Question

若sinα-cosα=(√2)/3,0<α<π/2,则sinα+cosα=?

若sinα-cosα=(√2)/3,0<α<π/2,则sinα+cosα=?

Answer 1

sinα-cosα=(√2)/3 平方可得 (sinα)^2+(cosα)^2 - 2sinαcosα=4/9
因为(sinα)^2+(cosα)^2 =1 所以可知2sinαcosα = 1- 4/9 = 5/9

所以
(sinα+cosα)^2 = (sinα)^2+(cosα)^2 + 2sinαcosα = 1 + 5/9 = 14/9
再根据0<α<π/2，知道0<sinα<1, 0<cosα<1
所以sinα+cosα = (√14)/3

Answer 2

sinα+cosα=√5/2 两边平方求得

Answer 3

∵sinα-cosα=√2sin(α-π/4)=√2/3,
∴sin(α-π/4)=1/3
cos(α-π/4)=2√2/3
sinα+cosα=√2sin(α+π/4)=√2sin(α-π/4+π/2)=√2cos(α-π/4)
=4/3