一道关于三角函数和向量的问题化简cos5°+cos77°+cos149°+cos221°+cos293°.
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/02 19:53:28
![一道关于三角函数和向量的问题化简cos5°+cos77°+cos149°+cos221°+cos293°.](/uploads/image/z/5161348-28-8.jpg?t=%E4%B8%80%E9%81%93%E5%85%B3%E4%BA%8E%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%E5%92%8C%E5%90%91%E9%87%8F%E7%9A%84%E9%97%AE%E9%A2%98%E5%8C%96%E7%AE%80cos5%C2%B0%2Bcos77%C2%B0%2Bcos149%C2%B0%2Bcos221%C2%B0%2Bcos293%C2%B0.)
一道关于三角函数和向量的问题化简cos5°+cos77°+cos149°+cos221°+cos293°.
一道关于三角函数和向量的问题
化简cos5°+cos77°+cos149°+cos221°+cos293°.
一道关于三角函数和向量的问题化简cos5°+cos77°+cos149°+cos221°+cos293°.
.解法1.令x = 18°
∴cos3x = sin2x
∴4(cosx)^3 - 3cosx = 2sinxcosx
∵cosx≠ 0
∴4(cosx)^2 - 3 = 2sinx
∴4sinx2 + 2sinx - 1 = 0,
又0 < sinx < 1
∴sinx = (√5 - 1)/4
即sin18° = (√5 - 1)/4.
解法2.作顶角为36°、腰长为1 的等腰三角形ABC,BD为其底角B的平分线,设AD = x
则AD = BD = BC = x,DC = 1 - x.
由相似三角形得:x2 = 1 - x
∴x = (√ 5 - 1)/2
∴sin18° = x/2 = (√5 - 1)/4.
.cos5+cos77+cos149+cos221+cos293
=cos5+cos77-cos31-cos41 +cos67
=2cos41*cos36 - 2cos36*cos5 +cos67
=2cos36* (cos41-cos5) +cos67
=-2cos36* 2sin23*sin18 +sin23
=-sin23 *(4sin18*cos18*cos36)/cos18 +sin23
=-sin23*(2sin36*cos36/cos18) +sin23
=-sin23* (sin72/sin72) +sin23
=-sin23+sin23
=0