证明是否存在非等腰三角形使 cosA+cosB=cosC
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解决时间 2021-02-24 20:30
- 提问者网友：最爱你的唇
- 2021-02-24 00:03
证明是否存在非等腰三角形使 cosA+cosB=cosC
最佳答案
- 五星知识达人网友：撞了怀
- 2021-02-24 00:09
存在这样的三角形, 但是构造不太直接.
考虑函数f(x) = cos(x)+cos(3x/2)+cos(5x/2).
可算得f(0) = 3 > 0,
而f(2π/5) = cos(2π/5)+cos(3π/5)+cos(π) = cos(2π/5)-cos(2π/5)-1 = -1 < 0.
由f(x)连续性, 存在θ ∈ (0,2π/5), 使f(θ) = 0.
取A = θ, B = 3θ/2, C = π-5θ/2, 可知A+B+C = π, 且A, B, C ∈ (0,π).
因此存在三角形以A, B, C为内角, 不妨记为△ABC.
而此时cos(A)+cos(B)-cos(C) = cos(θ)+cos(3θ/2)-cos(π-5θ/2)
= cos(θ)+cos(3θ/2)+cos(5θ/2) = f(θ) = 0.
即成立cos(A)+cos(B) = cos(C).
通过验证f(π/3) ≠ 0, 可知θ ≠ π/3, cos(B) = cos(3θ/2) ≠ 0.
于是cos(A) = cos(C)-cos(B) ≠ cos(C), 有A ≠ C.
而由0 < θ < 2π/5 < π/2, 有cos(A) = cos(θ) ≠ 0.
于是cos(B) = cos(C)-cos(A) ≠ cos(C), 有B ≠ C.
又A = θ ≠ 3θ/2 = B, 因此A, B, C两两不等, △ABC不为等腰三角形.
考虑函数f(x) = cos(x)+cos(3x/2)+cos(5x/2).
可算得f(0) = 3 > 0,
而f(2π/5) = cos(2π/5)+cos(3π/5)+cos(π) = cos(2π/5)-cos(2π/5)-1 = -1 < 0.
由f(x)连续性, 存在θ ∈ (0,2π/5), 使f(θ) = 0.
取A = θ, B = 3θ/2, C = π-5θ/2, 可知A+B+C = π, 且A, B, C ∈ (0,π).
因此存在三角形以A, B, C为内角, 不妨记为△ABC.
而此时cos(A)+cos(B)-cos(C) = cos(θ)+cos(3θ/2)-cos(π-5θ/2)
= cos(θ)+cos(3θ/2)+cos(5θ/2) = f(θ) = 0.
即成立cos(A)+cos(B) = cos(C).
通过验证f(π/3) ≠ 0, 可知θ ≠ π/3, cos(B) = cos(3θ/2) ≠ 0.
于是cos(A) = cos(C)-cos(B) ≠ cos(C), 有A ≠ C.
而由0 < θ < 2π/5 < π/2, 有cos(A) = cos(θ) ≠ 0.
于是cos(B) = cos(C)-cos(A) ≠ cos(C), 有B ≠ C.
又A = θ ≠ 3θ/2 = B, 因此A, B, C两两不等, △ABC不为等腰三角形.
全部回答
- 1楼网友：持酒劝斜阳
- 2021-02-24 01:27
cosa^2+cosb^2+cosc^2=1
cosa^2+cosb^2=sinc^2
1+(cos2a+os2b)/2=sinc^2
1+cos(a+b)cos(a-b)=sinc^2
cosccos(a-b)=cosc^2
cos(a-b)=cosc
a-b=c
a=b+c
a+b+c=180
a=90直角三角
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