Question

急！！求帮助已知向量a=(cosx+sinx,sinx),向量b=(cosx-sinx,2cosx)设f(x)=向量a乘向量b，

1）求函数f(x)的单调递增区间
2）设三角形ABC的三个角ABC所对的边分别是abc且满足A=π/3,f(B)=1,√3a+√2b=10求边c

Answer 1

(1)f(x)=a●b
= cos²x-sin²x+2sinxcosx
=cos2x+sin2x
=√2sin(2x+π/4)
2kπ-π/2≤ 2x+π/4≤ 2kπ+π/2
得 kπ-3π/8≤x≤kπ+π/8
∴f(x)单调递增区间为
[kπ-3π/8,kπ+π/8],k∈Z

(2)
∵f(B)=1 ∴√2sin(2B+π/4）=1
∴sin(2B+π/4）=√2/2
∵A=π/3 ∴0<B<2π/3
∴π/4<2B+π/4<19π/12
∴2B+π/4=3π/4
∴B=π/4
∴C=2π/3-π/4=5π/12
根据正弦定理
a=2RsinA=√3R,b=2RsinB=√2R
∵,√3a+√2b=10
∴3R+2R=10, ∴R=2
∴c=2Rsin5π/12
=4sin(π/4+π/6)
=√6+√2

Answer 2

(1)f(x)=cos^2x-sin^2x+2sinxcosx=cos2x+sin2x=√2sin(2x+π/4)其余自己做
(2)用余弦定理，其余自己做