2006-12-21 02:22:11 · 11 answers · asked by ~ ANGEL ~ 5 in Science & Mathematics ➔ Mathematics
as far as I know:
1-sinx=(sinx/2-cosx/2)^2
2006-12-22 03:10:17 · answer #1 · answered by Anonymous · 3⤊ 0⤋
There's no way to simplify this...you might be thinking of 1-sin(x)^2, which is cos(x)^2. The first answer is wrong, since sin(x) could be negative, which means that 0 <= 1-sin(x) <= 2
2006-12-21 10:48:46 · answer #2 · answered by mrsocialist 2 · 1⤊ 0⤋
0=<1- sin x=<1
2006-12-21 10:43:14 · answer #3 · answered by teco 2 · 3⤊ 0⤋
Depending on what X is the answer will correspond. However, if 1 - sinx = 0 or a certain real number, then it is solvable.
Example: 1 - sinx = 0
sinx = 1
x = 90 degree
2006-12-21 10:49:32 · answer #4 · answered by PIPI B 4 · 3⤊ 0⤋
Subtract 1 from both sides:
-sin x=-1
Divide -1 from both sides:
sin x=1
x=Ï/2 or 90°
2006-12-21 12:19:50 · answer #5 · answered by Anonymous · 2⤊ 0⤋
Everyone here is wrong, it's x = sin^(-1)(1)
2006-12-21 11:48:57 · answer #6 · answered by Anonymous · 2⤊ 0⤋
sin 90 - sin x
2006-12-21 12:31:29 · answer #7 · answered by Renaud 3 · 2⤊ 0⤋
1-sinx=sin^2x/2+cos^2x/2-2sinx/2cosx/2
=(sinx/2-cosx/2)^2 Ans.
2006-12-21 11:22:16 · answer #8 · answered by aminnyus 2 · 2⤊ 0⤋
1-sinx=sinpi/2-sinx=2sin(pi/4-x/2)cos(pi/4+x/2). OK???!!!!
2006-12-21 11:09:23 · answer #9 · answered by grassu a 3 · 2⤊ 0⤋
1-sinx=(sinx/2-cosx/2)^2 ok!!
2006-12-21 11:46:41 · answer #10 · answered by Grasu M 2 · 2⤊ 0⤋