(x 1 n x 1 n − 1 ⋯ 1 x 2 n x 2 n − 1 ⋯ 1 ⋮ ⋮ ⋱ ⋮ x m n x m n − 1 ⋯ 1) (p 1 p 2 ⋮ p n 1) = (y 1 y 2 ⋮ y m) , which polyfit solves with p = V\y Since the columns in the Vandermonde matrix are powers of the vector x , the condition number of V is often large for highorder fits, resulting in a singular coefficient matrixY = sin x and y = cos x look pretty similar;The graph of y=sin(x) is like a wave that forever oscillates between 1 and 1, in a shape that repeats itself every 2π units Specifically, this means that the domain of sin(x) is all real numbers, and the range is 1,1 See how we find the graph of y=sin(x) using the unitcircle definition of sin(x)
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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur He has been teaching from the past 9 years He provides courses for Maths and Science at TeachooD is the differential operator, int is the integration operator, C is the constant of integration Identities tan x = sin x/cos x equation 1 cot x = cos x/sin x equation 2 sec x = 1/cos x equation 3 csc x = 1/sin x equation 4X ï ì ô u o o ò v } x v p µ x À s z } o } u x w µ } z o } z } i } u w xz x &$*8$6 $&$'(0,$ &5,67,$1$ /2*26 '(
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