高中数学三角函数题求详解。

如题所述

 因为:

(a+b)(cosAcosB-sinAsinB) = -c(cosA+cosB)

(a+b)cos(A+B) = -c(cosA+cosB)

(a+b)cos(180°-C)=-c(cosA+cosB)

-(a+b)cosC = -c(cosA+cosB)

(a+b)cosC = c(cosA+cosB)

又因为:

a/sinA = b/sinB = c/sinC = 2R            注:R 为该三角形外接圆的半径

所以上式可以转化为:

(2RsinA+2RsinB)cosC = 2RsinC(cosA+cosB)

(sinA+sinB)cosC = sinC(cosA+cosB)

2sin[(A+B)/2]*cos[(A-B)/2] * cosC = sinC * 2cos[(A+B)/2]cos[(A-B)/2]

sin[(180°-C)/2] * cosC = sinC * cos[(180°-C)/2]

sin(90°-C/2) * cosC = sinC * cos(90° -C/2)

cos(C/2)*cosC = sinC * sin(C/2)

cos(C/2)*cosC = [2sin(C/2)*cos(C/2)] * sin(C/2)

cosC = 2sin²(C/2)

因为 cosC = 1 - 2sin²(C/2)

所以,2sin²(C/2) = 1 - 2sin²(C/2)

那么:cosC = 2sin²(C/2) = 1/2

所以 C = 60° = π/3

因此:

    sin(20C+π/4)

    = sin(20π/3 + π/4)

    = sin(6π + 2π/3 + π/4)

    = sin(2π/3 + π/4)

    =sin(2π/3)cos(π/4) +cos(2π/3)sin(π/4)

    =√3/2 * √2/2 - 1/2 * √2/2

    =(√6 - √2)/4

    因为 S = 1/2*ab*sinC

    则:

    4S/b + bc

    =2absinC/b + bc

    =2asinC + bc

    =2√3*a + b*√3

    =√3 * (2a + b)

    =√3 * (2*2RsinA + 2R*sinB)         注:2R = c/sinC =√3/(√3/2) = 2

    =√3 * (4sinA + 2sinB)

    =2√3 * (2sinA + sinB)

    =2√3 * [2sin(120°-B) + sinB]

    =2√3 * [2sin120°cosB - 2cos120°sinB + sinB]

    =2√3 * [√3 * cosB + sinB + sinB]

    =2√3 * [√3 cosB + 2sinB]

    =2√3 * √7 * [√3/√7 * cosB + 2/√7 * sinB]

    =2√21 * [sinαcosB + cosα*sinB]          注:sinα=√3/√7, α≈40.89°

    =2√21 * sin(α+B)

    可见,sinα < sin(α+B) ≤ sin90°

    因此:

    2√21 * (√3/√7) < 2√21 * sin(α+B) ≤ 2√21 * 1

    6 < 2√21 * sin(α+B) ≤ 2√21

    即 4S/b + bc 的取值范围为:(6, 2√21]

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第1个回答  2016-04-17