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Table 1 Values returned by the functional unit specified in [1] when the input I has exactly three bits (ϕ=π/23−1=π/4)

From: Using the complement of the cosine to compute trigonometric functions

   

x

πx

 

sin(πx)

cos(πx)

   

 

I

S

n

S/4

Sϕ

nϕ

sin(Sϕ)

cos(Sϕ)

      

      

sin(nϕ)

cos(nϕ)

000

0

0

0

0

0

0

1

001

1

1

\(\frac {1}{4}\)

\(\frac {\pi }{4}\)

\(\frac {\pi }{4}\)

\(\frac {1}{\sqrt {2}}\)

\(\frac {1}{\sqrt {2}}\)

010

2

2

\(\frac {2}{4}\)

\(\frac {2\pi }{4}\)

\(\frac {2\pi }{4}\)

1

0

011

3

3

\(\frac {3}{4}\)

\(\frac {3\pi }{4}\)

\(\frac {3\pi }{4}\)

\(\frac {1}{\sqrt {2}}\)

\(-\frac {1}{\sqrt {2}}\)

100

− 4

4

−1

\(-\frac {4\pi }{4}\)

\(\frac {4\pi }{4}\)

0

−1

101

− 3

5

\(-\frac {3}{4}\)

\(-\frac {3\pi }{4}\)

\(\frac {5\pi }{4}\)

\(-\frac {1}{\sqrt {2}}\)

\(-\frac {1}{\sqrt {2}}\)

110

− 2

6

\(-\frac {2}{4}\)

\(-\frac {2\pi }{4}\)

\(\frac {6\pi }{4}\)

−1

0

111

− 1

7

\(-\frac {1}{4}\)

\(-\frac {\pi }{4}\)

\(\frac {7\pi }{4}\)

\(-\frac {1}{\sqrt {2}}\)

\(\frac {1}{\sqrt {2}}\)