊ 0⤋
(sinx+cos)^2 + (sinx-cosx)^2
=sin^2x +2sinxcosx +cos^2x + sin^2x -2sinxcosx +cos^2x
1+1 =2
2006-12-19 14:38:19 · answer #10 · answered by ironduke8159 7 · 0⤊ 0⤋