cos^2(π/5)+sin^2(π/10) 求值

高一数学 cos²(π/5)+sin²(π/10) 求值
先求cos(π/5),即cos36⁰的值.
cos36⁰=1-2sin²18⁰；又cos36⁰=sin54⁰=3sin18⁰-4sin³18⁰；
故得1-2sin²18⁰=3sin18⁰-4sin³18⁰；
即有4sin³18⁰-2sin²18⁰-3sin18⁰+1=0
分解因式：4sin²18⁰(sin18⁰-1)+2sin18⁰(sin18⁰-1)-(sin18⁰-1)
=(sin18⁰-1)(4sin²18⁰+2sin18⁰-1)=0
sin18⁰-1≠0,故必有4sin²18⁰+2sin18⁰-1=0,
于是得sin18⁰=(-2+√20)/8=(-1+√5)/4；
故cos36⁰=cos(π/5)=1-2[(-1+√5)/4]²=1-(3-√5)/4=(1+√5)/4
∴cos²(π/5)+sin²(π/10)=cos²(π/5)+[1-cos(π/5)]/2
=cos²(π/5)-(1/2)cos(π/5)+1/2=[(1+√5)/4]²-(1+√5)/8+1/2
=(3+√5)/8-(1+√5)/8+1/2=1/4+1/2=3/4