求高数里面三角正反弦/余弦的转换公式？ 求高数里面三角正反弦/余弦的转换公式?
\u6c42\u9ad8\u6570\u91cc\u9762\u4e09\u89d2\u6b63\u53cd\u5f26/\u4f59\u5f26\u7684\u8f6c\u6362\u516c\u5f0f\uff1fsin\uff08\u03c0/2-\u03b1\uff09=cos\u03b1\uff0ccos\uff08\u03c0/2-\u03b1\uff09=sin\u03b1\uff0csin\uff08\u03c0/2+\u03b1\uff09=sin\u03b1\uff0ccos\uff08\u03c0/2+\u03b1\uff09=-cos\u03b1\uff1b
\u548c\u5dee\u5316\u79ef\u516c\u5f0f\uff1acos\uff08\u03b1+\u03b2\uff09=cos\u03b1cos\u03b2-sin\u03b1sin\u03b2\uff0ccos\uff08\u03b1-\u03b2\uff09=cos\u03b1cos\u03b2+sin\u03b1sin\u03b2\uff0csin\uff08\u03b1+\u03b2\uff09=sin\u03b1cos\u03b2+cos\u03b1sin\u03b2\uff0csin\uff08\u03b1-\u03b2\uff09=sin\u03b1cos\u03b2-cos\u03b1sin\u03b2
tan\uff08\u03b1+\u03b2\uff09=\uff08tan\u03b1+tan\u03b2\uff09/\uff081-tan\u03b1tan\u03b2\uff09
\u79ef\u5316\u548c\u5dee\u516c\u5f0f\uff1acos\u03b1cos\u03b2=1/2\u3010cos\uff08\u03b1+\u03b2\uff09+cos\uff08\u03b1-\u03b2\uff09\u3011\uff0csin\u03b1sin\u03b2=-1/2\u3010cos\uff08\u03b1+\u03b2\uff09-cos\uff08\u03b1-\u03b2\uff09\u3011\uff0c
\u500d\u89d2\u516c\u5f0f\uff1acos\uff082\u03b1\uff09=cos^2(\u03b1)-sin^2(\u03b1)=2cos^2(\u03b1)-1=1-sin^2(\u03b1)
sin(2\u03b1)=2sin\u03b1cos\u03b1,tan(2\u03b1)=2tan\u03b1/(1-2tan^2(\u03b1))
\u534a\u89d2\u516c\u5f0f\uff1asin^2\uff08\u03b1/2\uff09=(1-cos\u03b1)/2,cos^2(\u03b1/2)=(1+cos\u03b1)/2,tan(\u03b1/2)=(1-cos\u03b1)/sin\u03b1=sin\u03b1/(1+cos\u03b1)
\u6b63\u5272sec\u03b1=1/cos\u03b1\uff0c\u4f59\u5272csc\u03b1=1/sin\u03b1
\u53cc\u66f2\u6b63\u5207\uff1asinhx=(e^x-e^(-x))/2\uff0ccoshx=(e^x+e^(-x))/2\uff0ctanhx=sinhx/coshx
\u4e0d\u77e5\u9053\u8fd9\u4e9b\u662f\u4e0d\u662f\u4f60\u60f3\u95ee\u7684\uff0c\u5982\u679c\u662f\u53cd\u4e09\u89d2\u51fd\u6570\u7684\u8bdd\uff0c\u5c31\u4e0d\u77e5\u9053\u4f60\u60f3\u77e5\u9053\u4ec0\u4e48\u4e86
sin\uff08\u03c0/2-\u03b1\uff09=cos\u03b1,cos\uff08\u03c0/2-\u03b1\uff09=sin\u03b1,sin\uff08\u03c0/2+\u03b1\uff09=sin\u03b1,cos\uff08\u03c0/2+\u03b1\uff09=-cos\u03b1\uff1b
\u548c\u5dee\u5316\u79ef\u516c\u5f0f\uff1acos\uff08\u03b1+\u03b2\uff09=cos\u03b1cos\u03b2-sin\u03b1sin\u03b2,cos\uff08\u03b1-\u03b2\uff09=cos\u03b1cos\u03b2+sin\u03b1sin\u03b2,sin\uff08\u03b1+\u03b2\uff09=sin\u03b1cos\u03b2+cos\u03b1sin\u03b2,sin\uff08\u03b1-\u03b2\uff09=sin\u03b1cos\u03b2-cos\u03b1sin\u03b2
tan\uff08\u03b1+\u03b2\uff09=\uff08tan\u03b1+tan\u03b2\uff09/\uff081-tan\u03b1tan\u03b2\uff09
\u79ef\u5316\u548c\u5dee\u516c\u5f0f\uff1acos\u03b1cos\u03b2=1/2\u3010cos\uff08\u03b1+\u03b2\uff09+cos\uff08\u03b1-\u03b2\uff09\u3011,sin\u03b1sin\u03b2=-1/2\u3010cos\uff08\u03b1+\u03b2\uff09-cos\uff08\u03b1-\u03b2\uff09\u3011,
\u500d\u89d2\u516c\u5f0f\uff1acos\uff082\u03b1\uff09=cos^2(\u03b1)-sin^2(\u03b1)=2cos^2(\u03b1)-1=1-sin^2(\u03b1)
sin(2\u03b1)=2sin\u03b1cos\u03b1,tan(2\u03b1)=2tan\u03b1/(1-2tan^2(\u03b1))
\u534a\u89d2\u516c\u5f0f\uff1asin^2\uff08\u03b1/2\uff09=(1-cos\u03b1)/2,cos^2(\u03b1/2)=(1+cos\u03b1)/2,tan(\u03b1/2)=(1-cos\u03b1)/sin\u03b1=sin\u03b1/(1+cos\u03b1)
\u6b63\u5272sec\u03b1=1/cos\u03b1,\u4f59\u5272csc\u03b1=1/sin\u03b1
\u53cc\u66f2\u6b63\u5207\uff1asinhx=(e^x-e^(-x))/2,coshx=(e^x+e^(-x))/2,tanhx=sinhx/coshx
\u4e0d\u77e5\u9053\u8fd9\u4e9b\u662f\u4e0d\u662f\u4f60\u60f3\u95ee\u7684,\u5982\u679c\u662f\u53cd\u4e09\u89d2\u51fd\u6570\u7684\u8bdd,\u5c31\u4e0d\u77e5\u9053\u4f60\u60f3\u77e5\u9053\u4ec0\u4e48\u4e86
和差化积公式：cos（α+β）=cosαcosβ-sinαsinβ，cos（α-β）=cosαcosβ+sinαsinβ，sin（α+β）=sinαcosβ+cosαsinβ，sin（α-β）=sinαcosβ-cosαsinβ
tan（α+β）=（tanα+tanβ）/（1-tanαtanβ）
积化和差公式：cosαcosβ=1/2【cos（α+β）+cos（α-β）】，sinαsinβ=-1/2【cos（α+β）-cos（α-β）】，
倍角公式：cos（2α）=cos^2(α)-sin^2(α)=2cos^2(α)-1=1-sin^2(α)
sin(2α)=2sinαcosα,tan(2α)=2tanα/(1-2tan^2(α))
半角公式：sin^2（α/2）=(1-cosα)/2,cos^2(α/2)=(1+cosα)/2,tan(α/2)=(1-cosα)/sinα=sinα/(1+cosα)
正割secα=1/cosα，余割cscα=1/sinα
双曲正切：sinhx=(e^x-e^(-x))/2，coshx=(e^x+e^(-x))/2，tanhx=sinhx/coshx
不知道这些是不是你想问的，如果是反三角函数的话，就不知道你想知道什么了
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绛旓細瑕佽瘉arcsinx+arccosx=蟺/2 arcsinx=蟺/2-arccosx 2杈瑰彇姝ｅ鸡 宸﹁竟=sin(arcsinx)=x 鍙宠竟=sin(蟺/2-arccosx)=cos(arccosx)=x (鍒╃敤浜唖inx=cos(蟺/2-x))宸﹁竟=鍙宠竟 鍗宠瘉
