2007-11-09 02:23:35 · 12 answers · asked by waznnathan 3 in Science & Mathematics ➔ Mathematics
I might also need help with adding sinx to cosx in terms of sinx, so really I need sinx + cosx in terms of sinx, but maybe just cosx will get me in the right direction.
2007-11-09 02:31:30 · update #1
options:
cosx = sin(pi/2 - x)
cos x = sqrt[1 - sin^2 x]
2007-11-09 02:28:05 · answer #1 · answered by Linda K 5 · 1⤊ 0⤋
Since (cosx)^2=1-(sinx)^2, we get:
cosx=+-sqrt[1-(sinx)^2].
2007-11-10 15:51:41 · answer #2 · answered by Arc 2 · 0⤊ 0⤋
Sinx = O/H whereas cosx = A/H
2007-11-09 10:29:37 · answer #3 · answered by SAD 2 · 0⤊ 0⤋
cosx = (1-sin^2x)^1/2
so sinx + cosx = sinx + (1-sin^2x)^1/2
2007-11-09 13:24:03 · answer #4 · answered by rajesh v 2 · 0⤊ 0⤋
cos x=1/sq.root of 2 ,x=45 degrees
sinx= 1/sq.root of 2 ,x=45 degrees
cosx=sinx when x=45degrees
2007-11-10 02:22:51 · answer #5 · answered by Ashok B 3 · 0⤊ 0⤋
If you construct a right triangle, you can easily see (using Pythagoras) that
sin^2x + cos^2 x = 1
Therefore
cosx = sqrt(1 - sin^2x)
2007-11-09 10:32:19 · answer #6 · answered by Joe L 5 · 0⤊ 0⤋
(sinx)^2+(cosx)^2=1
cosx=sqrt(1-(sinx)^2)
2007-11-11 01:29:52 · answer #7 · answered by Anonymous · 0⤊ 0⤋
cosx = sin(90 - x)
2007-11-09 19:30:38 · answer #8 · answered by cinash 2 · 0⤊ 0⤋
cosx=sin(90-x)
2007-11-12 06:28:53 · answer #9 · answered by parta100 1 · 0⤊ 0⤋
cos x = â(1-sin² x)
To get sin x + cos x in terms of sin x,
note that sin Ï/4 = cos Ï/4.
So sin(x + Ï/4) = sin x cos Ï/4 + cos x sin Ï/4
= sin Ï/4(sin x + cos x) = â2/2(sin x + cos x)
and sin x + cos x = â2*sin(x + Ï//4)
2007-11-09 10:55:17 · answer #10 · answered by steiner1745 7 · 0⤊ 0⤋