Question

已知5根号3sinx+5cosx=6,根号2siny+根号6cosy=1且x属于(0,π/3),y属于(π/6,π/2)，求cos（x+y）

Answer 1

5√3sinx+5cosx=6,
5√3sinx/10+5cosx/10=6/10
√3/2*sinx+1/2*cosx=3/5
sinxcosπ/6+cosxsinπ/6=3/5
sin(x+π/6)=3/5
x∈(0,π/3)
x+π/6∈(π/6,π/2)
cos(x+π/6)=4/5

√2siny+√6cosy=1
√2siny/√2+√6cosy/√2=1/√2
siny+√3cosy=√2/2
1/2*siny+√3/2*cosy=√2/4
sinysinπ/6+cosycosπ/6=√2/4
cos(y-π/6)=√2/4
y∈(π/6,π/2)
y-π/6∈(0,π/3)
sin(y-π/6)=√14/4

cos（x+y）
=cos[(x+π/6)+(y-π/6)]
=cos(x+π/6)cos(y-π/6)-sin(x+π/6)sin(y-π/6)
=4/5*√2/4-3/5*√14/4
=√2/5-3√14/12