三角函數

[MathPower]

請問
(sin pi/7 )^2 + (sin 2pi/7 )^2 +...+(sin 6pi/7 )^2 = ?
和
(cos pi/7 )^2 + (cos 2pi/7 )^2 +...+(cos 6pi/7 )^2 = ?
另外次方改為n次方會有一般項嗎?
謝謝

[thepiano]

(sinx)^2 = (1 - cos2x)/2

[sin(π/7)]^2 + [sin(2π/7)]^2 + [sin(3π/7)]^2 + ...... + [sin(6π/7)]^2
= (1/2){6 - [cos(2π/7) + cos(4π/7) + cos(6π/7) + ...... + cos(12π/7)]}
= 3 - [cos(2π/7) + cos(4π/7) + cos(6π/7)]
= 7/2


[cos(π/7)]^2 + [cos(2π/7)]^2 + [cos(3π/7)]^2 + ...... + [cos(6π/7)]^2
= 6 - {[sin(π/7)]^2 + [sin(2π/7)]^2 + [sin(3π/7)]^2 + ...... + [sin(6π/7)]^2}
= 5/2