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小 發表於 2016-5-2 07:07 PM (第 2887 天)
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How would you find cos(3pi/5)?
This is referring to the 2016 DSE question.As the question is asking for cos(3pi/5), naturally the substitution x=cos(3pi/5) should be attempted.(It is not difficult to show that 4x^3+2x^2-3x-1=(4x^3-3x)+(2x^2-1)=cos(9pi/5)+cos(6pi/5)=cos(pi/5)-cos(pi/5)=0.)
It seems that some candidates prefer to let x=cos(pi/5) and then use double/triple angle formula to get back x=cos(3pi/5).
I am a bit curious what is the motivation behind your choice. Feel free to share your opinion.
[ 本帖最後由 桃子 於 2016-5-2 07:21 PM 編輯 ]