三角恒等变换公式

诱导公式:奇变偶不变,符号看象限

六边形:

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其中有三组关系:

  • 边上的三角函数两边相乘等于中间
  • 染了色的三角形上面两个三角函数的平方和等于下面的
  • 相对的三角函数是倒数关系

和差角公式:

  • \(\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\sin\beta\cos\alpha\)
  • \(\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta\)
  • \(\tan(\alpha\pm\beta)=\dfrac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha\tan\beta}\)

二倍角公式:

  • \(\sin2\alpha=2\sin\alpha\cos\alpha\)
  • \(\cos2\alpha=\cos^2\alpha-\sin^2\alpha=2\cos^2\alpha-1=1-2\sin^2\alpha\)
  • \(\tan2\alpha=\dfrac{2\tan\alpha}{1-\tan^2\alpha}\)

三倍角公式:

  • \(\sin3\alpha=3\sin\alpha\cos^2\alpha-\sin^3\alpha\)
  • \(\cos3\alpha=\cos^3\alpha-3\sin^2\alpha\cos\alpha\)

半角公式:

  • \(\sin\dfrac{\alpha}2=\pm\sqrt{\dfrac{1-\cos\alpha}2}\),符号看象限
  • \(\cos\dfrac{\alpha}2=\pm\sqrt{\dfrac{1+\cos\alpha}2}\),符号看象限
  • \(\tan\dfrac{\alpha}2=\pm\sqrt{\dfrac{1-\cos\alpha}{1+\cos\alpha}}\),符号看象限
  • \(\tan\dfrac{\alpha}2=\dfrac{\sin\alpha}{1+\cos\alpha}=\dfrac{1-\cos\alpha}{\sin\alpha}\)

点鞭炮公式:

\[\cos\theta\cos2\theta\cos4\theta\cdots\cos2^n\theta=\prod_{i=0}^n\cos2^i\theta=\dfrac{\sin2^{n+1}\alpha}{2^{n+1}\sin\alpha} \]

降幂公式:

  • \(\sin\alpha\cos\alpha=\dfrac{\sin2\alpha}2\)
  • \(\sin^2\alpha=\dfrac{1-\cos2\alpha}2\)
  • \(\cos^2\alpha=\dfrac{1+\cos2\alpha}2\)

\(\lambda\) 等分点:

\(\overrightarrow{p_1}\overrightarrow p=\lambda\overrightarrow p\overrightarrow{p_2}\),其中 \(p_1(a_1,a_2,\cdots,a_n),p_2(b_1,b_2,\cdots,b_n)\)

\[p=\left(\dfrac{a_1+\lambda b_1}{1+\lambda},\dfrac{a_2+\lambda b_2}{1+\lambda},\cdots,\dfrac{a_n+\lambda b_n}{1+\lambda}\right) \]

辅助角公式:

\[a\sin\theta+b\cos\theta=\sqrt{a^2+b^2}\sin(\theta+\varphi) \]

其中 \(\tan\varphi=\dfrac ba\)

和差化积 & 积化和差 公式:

  • \(\sin\alpha\cos\beta=\dfrac 12(\sin(\alpha+\beta)+\sin(\alpha-\beta))\)
  • \(\cos\alpha\sin\beta=\dfrac 12(\sin(\alpha+\beta)-\sin(\alpha-\beta))\)
  • \(\cos\alpha\cos\beta=\dfrac 12(\cos(\alpha+\beta)+\cos(\alpha-\beta))\)
  • \(\sin\alpha\sin\beta=-\dfrac 12(\cos(\alpha+\beta)-\cos(\alpha-\beta))\)
  • \(\sin\theta+\sin\varphi=2\sin\dfrac{\theta+\varphi}2\cos\dfrac{\theta-\varphi}2\)
  • \(\sin\theta-\sin\varphi=2\cos\dfrac{\theta+\varphi}2\sin\dfrac{\theta-\varphi}2\)
  • \(\cos\theta+\cos\varphi=2\cos\dfrac{\theta+\varphi}2\cos\dfrac{\theta-\varphi}2\)
  • \(\cos\theta-\cos\varphi=-2\sin\dfrac{\theta+\varphi}2\sin\dfrac{\theta-\varphi}2\)
posted @ 2020-11-07 21:23  yspm  阅读(8278)  评论(11)    收藏  举报
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