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倍角公式(II) (Double-angle Formulas)

倍角公式(II) (Double-angle Formulas)
國立蘭陽女中數學科陳敏晧老師

連結:倍角公式(I)

三角函數中的倍角公式,主要有兩類,一類是二倍角公式,一類是三倍角公式,其中二倍角公式主要有:

  1. \(\sin 2\theta= 2\sin \theta \cos \theta\)
  2. \(\cos 2\theta= {\cos ^2}\theta- {\sin ^2}\theta= 2{\cos ^2}\theta- 1 = 1 – 2{\sin ^2}\theta\)
  3. \(\displaystyle\tan 2\theta= \frac{{2\tan \theta }}{{1 – {{\tan }^2}\theta }}\)

這些公式的證明主要是利用正弦與餘弦的和差角公式:

\(\sin \left( {\alpha+ \beta } \right)= \sin \alpha\cos \beta+ \cos \alpha \sin \beta \\\cos \left( {\alpha+ \beta } \right) = \cos \alpha \cos \beta- \sin \alpha \sin \beta\)

若 \(\alpha=\beta=\theta\),則 \(\sin 2\theta= \sin \theta \cos \theta+ \cos \theta \sin \theta= 2\sin \theta \cos \theta\),

\(\begin{array}{ll}\cos 2\theta &= \cos \theta \cos \theta- \sin \theta \sin \theta\\&= {\cos ^2}\theta- {\sin ^2}\theta= {\cos ^2}\theta- (1 – {\cos ^2}\theta ) \\&= 2{\cos ^2}\theta- 1 = 2(1- {\sin ^2}\theta ) – 1 = 1 – 2{\sin ^2}\theta\end{array}\)