不用计算器,求(cos(π/8))^4+(cos(3π/8))^4+(cos(5π/8))^4+(cos(7π/8))^4的值

不用计算器,求(cos(π/8))^4+(cos(3π/8))^4+(cos(5π/8))^4+(cos(7π/8))^4的值
不用计算器,求(cos(π/8))^4+(cos(3π/8))^4+(cos(5π/8))^4+(cos(7π/8))^4的值

不用计算器,求(cos(π/8))^4+(cos(3π/8))^4+(cos(5π/8))^4+(cos(7π/8))^4的值
(cosa)^2=(cos2a+1)/2
(cosa)^2=(cos2a+1)^2/4
cos(π/8))^4=(cosπ/4+1)^2/4=[3+2√2]/8
cos(3π/8))^4=(cos3π/4+1)^2/4=[3-2√2]/8
cos(5π/8))^4=(cos5π/4+1)^2/4=[3-2√2]/8
cos(7π/8))^4=(cos7π/4+1)^2/4=[3+2√2]/8
(cos(π/8))^4+(cos(3π/8))^4+(cos(5π/8))^4+(cos(7π/8))^4
=3/2

(cos(5π/8))^4=(cos(3π/8))^4，(cos(π/8))^4=(cos(7π/8))^4
所以原式=2[(cos(π/8))^4+(cos(3π/8))^4]
又cos(3π/8)=sin(π/2-3π/8)=sin(π/8)
2[(cos(π/8))^4+sin(π/8)^4]
=2{【(cos(π/8))^2+sin(π/8)^2】^2-2cos(π/8)^2sin(π/8)^2}
=2/3 哪一步不懂？

(cos(π/8))^4+(cos(3π/8))^4+(cos(5π/8))^4+(cos(7π/8))^4
=[(cosπ/4+1)/2]²+{[cos(6π/8)+1]/2}²+{[cos(10π/8)+1)]/2}²+{[cos(14π/8)+1]/2}²
=[(√2/2+1)/2]²+[(1-√2/2)/2]²+[(1-√2/2)/2]²+[(√2/2+1)/2]²
=3/2