已知α,β为锐角,且3sin2α+2sin2β=1,3sin2α-2sin2β=0,试求cos(π/3+α+2β)的值。

α,β为锐角

0 < sinα < 1, 0 < cosα < 1.
0 < sinβ < 1, 0 < cosβ < 1.

3sin2α+2sin2β=1,
3sin2α-2sin2β=0

sin2α = 1/6,
sin2β = 1/4.

(sinα + cosα)^2 = 1 + sin2α = 1 + 1/6 = 7/6,
sinα + cosα = (7/6)^(1/2),

(sinα - cosα)^2 = 1 - sin2α = 1 - 1/6 = 5/6,
sinα - cosα = (5/6)^(1/2),

sinα = [7^(1/2) + 5^(1/2)]/[2(6)^(1/2)]
cosα = [7^(1/2) - 5^(1/2)]/[2(6)^(1/2)].

同理，
sinβ = [5^(1/2) + 3^(1/2)]/4

cos2β = 1 - 2(sinβ)^2 = 1 - 2[5^(1/2) + 3^(1/2)]^2/16
= 1 - [8 + 2(15)^(1/2)]/8
= -(15)^(1/2)/4

sin(π/3 + 2β) = sin(π/3)cos(2β) + cos(π/3)sin(2β)

= 3^(1/2)/2[-(15)^(1/2)/4] + (1/2)(1/4)

= [1 - 3(5)^(1/2)]/8

cos(π/3 + 2β) = cos(π/3)cos(2β) - sin(π/3)sin(2β)

= (1/2)[-(15)^(1/2)/4] - [3^(1/2)/2](1/4)

= -3^(1/2)[1 + 5^(1/2)]/8

cos(π/3 + α + 2β) = cos(π/3 + 2β)cosα - sin(π/3 + 2β)sinα