2007-05-24 15:41:55 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
If you draw a diagram, it's obvious.
Draw a right triangle.
Let one of the acute angles be x
Then the other acute angle must be (90-x), since the sum of the degrees in the triangle must be 180.
By definition of sine and cosine,
sin x = opposite(x)/hypotenuse = adjacent(90-x)/hypotenuse = cos (90-x).
Hope that helps!
2007-05-24 15:48:22 · answer #1 · answered by Bramblyspam 7 · 1⤊ 1⤋
Cos 90-x
2016-09-30 00:47:52 · answer #2 · answered by Anonymous · 0⤊ 0⤋
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RE:
prove that cos(90-x)=sinx?
2015-08-18 16:13:03 · answer #3 · answered by ? 1 · 0⤊ 0⤋
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cos( 90+x) = cos90cosx - sin 90 sin x =0*cosx -1*sinx = -sinx
2016-03-27 04:19:02 · answer #4 · answered by ? 4 · 0⤊ 0⤋
We will use the difference formula
cos(A-B) = cosA cosB + sinA sinB
where
cos (90 - x) ===> A=90 and B = x
also, recall cos90 = 0 and sin90 = 1
cos(A-B) = cosA cosB + sinA sinB
cos(90 - x) = cos90cosx + sin90 sinx
cos(90 - x) = (0)cosx + (1) sinx
cos(90-x) = sin x
=]
2007-05-24 15:45:53 · answer #5 · answered by Anonymous · 5⤊ 0⤋
LHS
= cos 90° cos x + sin 90° .sin x
= 0 + sin x
= sin x = RHS
2007-05-24 19:54:00 · answer #6 · answered by Como 7 · 0⤊ 1⤋
cos(90 - x) = cos(90)*cos(x) + sin(90)*sin(x)
=0*cos(x) + 1*sin(x)
=sin(x)
2007-05-24 15:45:02 · answer #7 · answered by Anonymous · 0⤊ 1⤋