已知α为锐角,且tanα=1/2,求(sin2αcos2α-sinα)/(sin2αcos2α) 的值 已知α为锐角,且tanα=1/3,求sin2α的值
\u5df2\u77e5\u03b1\u4e3a\u9510\u89d2\uff0c\u4e14tan\u03b1= 1 2 \uff0c\u6c42 sin2\u03b1cos\u03b1-sin\u03b1 sin2\u03b1cos2\u03b1 \u7684\u503c\u2235 tan\u03b1= 1 2 \uff0c\u03b1\u4e3a\u9510\u89d2\u2234 cos\u03b1= 2 5 \u2234 sin2\u03b1cos\u03b1-sin\u03b1 sin2\u03b1cos2\u03b1 = sin\u03b1(2 cos 2 \u03b1-1) 2sin\u03b1cos\u03b1cos2\u03b1 = 1 2cos\u03b1 = 5 4 \uff0e
\u2235\u03b1\u4e3a\u9510\u89d2\uff0c\u4e14tan\u03b1=1/3
\u2234sin\u03b1=1/\u221a(1²+3²)=\u221a10/10
cos\u03b1=sin\u03b1/tan\u03b1=3\u221a10/10
\u5219sin2\u03b1=2sin\u03b1cos\u03b1=3/5
∴sinx=1\√5;cosx=2\√5
有原始式子=1-sinx\sinx.cosx
( 由公式:sin2x=2sinx.cosx)
=1-sinx除以1\2.sin4x(分子不变)
=1-2sinx\sin4x
=1-2sinx\sin(2.2x)
(再用一次上述公式)
=1-2sinx\2sin2x.cos2x(分子不变)
=1-sinx\sin2x.cos2x
(由sin2x=2sinx.cosx和cos2x=2cos^2-1)
=1-sinx\(2sinx.cosx)(2cosx的平方-1)
(把开始算的sinx和cosx的值代入)
=1-(1\√5)\{(2.1\√5)(2\√5)2(2\√5)^2-1}
=1-5√5.7
我的计算可能有错,不过方法应该没错。(有的符号不打,所以见谅,仔细看哈) 希望对你有些帮助!
tanα=1/2===>sinα=1/√5, cosα=2/√5
∴原式=1-sinα/[2sinαcosα(1-2sin²α)]
=1-(1/√5)/[2(1/√5)(2/√5)(1-2/5)]
=1-(1/√5)/[(4/5)*(3/5)]
=1-(1/√5)(25/12)
=1-5√5/12
tanA=sinA/cosA=1/2锛屼笖sinA2+cosA2=1锛岃В寰梥inA=镙瑰佛1/5锛宑os=镙瑰佛4/5锛屾墍浠�in2A=4/5锛宑oS2A=3/5锛屼唬鍏ュ缑(4/5脳3/5-镙瑰佛1/5)梅(4/5脳3/5)=(12-镙瑰佛75)/12
绛旓細鈭滴辨槸閿愯锛tan45掳=1锛屸埓伪=45掳锛庢晠閫塀锛
绛旓細sin^2伪锛媠in伪cos伪-2cos^2=0 (sin伪+2cos伪)(sin伪-cos伪)=0 鍥犱负伪涓閿愯 鎵浠 sin伪=cos伪 tan伪=1 鎵浠 伪=蟺/4 (2)sin(伪-蟺/3)=sin伪cos蟺/3-cos伪sin蟺/3 =鈭2/2*1/2-鈭2/2*鈭3/2 =(鈭2-鈭6)/4
绛旓細tan伪=1/tan锛30锛=鏍瑰彿3 銆傛槗寰椢=60.鈭磗in伪=鏍瑰彿3/2,cos伪=0.5 鈭磗in+cos=锛1+鏍瑰彿3锛/2
绛旓細鈭滴便佄插潎涓閿愯锛屸埓cos尾+sin尾锛0锛屸埓sin伪-cos伪=0锛屸埓tan伪=1锛庢晠绛旀涓猴細1
绛旓細tan伪=1 涓斿張鏄涓夎薄闄愮殑瑙 伪=蟺/4卤n蟺锛坣涓哄鏁帮級鍗宠繖涓鍦ㄧ涓夎薄闄 鍥犱负姝e鸡鍊煎拰浣欏鸡鍊煎湪绗笁璞¢檺閮芥槸璐熺殑鎵浠 sin伪=cos伪=-鈭2/2
绛旓細锛涳紙2锛夋墍姹傚紡瀛愯緝澶嶆潅锛岄渶鍏堝寲绠.鍏堢粺涓瑙掞紝 = 锛屽洜姝ゅ彧闇姹 鐨勫煎嵆鍙.鐢卞悓瑙掍笁瑙掑嚱鏁板叧绯诲緱sin伪= 锛宑os伪= 锛屽洜姝ゅ師寮忎负 锛庤瘯棰樿В鏋愶細锛1锛夆埖 鈭 锛屽嵆 锛岃В涔嬪緱tan伪= 锛涳紙2锛 鈭伪涓洪攼瑙掍笖tan伪= 鈭磗in伪= 锛宑os伪= 锛屽彲寰梒os伪+sin伪= 锛
绛旓細鍥犱负sin(伪-尾)=sin伪*cos尾-cos伪*sin尾 cos(伪+尾)=cos伪*cos尾-sin伪*sin尾)鎵浠in伪*cos尾-cos伪*sin尾=cos伪*cos尾-sin伪*sin尾 鎵浠in伪*(sin尾+cos尾)=cos伪*(sin尾+cos尾)鎵浠in伪=cos伪 鎵浠tan伪=sin伪/cos伪=1 ...
绛旓細tan伪cot伪=1 鎵浠ot伪=1/tan伪=1/(1/a)=a
绛旓細鈭伪涓洪攼瑙掞紝涓攖an伪=1/3 鈭磗in伪=1/鈭(1²+3²)=鈭10/10 cos伪=sin伪/tan伪=3鈭10/10 鍒檚in2伪=2sin伪cos伪=3/5
绛旓細鐢讳釜鍥 伪瀵硅竟涓1锛涓磋竟涓2锛屽垏杈逛负鏍瑰彿5 sin伪=鈭5/5