2007-01-13 13:42:02 · 9 answers · asked by Princess Jasmine 2 in Science & Mathematics ➔ Mathematics
cos( 90+x)
= cos90cosx - sin 90 sin x
=0*cosx -1*sinx
= -sinx
2007-01-13 13:51:56 · answer #1 · answered by ironduke8159 7 · 2⤊ 0⤋
cos90+x or cos pi/2+x is arc whose the difference is pi/2.The cosines is negative and change in sines in the 2 part when we divide the trigo. circle in 4 equal parts.
we know that:
cos(a+b)=cosacosb-sinasinb
or
cos(90+x)=cos90cosx-sinxsin90
'' =0*cosx-1*sinx
cos(90+x)=-sinx
cos90 =0 is in the 2 parts of circle .
2007-01-13 14:33:52 · answer #2 · answered by Johnny 2 · 0⤊ 0⤋
cos (A+B) = cosA cosB - sinA sinB.
Also cos 90 = 0 and sin 90 = 1.
cos (90+x) = cos90 cosx - sin90 sinx.
cos (90+x) = 0 * cosx - 1 * sinx.
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Draw a rectangle OABC,
with vertical sides OA & CB and
with horizontal sides AB & OC.
BC is in the left side and
OA is in the right side.
AB is at the top side.
O is the origin of the coordinate axes.
Draw the diagonal OB. Extend CO to E.
The angle EOB is (90 + x). The angle AOB is x.
By definition cos (90+x) = minus OC/ OB.
Since, minus OC = minus AB
cos (90+x) = minus AB/ OB.
From the right angled triangle OCB,
AB/OB = sin x.
Therefore
cos (90+x) = minus sinx.
2007-01-13 14:07:43 · answer #3 · answered by Pearlsawme 7 · 0⤊ 0⤋
Expand it by the cosine addition formula:
cos(90 + x) = cos 90 cos x - sin 90 sin x.
But cos 90 = 0 and sin 90 = 1,
so all you're left with is -sin x.
2007-01-13 14:15:39 · answer #4 · answered by steiner1745 7 · 0⤊ 0⤋
sin 90° cos x - cos 90° sin x (1) cosx - 0 cos x
2016-05-23 22:42:42 · answer #5 · answered by Anonymous · 0⤊ 0⤋
Remember that brackets are totally important when conveying your question to readers. I assume you mean
cos(90 + x) = -sin(x)
By the sine addition formula,
cos(a + b) = cos(a)cos(b) - sin(a)sin(b)
We apply this to the left hand side.
cos(90 + x) = cos(90)cos(x) - sin(90)sin(x)
Note that cos(90) = cos(pi/2) = 0
sin(90) = sin(pi/2) = 1.
cos(90 + x) = [0]cos(x) - (1)sin(x)
cos(90 + x) = 0 - sin(x)
cos(90 + x) = -sin(x)
2007-01-13 14:33:45 · answer #6 · answered by Puggy 7 · 0⤊ 0⤋
The sin and cosine functions are 90 degrees out of phase so when you add 90 degrees to the cosine, they are identical.
2007-01-13 13:52:07 · answer #7 · answered by The answer guy 3 · 1⤊ 1⤋
it doesnt always...cos 90 = 0, so basically the equation becomes x = -sin x, for which there is no solution
2007-01-13 13:53:03 · answer #8 · answered by Anonymous · 0⤊ 1⤋
6^.01
2007-01-13 13:48:33 · answer #9 · answered by Anonymous · 0⤊ 3⤋