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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 \(-\frac {1}{4}\) \(-\frac {\pi }{4}\) \(\frac {7\pi }{4}\) \(-\frac {1}{\sqrt {2}}\) \(\frac {1}{\sqrt {2}}\)