2007-09-06 03:10:57 · 7 answers · asked by 15fsg546rge1rrheljh45hjr90459ty3 3 in Science & Mathematics ➔ Mathematics
To PROVE sin x° = cos (90 - x)°:-
cos (90 - x)° = cos 90° cos x° + sin 90° sin x°
cos (90 - x)° = 0 + 1 (sin x°)
cos (90 - x)° = sin x°
2007-09-09 20:32:26 · answer #1 · answered by Como 7 · 0⤊ 0⤋
We have this formulae :
Cos (a-b) = cos a . cos b + sin a . sin b
so
Cos (90-x) = cos 90 . cos x + sin 90 . sin x
= 0 . cos x + 1 . sin x
= sin x
2007-09-06 03:20:58 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Assuming you mean prove not solve...
using the formula cos(A-B) = cosAcosB+sinAsinB
A=90 and B = x
cos90cos(x) + sin90sin(x)
Knowing cos90 = 0 and sin90 = 1
0*cos(x) + 1*sin(x) = sin(x)
QED
2007-09-06 03:18:50 · answer #3 · answered by SS4 7 · 0⤊ 0⤋
let in triangle ABC,
C = 90 deg, A = x deg, B = (90 -- x) deg
then sin x = sinA = BC/AB = cosB = cos(90 -- x)
2007-09-06 03:26:57 · answer #4 · answered by sv 7 · 0⤊ 0⤋
cos(a-b) = cosa*cosb + sina*sinb
cos(90-x) = cos90*cosx + sin90*sinx
sin90 = 1 cos90 = 0
cos(90-x) = cos90*0 + 1*sinx
cos(90-x) = sinx
2007-09-06 03:19:15 · answer #5 · answered by sdlei 2 · 0⤊ 0⤋
The thing is true for all angles in the first quadrant, so there are infinitely many solutions.
2007-09-06 03:19:08 · answer #6 · answered by Anonymous · 0⤊ 0⤋
sin x = sin x
can not be solved
2007-09-06 03:20:33 · answer #7 · answered by CPUcate 6 · 0⤊ 0⤋