cos^2x化为tan是什么
\u4e3a\u4ec0\u4e481/cos^2x=tan^2x?\u8fd9\u4e2a\u7ed3\u8bba\u662f\u9519\u7684
1/cos²x
=(sin²x+cos²x)/cos²x
=sin²x/cos²x+cos²x/cos²x
=tan²x+1
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\u662f [ 1 -(tan x)^2 ] / [ 1 +(tan x)^2 ] = (cos x)^2 -(sin x)^2 \u5417\uff1f
= = = = = = = = =
\u8bc1\u660e\uff1a[ 1 -(tan x)^2 ] / [ 1 +(tan x)^2 ]
= { [ 1 -(tan x)^2 ] *(cos x)^2 } / { [ 1 +(tan x)^2 ] *(cos x)^2 }
= [ (cos x)^2 -(sin x)^2 ] / [ (cos x)^2 +(sin x)^2 ]
= (cos x)^2 -(sin x)^2.
= = = = = = = = =
\u5206\u5b50\u5206\u6bcd\u540c\u65f6\u4e58\u4ee5 (cos x)^2 \uff0c\u4f20\u8bf4\u4e2d\u7684\u5207\u5272\u5316\u5f26\u3002
=cos^2x/1
=cos^2x/(sin^2x+cos^2x)(分子分母同时除以cos^2x)
=(cos^2x/cos^2x)/(sin^2x/cos^2x+cos^2x/cos^2x)
=1/(tan^2x+1)
绛旓細=cos^2x/(sin^2x+cos^2x)(鍒嗗瓙鍒嗘瘝鍚屾椂闄や互cos^2x)=(cos^2x/cos^2x)/(sin^2x/cos^2x+cos^2x/cos^2x)=1/(tan^2x+1)
绛旓細=cos^2x/(sin^2x+cos^2x)(鍒嗗瓙鍒嗘瘝鍚屾椂闄や互cos^2x)=(cos^2x/cos^2x)/(sin^2x/cos^2x+cos^2x/cos^2x)=1/(tan^2x+1)
绛旓細鐢变竾鑳藉叕寮忥紝寰楋細cos(2x)=[1-(tanx)^2]/[1+(tanx)^2]锛屽洜涓篶os(2x)=tanx锛屾墍浠 1-(tanx)^2=tanx+(tanx)^3 锛(tanx)^3+(tanx)^2+tanx+1=0锛(tanx+1)[(tanx)^2+1]=0锛岃(tanx)^2+1>0锛屾墍浠anx+1=0锛宼anx=-1锛屽張0<=x<=2蟺锛屾墍浠=3蟺/4锛屾垨 x=7蟺/4銆
绛旓細cos2x=cos²x-sin²x=cos²x锛1-tan²x锛=锛1-tan²x锛/sec²x=锛1-tan²x锛/锛1+tan²x锛
绛旓細2銆cos2x=锛1-tan²x锛/锛1+tan²x锛夎В鏋愯繃绋嬪涓嬶細璁総an(A/2)=t sinA=2t/(1+t^2) 锛圓鈮2k蟺+蟺锛宬鈭圸锛塼anA=2t/(1-t^2) 锛圓鈮2k蟺+蟺锛宬鈭圸锛塩osA=(1-t^2)/(1+t^2) 锛圓鈮2k蟺+蟺 k鈭圸锛夋墍浠ワ細sin2x=2tanx/锛1+tan²x锛塩os2x=锛1-tan²...
绛旓細涓夎鍑芥暟浠ｆ崲锛cos2X=(cosX)^2-(sinX)^2=2*(cosX)^2-1=1-2*(sinX)^2 鍗筹細cos2x=2cosx鐨勫钩鏂-1=cosx鐨勫钩鏂-sinx骞虫柟=1-2sinx鐨勫钩鏂 cos2x鐨勫嚱鏁板浘鍍:
绛旓細cos2x涓巘anx閮藉睘浜庝笁瑙掑嚱鏁扮殑璇卞鍏紡锛屼簩鑰呯殑鍏崇郴濡備笅锛1銆乻in²x+cos²x=1锛2銆乼anx=sinx/cosx锛3銆乧os2x=(1-tan²x)/(1+tan²x)锛4銆乧os2x=1-tan²x/tan²x+1銆備笁瑙掑嚱鏁拌瀵煎叕寮忕殑浣滅敤涓哄彲浠ュ皢浠绘剰瑙掔殑涓夎鍑芥暟杞寲涓洪攼瑙掍笁瑙掑嚱鏁般
绛旓細cos鐨勫钩鏂箈绛変簬濡備笅锛歝os²x=1-sin²x cos²x=(cos2x+1)/2 cos²x=cos2x+sin²x cos²x=sin²x/tan²x
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