已知cos(1/4π+α)=-3/5,α∈(0,1/2π),求cosα的值.

问题描述:

已知cos(1/4π+α)=-3/5,α∈(0,1/2π),求cosα的值.

由cos(1/4π+α)=-3/5得:√2(cosa-sina)/2=-3/5,所以:cosa-sina=-3√2/5,
所以:1-2sinacosa=18/25,2sinacosa=7/25,
1+2sinacosa=32/25,即:(cosa+sina)²=32/25,因为a∈(0,1/2π),
所以cosa+sina=4√2/5,解由cosa+sina=4√2/5和cosa-sina=-3√2/5
组成的方程组得:cosa=√2/10


a∈(0,π/2)
∴a+π/4∈(π/4,3π/4)
∴sin(a+π/4)>0
∵cos(π/4+a)=-3/5
∴sina=√1-(-3/5)²=4/5
∴cosa
=cos[(a+π/4)-π/4]
=cos(a+π/4)cosπ/4+sin(a+π/4)sinπ/4
=√2/2×[(-3/5)+(4/5)]
=√2/2×(1/5)
=√2/10