Let f (x) be a function of R with a minimum positive period of 3 π / 2, if f (x) = cosx, (- π / 2 ≤ x0) SiNx, (0 ≤ x)

Let f (x) be a function of R with a minimum positive period of 3 π / 2, if f (x) = cosx, (- π / 2 ≤ x0) SiNx, (0 ≤ x)

Because - 15 π / 4 = 3 π / 4-3 * 3 π / 2, t = 3 π / 2,
Thus f (- 15 π / 4) = f (3 π / 4) = sin (3 π / 4) = √ 2 / 2