Trigonometry: Angles with sine, cosine and tangent
Addition formulas for trigonometric functions
Addition formulas for trigonometric functions
The sine and cosine satisfy the following rules. Here #\blue \alpha# and #\green \beta# are arbitrary numbers.
\[\sin(\blue \alpha+\green \beta)=\sin(\blue \alpha) \cos(\green \beta)+\cos(\blue \alpha)\sin(\green \beta)\]
\[\sin(\blue \alpha-\green \beta)=\sin(\blue \alpha) \cos(\green \beta)-\cos(\blue \alpha)\sin(\green \beta)\]
\[\cos(\blue \alpha+\green \beta)=\cos(\blue \alpha) \cos(\green \beta)-\sin(\blue \alpha)\sin(\green \beta)\]
\[\cos(\blue \alpha-\green \beta)=\cos(\blue \alpha) \cos(\green \beta)+\sin(\blue \alpha)\sin(\green \beta)\]
Example
\[\begin{array}{rcl}\sin(\frac{5 \pi}{12})&=&\sin(\blue{\frac{\pi}{6}}+\green{\frac{\pi}{4}})\\&=&\sin(\blue{\frac{\pi}{6}}) \cos(\green{\frac{\pi}{4}})+\\&&\cos(\blue{\frac{\pi}{6}})\sin(\green {\frac{\pi}{4}})\\&=&\frac{1}{2} \cdot \frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}}\\&=&\frac{1+\sqrt{3}}{2 \sqrt{2}} \end{array}\]
By immediately choosing #\blue \alpha# and #\green \beta# in the addition formulas, we find the so-called double-angle formulas.
Double-angle formulas
The following rules apply to the sine and cosine:
\[\sin(2 \blue \alpha)=2\sin(\blue \alpha)\cos(\blue \alpha)\]
\[\cos(2 \blue \alpha)=\cos^2(\blue \alpha)-\sin^2(\blue \alpha)=2 \cos^2(\blue \alpha)-1\]
Example
\[\begin{array}{rcl}\cos(\frac{2 \pi}{3})&=&\cos(2 \cdot \blue{\frac{\pi}{3}})\\&=&2 \cos^2(\blue{\frac{\pi}{3}})-1\\&=&2 \cdot \left(\frac{1}{2}\right)^2-1=-\frac{1}{2}\end{array}\]
If we write #\dfrac{7\pi}{12}# as #\dfrac{\pi}{3}+\dfrac{\pi}{4}# we can use the addition formula #\cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)# with #\alpha=\dfrac{\pi}{3}# and #\beta=\dfrac{\pi}{4}#:
\[\begin{array}{rcl}
\cos\left(\dfrac{7\pi}{12}\right)&=&\cos\left(\dfrac{\pi}{3}+\dfrac{\pi}{4}\right)\\
&=&\cos\left(\dfrac{\pi}{3}\right)\cos\left(\dfrac{\pi}{4}\right)-\sin\left(\dfrac{\pi}{3}\right)\sin\left(\dfrac{\pi}{4}\right)\\
&=&\dfrac{1}{2}\cdot\dfrac{1}{\sqrt{2}}-\dfrac{\sqrt{3}}{2}\cdot\dfrac{1}{\sqrt{2}}\\
&=&\dfrac{1-\sqrt{3}}{2\sqrt{2}}\tiny.
\end{array}\]
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