✔ 最佳答案
(cosx+isinx)^6=cos6x+isin6x
LHS=cos^6x+6cos^5x(isinx)+15cos^4x(i²sin²x)+20cos³x(i³sin³x)+
15cos²x(i^4sin^4x)+6cosx(i^5sin^5x)+i^6sin^6x
=cos^6x-15cos^4xsin²x+15cos²xsin^4x-sin^6x+imaginary terms
∴cos6x=cos^6x-15cos^4xsin²x+15cos²xsin^4x-sin^6x
=(1-sin²x)³-15(1-sin²x)²sin²x+15(1-sin²x)sin^4x-sin^6x
=1-3sin²x+3sin^4x-sin^6x-15sin²x(1-2sin²x+sin^4x)+15sin^4x-
15sin^6x-sin^6x
=1-3sin²x+3sin^4x-sin^6x-15sin²x+30sin^4x-15sin^6x+15sin^4x-
15sin^6x-sin^6x
=1-18sin²x+48sin^4x-32sin^6x
=2-18sin²x+48sin^4x-32sin^6x-1
=(1-sin²x)(2-16sin²x+32sin^4x)-1
∴cos6x+1=(1-sin²x)(2-16sin²x+32sin^4x)
For x=10°
3/4=(1-sin²10°)(1-4sin²10°)²
3/4=[(1-sin²10°)(1-4sin²10°)]²/cos²10°
(sqrt3/2)(cos²10°)=cos²10°+sin²20°
(sqrt3/2)(cos²10°)=cos²10°+(1-cos²20°)
sqrt(3/2)(cos²10°)=cos²10°+[1-(2cos²10-1)²]
sqrt(3/2)(cos²10°)=cos²10°-4cos^4(10°)+4cos²10°
sqrt(3/2)=5-4cos²10°-------(1)
cos3x=4cos³x-3cosx
∴cos30°=4cos³10°-3cos10°--------(2)
Sub(1)in (2)
5-4cos²10°=4cos³10°-3cos10°
4cos³10°+4cos²10°-3cos10°-5=0
(cos10°-1)(4cos²10°+8cos10°+5)=0
4cos²10°+8cos10°+5=0
(cos²10°+2cos10°+1)=-1/4
(cos10°+1)=i/2
cos10°=i/2-1
cos²10°=i²/4-i+1
sin²10°=i-i²/4
sin²10°=i+1/4, which is not in real surd form
∴sin10° cannot be represented in surd form
參考: Silly method......