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三角恒等变换公式

2022-07-31 14:13:00 【AoBeiChuan】

1.和角公式

  1. sin ( A + B ) = sin A cos B + cos A sin B \text{sin}{(A+B)}=\text{sin}A\text{cos}B+\text{cos}A\text{sin}B sin(A+B)=sinAcosB+cosAsinB
  2. cos ( A + B ) = cos A cos B − sin A sin B \text{cos}{(A+B)}=\text{cos}A\text{cos}B-\text{sin}A\text{sin}B cos(A+B)=cosAcosB−sinAsinB
  3. tan ( A + B ) = t a n A + t a n B 1 − t a n A t a n B \text{tan}(A+B)=\cfrac{tanA+tanB}{1-tanAtanB} tan(A+B)=1−tanAtanBtanA+tanB​

2.差角公式

  1. sin ( A − B ) = sin A cos B − cos A sin B \text{sin}{(A-B)}=\text{sin}A\text{cos}B-\text{cos}A\text{sin}B sin(A−B)=sinAcosB−cosAsinB
  2. cos ( A − B ) = cos A cos B + sin A sin B \text{cos}{(A-B)}=\text{cos}A\text{cos}B+\text{sin}A\text{sin}B cos(A−B)=cosAcosB+sinAsinB
  3. tan ( A − B ) = t a n A − t a n B 1 + t a n A t a n B \text{tan}(A-B)=\cfrac{tanA-tanB}{1+tanAtanB} tan(A−B)=1+tanAtanBtanA−tanB​

3.倍角公式

  1. sin 2 a = 2 sin a cos a \text{sin}{2a}=2\text{sin}a\text{cos}a sin2a=2sinacosa
  2. cos 2 a = cos 2 a − sin 2 a = 2 cos 2 a − 1 = 1 − 2 sin 2 a \text{cos}{2a}=\text{cos}^2a-\text{sin}^2a=2\text{cos}^2a-1=1-2\text{sin}^2a cos2a=cos2a−sin2a=2cos2a−1=1−2sin2a
  3. tan 2 a = 2 tan a 1 − tan 2 a \text{tan}{2a}=\cfrac{2\text{tan}a}{1-\text{tan}^2a} tan2a=1−tan2a2tana​

4.辅助角公式

  1. a sin x + b cos x = ( a 2 + b 2 sin ( x + ψ ) a\text{sin}x+b\text{cos}x=\sqrt{\mathstrut {a^2+b^2}}\text{sin}(x+\psi) asinx+bcosx=(a2+b2​sin(x+ψ)
    其中: cos ψ = 1 ( a 2 + b 2 \text{cos}{\psi}=\cfrac{1}{\sqrt{\mathstrut {a^2+b^2}}} cosψ=(a2+b2​1​

5.积化和差公式

  1. sin A cos B = 1 2 [ sin ( A + B ) + sin ( A − B ) ] \text{sin}A\text{cos}B=\cfrac{1}{2}[\text{sin}(A+B)+\text{sin}(A-B)] sinAcosB=21​[sin(A+B)+sin(A−B)]
  2. cos A cos B = 1 2 [ cos ( A + B ) + cos ( A − B ) ] \text{cos}A\text{cos}B=\cfrac{1}{2}[\text{cos}(A+B)+\text{cos}(A-B)] cosAcosB=21​[cos(A+B)+cos(A−B)]
  3. sin A sin B = 1 2 [ cos ( A + B ) − cos ( A − B ) ] \text{sin}A\text{sin}B=\cfrac{1}{2}[\text{cos}(A+B)-\text{cos}(A-B)] sinAsinB=21​[cos(A+B)−cos(A−B)]

6.和差化积公式

  1. sin A + sin B = 2 sin A + B 2 cos A − B 2 \text{sin}A+\text{sin}B=2\text{sin}\cfrac{A+B}{2}\text{cos}\cfrac{A-B}{2} sinA+sinB=2sin2A+B​cos2A−B​
  2. sin A − sin B = 2 cos A + B 2 sin A − B 2 \text{sin}A-\text{sin}B=2\text{cos}\cfrac{A+B}{2}\text{sin}\cfrac{A-B}{2} sinA−sinB=2cos2A+B​sin2A−B​
  3. cos A + cos B = 2 cos A + B 2 cos A − B 2 \text{cos}A+\text{cos}B=2\text{cos}\cfrac{A+B}{2}\text{cos}\cfrac{A-B}{2} cosA+cosB=2cos2A+B​cos2A−B​
  4. cos A − cos B = 2 sin A + B 2 sin A − B 2 \text{cos}A-\text{cos}B=2\text{sin}\cfrac{A+B}{2}\text{sin}\cfrac{A-B}{2} cosA−cosB=2sin2A+B​sin2A−B​
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本文为[AoBeiChuan]所创，转载请带上原文链接，感谢
https://blog.csdn.net/kitagawa_myouya/article/details/125933033

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