cos cos2 cos3cosn化簡後結果是什麼啊？高手幫忙啊

1樓：

cos@+cos2@+cos3@+....+cosn@=(2sinacosa+2sinacos2a+...+2sinacosna)*1/2sina

=(sin2a+(sin3a-sina)+(sin4a-sin2a)+...+(sin(n+1)a-sin(n-1)a))*1/2sina

=(sin(n+1)a+sinna-sina)*1/2sina=2sin(2n+1)a/2cosa/2*1/2sina-1/2=sin(2n+1)a/2 / 2sina/2 -1/2

2樓：

2sin(θ/2)×(cosθ＋cos2θ＋……＋cosnθ)＝sin(3θ/2)－sin(θ/2)＋sin(5θ/2)－sin(3θ/2)＋……＋sin((2n+1)θ/2)－sin((2n-1)θ/2)

＝sin((2n+1)θ/2)－sin(θ/2)所以，cosθ＋cos2θ＋……＋cosnθ＝[sin((2n+1)θ/2)－sin(θ/2)]/[2sin(θ/2)]

cos10cos30cos50cos70 化簡後是什麼答案??

3樓：吉祥如意

cos10 cos30 cos50 cos70 ＝3/16cos10 cos70 cos50 cos30= 1/2(cos80 + cos60) cos50 cos30= 1/2(cos80 cos50 + cos60 cos50) cos30

= 1/2[(cos130 + cos30)/2 + cos50 /2)] cos30

= 1/4[- cos50 + cos30 + cos50] cos30

= 1/4cos30 cos30

= 3/16

(sinα+sin2α+...+sin(nα))/(coaα+cos2α+...+cosnα)化簡

4樓：

分子分母同乘以 sin(a/2)

分子變成：½ ( cos(a/2)-cos(3a/2) + cos(3a/2)-cos(5a/2) + ... +cos((2n-1)a/2)-cos((2n+1)a/2)) = ½ (cos(a/2)-cos((2n+1)a/2)) = sin(na/2) sin((n+1)a/2)

分母變成：½ ( -sin(a/2)+sin(3a/2) -sin(3a/2)+sin(5a/2) -... -sin((2n-1)a/2)+sin((2n+1)a/2)) = ½ (-sin(a/2)+sin((2n+1)a/2)) = sin(na/2) cos((n+1)a/2)

所以原式 = tan((n+1)a/2)