假設 0 < A < 丌/2 及 sinA - cosA = m
(a)以 m 表示 sin³A - cos³A
(b)由此,求以sin³A 和 -cos³A 為根的二次方程
三角比方程
2009-06-12 10:28 pm
回答 (2)
2009-06-12 10:45 pm
✔ 最佳答案
a)sinA-cosA=m
(sinA-cosA)^2=m^2
1-2sinAcosA=m^2
sinAcosA=(1-m^2)/2
sin^3A-cos^3A
=(sinA-cosA)(sin^2A+sinAcosA+cos^2A)
=m(1+(1-m^2)/2)
=(m/2)(3-m^2)
b)
Given :
sin^3A+(-cos^3A)=(m/2)(3-m^2) and
sin^3Acos^3A=[(1-m^2)/2]^3
The required quadratic equation :
x^2-(m/2)(3-m^2)x+[(1-m^2)/2]^3=0
x^2-(m/2)(3-m^2)x+(1-3m^2+3m^4-m^6)/8=0
收錄日期: 2021-04-13 16:40:09
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