(cos1+isin1)(cos2+isin2)=cos3+isin3 (con3+isin3)(cos5+isin5)

1个回答

  • (1) 对f(x) = cos(x)+isin(x),成立f(x+y) = f(x)f(y).

    f(x)f(y) = (cos(x)+isin(x))(cos(y)+isin(y))

    = (cos(x)cos(y)-sin(x)sin(y))+i(cos(x)sin(y)+sin(x)cos(y))

    = cos(x+y)+isin(x+y)

    = f(x+y).

    (2) (f(x))^n = f(nx).

    证明使用数学归纳法.

    首先对n = 1显然成立.

    假设对n = k成立,即有(f(x))^k = f(kx).

    则(f(x))^(k+1) = f(x)·(f(x))^k = f(x)f(kx) = f((k+1)x),即得n = k+1时成立.

    因此结论对任意正整数k成立.

    (3) 所求式 = f(π/12)f(5π/12)+(f(π/6))^2007

    = f(π/12+5π/12)+f(2007π/6)

    = f(π/2)+f(669π/2)

    = cos(π/2)+isin(π/2)+cos(669π/2)+isin(669π/2)

    = 2i.