余弦正弦转换公式 三角函数正弦和余弦的转换公式?

\u6b63\u4f59\u5f26\u8f6c\u5316\u516c\u5f0f

\u8bf1\u5bfc\u516c\u5f0f\uff08\u53e3\u8bc0:\u5947\u53d8\u5076\u4e0d\u53d8,\u7b26\u53f7\u770b\u8c61\u9650.\uff09
sin\uff08\uff0d\u03b1\uff09\uff1d\uff0dsin\u03b1
cos\uff08\uff0d\u03b1\uff09\uff1dcos\u03b1 tan\uff08\uff0d\u03b1\uff09\uff1d\uff0dtan\u03b1
cot\uff08\uff0d\u03b1\uff09\uff1d\uff0dcot\u03b1
sin\uff08\u03c0/2\uff0d\u03b1\uff09\uff1dcos\u03b1
cos\uff08\u03c0/2\uff0d\u03b1\uff09\uff1dsin\u03b1
tan\uff08\u03c0/2\uff0d\u03b1\uff09\uff1dcot\u03b1
cot\uff08\u03c0/2\uff0d\u03b1\uff09\uff1dtan\u03b1
sin\uff08\u03c0/2\uff0b\u03b1\uff09\uff1dcos\u03b1
cos\uff08\u03c0/2\uff0b\u03b1\uff09\uff1d\uff0dsin\u03b1
tan\uff08\u03c0/2\uff0b\u03b1\uff09\uff1d\uff0dcot\u03b1
cot\uff08\u03c0/2\uff0b\u03b1\uff09\uff1d\uff0dtan\u03b1
sin\uff08\u03c0\uff0d\u03b1\uff09\uff1dsin\u03b1
cos\uff08\u03c0\uff0d\u03b1\uff09\uff1d\uff0dcos\u03b1
tan\uff08\u03c0\uff0d\u03b1\uff09\uff1d\uff0dtan\u03b1
cot\uff08\u03c0\uff0d\u03b1\uff09\uff1d\uff0dcot\u03b1
sin\uff08\u03c0\uff0b\u03b1\uff09\uff1d\uff0dsin\u03b1
cos\uff08\u03c0\uff0b\u03b1\uff09\uff1d\uff0dcos\u03b1
tan\uff08\u03c0\uff0b\u03b1\uff09\uff1dtan\u03b1
cot\uff08\u03c0\uff0b\u03b1\uff09\uff1dcot\u03b1
sin\uff083\u03c0/2\uff0d\u03b1\uff09\uff1d\uff0dcos\u03b1
cos\uff083\u03c0/2\uff0d\u03b1\uff09\uff1d\uff0dsin\u03b1
tan\uff083\u03c0/2\uff0d\u03b1\uff09\uff1dcot\u03b1
cot\uff083\u03c0/2\uff0d\u03b1\uff09\uff1dtan\u03b1
sin\uff083\u03c0/2\uff0b\u03b1\uff09\uff1d\uff0dcos\u03b1
cos\uff083\u03c0/2\uff0b\u03b1\uff09\uff1dsin\u03b1
tan\uff083\u03c0/2\uff0b\u03b1\uff09\uff1d\uff0dcot\u03b1
cot\uff083\u03c0/2\uff0b\u03b1\uff09\uff1d\uff0dtan\u03b1
sin\uff082\u03c0\uff0d\u03b1\uff09\uff1d\uff0dsin\u03b1
cos\uff082\u03c0\uff0d\u03b1\uff09\uff1dcos\u03b1
tan\uff082\u03c0\uff0d\u03b1\uff09\uff1d\uff0dtan\u03b1
cot\uff082\u03c0\uff0d\u03b1\uff09\uff1d\uff0dcot\u03b1
sin\uff082k\u03c0\uff0b\u03b1\uff09\uff1dsin\u03b1
cos\uff082k\u03c0\uff0b\u03b1\uff09\uff1dcos\u03b1
tan\uff082k\u03c0\uff0b\u03b1\uff09\uff1dtan\u03b1
cot\uff082k\u03c0\uff0b\u03b1\uff09\uff1dcot\u03b1
(\u5176\u4e2dk\u2208Z)
\u4e24\u89d2\u548c\u4e0e\u5dee\u7684\u4e09\u89d2\u51fd\u6570\u516c\u5f0f \u4e07\u80fd\u516c\u5f0f
sin\uff08\u03b1\uff0b\u03b2\uff09\uff1dsin\u03b1cos\u03b2\uff0bcos\u03b1sin\u03b2
sin\uff08\u03b1\uff0d\u03b2\uff09\uff1dsin\u03b1cos\u03b2\uff0dcos\u03b1sin\u03b2
cos\uff08\u03b1\uff0b\u03b2\uff09\uff1dcos\u03b1cos\u03b2\uff0dsin\u03b1sin\u03b2
cos\uff08\u03b1\uff0d\u03b2\uff09\uff1dcos\u03b1cos\u03b2\uff0bsin\u03b1sin\u03b2
tan\uff08\u03b1\uff0b\u03b2\uff09\uff1d(tan\u03b1\uff0btan\u03b2)/(1\uff0dtan\u03b1 \u00b7tan\u03b2)
\u3000\u3000\u3000\u3000\u3000 tan\u03b1\uff0dtan\u03b2
tan\uff08\u03b1\uff0d\u03b2\uff09\uff1d\u2014\u2014\u2014\u2014\u2014\u2014
\u3000\u3000\u3000\u3000\u3000\u30001\uff0btan\u03b1 \u00b7tan\u03b2
\u3000\u3000\u30002tan(\u03b1/2)
sin\u03b1\uff1d\u2014\u2014\u2014\u2014\u2014\u2014
\u3000\u3000\u3000\u30001\uff0btan2(\u03b1/2)
\u3000\u3000 1\uff0dtan^2(\u03b1/2)
cos\u03b1\uff1d\u2014\u2014\u2014\u2014\u2014\u2014
\u3000\u3000\u30001\uff0btan^2(\u03b1/2)
\u3000\u3000\u30002tan(\u03b1/2)
tan\u03b1\uff1d\u2014\u2014\u2014\u2014\u2014\u2014
\u3000\u3000\u30001\uff0dtan^2(\u03b1/2)
\u534a\u89d2\u7684\u6b63\u5f26\u3001\u4f59\u5f26\u548c\u6b63\u5207\u516c\u5f0f \u4e09\u89d2\u51fd\u6570\u7684\u964d\u5e42\u516c\u5f0f
\u4e8c\u500d\u89d2\u7684\u6b63\u5f26\u3001\u4f59\u5f26\u548c\u6b63\u5207\u516c\u5f0f \u4e09\u500d\u89d2\u7684\u6b63\u5f26\u3001\u4f59\u5f26\u548c\u6b63\u5207\u516c\u5f0f
sin2\u03b1\uff1d2sin\u03b1cos\u03b1
cos2\u03b1\uff1dcos2\u03b1\uff0dsin2\u03b1\uff1d2cos2\u03b1\uff0d1\uff1d1\uff0d2sin2\u03b1
\u3000\u3000\u3000\u30002tan\u03b1
tan2\u03b1\uff1d\u2014\u2014\u2014\u2014\u2014
\u3000\u3000\u3000 1\uff0dtan^2\u03b1
sin3\u03b1\uff1d3sin\u03b1\uff0d4sin^3\u03b1
cos3\u03b1\uff1d4cos^3\u03b1\uff0d3cos\u03b1
\u3000\u3000\u30003tan\u03b1\uff0dtan^3\u03b1
tan3\u03b1\uff1d\u2014\u2014\u2014\u2014\u2014\u2014
\u3000\u3000\u3000\u30001\uff0d3tan^2\u03b1
\u4e09\u89d2\u51fd\u6570\u7684\u548c\u5dee\u5316\u79ef\u516c\u5f0f \u4e09\u89d2\u51fd\u6570\u7684\u79ef\u5316\u548c\u5dee\u516c\u5f0f
\u3000\u3000\u3000\u3000\u3000\u3000\u3000\u3000\u3000 \u03b1\uff0b\u03b2\u3000\u3000 \u03b1\uff0d\u03b2
sin\u03b1\uff0bsin\u03b2\uff1d2sin\u2014\u2014\u2014\u00b7cos\u2014\u2014\u2014
\u3000\u3000\u3000\u3000\u3000\u3000\u3000\u3000\u3000\u3000 2\u3000\u3000\u3000\u30002
\u3000\u3000\u3000\u3000\u3000\u3000\u3000\u3000 \u03b1\uff0b\u03b2\u3000\u3000 \u03b1\uff0d\u03b2
sin\u03b1\uff0dsin\u03b2\uff1d2cos\u2014\u2014\u2014\u00b7sin\u2014\u2014\u2014
\u3000\u3000\u3000\u3000\u3000\u3000\u3000\u3000\u3000\u3000 2\u3000\u3000\u3000\u30002
\u3000\u3000\u3000\u3000\u3000\u3000\u3000\u3000\u03b1\uff0b\u03b2\u3000\u3000\u3000\u03b1\uff0d\u03b2
cos\u03b1\uff0bcos\u03b2\uff1d2cos\u2014\u2014\u2014\u00b7cos\u2014\u2014\u2014
\u3000\u3000\u3000\u3000\u3000\u3000\u3000\u3000\u3000\u30002\u3000\u3000\u3000\u3000 2
\u3000\u3000\u3000\u3000\u3000\u3000\u3000\u3000\u3000 \u03b1\uff0b\u03b2\u3000\u3000 \u03b1\uff0d\u03b2
cos\u03b1\uff0dcos\u03b2\uff1d\uff0d2sin\u2014\u2014\u2014\u00b7sin\u2014\u2014\u2014
\u3000\u3000\u3000\u3000\u3000\u3000\u30001\u3000\u3000\u3000\u30002\u3000\u3000\u3000\u3000 2\u3000\u3000\u3000\u3000
sin\u03b1 \u00b7cos\u03b2\uff1d-[sin\uff08\u03b1\uff0b\u03b2\uff09\uff0bsin\uff08\u03b1\uff0d\u03b2\uff09]
\u3000\u3000\u3000\u3000\u3000\u30002
\u3000\u3000\u3000\u3000\u3000\u30001
cos\u03b1 \u00b7sin\u03b2\uff1d-[sin\uff08\u03b1\uff0b\u03b2\uff09\uff0dsin\uff08\u03b1\uff0d\u03b2\uff09]
\u3000\u3000\u3000\u3000\u3000\u30002
\u3000\u3000\u3000\u3000\u3000\u30001
cos\u03b1 \u00b7cos\u03b2\uff1d-[cos\uff08\u03b1\uff0b\u03b2\uff09\uff0bcos\uff08\u03b1\uff0d\u03b2\uff09]
\u3000\u3000\u3000\u3000\u3000\u30002
\u3000\u3000\u3000\u3000\u3000\u3000\u3000 1
sin\u03b1 \u00b7sin\u03b2\uff1d\u2014 -[cos\uff08\u03b1\uff0b\u03b2\uff09\uff0dcos\uff08\u03b1\uff0d\u03b2\uff09]
\u3000\u3000\u3000\u3000\u3000\u3000\u3000 2

这是简单的变形。sin a=cos (pi/2-a) 然后乘以sin20再除以sin20
就变成4sin20*cos20*cos40*cos80/sin20
再利用2sina*cosa=sin2a变形得
2*sin40*cos40*cos80/sin20接着变形
sin80*cos80/sin20继续
sin160/2sin20再由sina=sin(pi-a)得
sin20/2sin20=1/2

一个角的正弦值等于这个角余角的余弦值,反之亦然。所以sin10=cos80,sin50=cos40,sin70=cos20
观察发现20度40度80度依次为前者的2倍。
2sinAcosA=sin2A
从第二行通过乘以一个2sin20再除以一个sin20,反复使用该公式将20度变为40度后变为80度从而是度数统一而简化运算

  • 浣欏鸡姝e鸡杞崲鍏紡
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