三角函式/愛好者/不用正弦

想法

我們知道

$\sin ^{2}\theta =1-\cos ^{2}\theta \,$

那麼我們真的需要$\sin \,$函式嗎？

或者換句話說，我們是否可以用 $\cos \,$來計算出像 $\cos(a+b)\,$這樣的所有有趣的公式，然後從那裡推匯出所有包含$\sin \,$的公式？

答案是肯定的。

我們不需要為$\cos(a+b)\,$進行一個幾何論證，然後對$\sin(a+b)\,$進行另一個幾何論證。我們可以直接從 $\cos \,$ 的公式中得到關於 $\sin \,$ 的公式。

角加減公式

To find a formula for $\cos(\theta _{1}+\theta _{2})\,$ in terms of $\cos \left(\theta _{1}\right)$ and $\cos \left(\theta _{2}\right)$: construct two different right angle triangles each drawn with side $c\,$ having the same length of one, but with $\theta _{1}\neq \theta _{2}$, and therefore angle $\psi _{1}\neq \psi _{2}$. Scale up triangle two so that side $a_{2}\,$ is the same length as side $c_{1}\,$. Place the triangles so that side $c_{1}\,$ is coincidental with side $a_{2}\,$, and the angles $\theta _{1}\,$ and $\theta _{2}\,$ are juxtaposed to form angle $\theta _{3}=\theta _{1}+\theta _{2}\,$ at the origin. The circumference of the circle within which triangle two is embedded (circle 2) crosses side $a_{1}\,$ at point $g\,$, allowing a third right angle to be drawn from angle $\varphi _{2}$ to point $g\,$ . Now reset the scale of the entire figure so that side $c_{2}\,$ is considered to be of length 1. Side $a_{2}\,$ coincidental with side $c_{1}\,$ will then be of length $\cos \left(\theta _{1}\right)$, and so side $a_{1}\,$ will be of length $\cos \left(\theta _{1}\right)\cdot \cos \left(\theta _{2}\right)$ in which length lies point $g\,$. Draw a line parallel to line $a_{1}\,$ through the right angle of triangle two to produce a fourth right angle triangle, this one embedded in triangle two. Triangle 4 is a scaled copy of triangle 1, because

 (1) it is right angled, and 
 (2) $\theta _{4}+(\pi -{\frac {\pi }{2}}-\theta _{1}-\theta _{2})=\varphi _{2}=\pi -{\frac {\pi }{2}}-\theta _{2}\Rightarrow \theta _{4}=\theta _{1}$. 

邊$b_{4}\,$的長度為$cos(\varphi _{2})cos(\varphi _{4})=cos(\varphi _{2})cos(\varphi _{1})$，因為 $\theta _{4}\,=\theta _{1}\,$。因此，點$g\,$位於長度

$cos(\theta _{1}+\theta _{2})=cos(\theta _{1})cos(\theta _{2})-cos(\varphi _{1})cos(\varphi _{2}),$ 其中 $\theta _{1}+\varphi _{1}=\theta _{2}+\varphi _{2}={\frac {\pi }{2}}$

這給了我們“餘弦角和公式”。

證明角和公式和二倍角公式是一致的

我們可以立即將此公式應用於將兩個相等的角相加

   $cos(2\theta )=cos(\theta +\theta )=cos(\theta )cos(\theta )-cos(\varphi )cos(\varphi )=cos(\theta )^{2}-cos(\varphi )^{2}$           (I)
    where $\theta +\varphi ={\frac {\pi }{2}}$

根據畢達哥拉斯定理，我們知道

  $a^{2}+b^{2}=c^{2}\,$

在這種情況下

   $cos(\theta )^{2}+cos(\varphi )^{2}=1^{2}$
   $\Rightarrow cos(\varphi )^{2}=1-cos(\theta )^{2}$
   where $\theta +\varphi ={\frac {\pi }{2}}$

代入 (I) 得

   $cos(2\theta )=cos(\theta )^{2}-cos(\varphi )^{2}$
   $=cos(\theta )^{2}-(1-cos(\theta )^{2})\,$
   $=2cos(\theta )^{2}-1\,$
   where $\theta +\varphi ={\frac {\pi }{4}}$

這與“餘弦二倍角和公式”相同

   $2cos(\delta )^{2}-1=cos(2\delta )\,$

畢達哥拉斯恆等式

有了這個 $sin()\,$ 函式的定義，我們可以重新表述畢達哥拉斯定理，對於一個邊長為 1 的直角三角形，從

    $cos(\theta )^{2}+cos(\varphi )^{2}=1^{2}$  where $\theta +\varphi ={\frac {\pi }{2}}$

到

    $cos(\theta )^{2}+sin(\theta )^{2}=1\,$

我們也可以從

 $cos(\theta _{1}+\theta _{2})=cos(\theta _{1})cos(\theta _{2})-cos(\varphi _{1})cos(\varphi _{2})$ where $\theta _{1}+\varphi _{1}=\theta _{2}+\varphi _{2}={\frac {\pi }{2}}$

到

 $cos(\theta _{1}+\theta _{2})=cos(\theta _{1})cos(\theta _{2})-sin(\theta _{1})sin(\theta _{2})\,$

正弦公式

我們為這個新函式 $sin()\,$ 的符號便利性付出的代價是，我們現在必須回答諸如：是否存在“正弦角和公式”之類的問題。這些問題總是可以透過採用 $cos()\,$ 形式並有選擇地替換 $cos(\theta )^{2}\,$ 為 $1-sin(\theta )^{2}\,$，然後使用代數來簡化所得方程。將此技術應用於“餘弦角和公式”得到

   $cos(\theta _{1}+\theta _{2})=cos(\theta _{1})cos(\theta _{2})-sin(\theta _{1})sin(\theta _{2})\,$
   $\Rightarrow cos(\theta _{1}+\theta _{2})^{2}=(cos(\theta _{1})cos(\theta _{2})-sin(\theta _{1})sin(\theta _{2}))^{2}\,$
   $\Rightarrow 1-cos(\theta _{1}+\theta _{2})^{2}=1-(cos(\theta _{1})cos(\theta _{2})-sin(\theta _{1})sin(\theta _{2}))^{2}\,$
   $\Rightarrow sin(\theta _{1}+\theta _{2})^{2}=1-(cos(\theta _{1})^{2}cos(\theta _{2})^{2}+sin(\theta _{1})^{2}sin(\theta _{2})^{2}-2cos(\theta _{1})cos(\theta _{2})sin(\theta _{1})sin(\theta _{2}))\,$
   -- Pythagoras on left, multiply out right hand side

                     $=1-(cos(\theta _{1})^{2}(1-sin(\theta _{2})^{2})+sin(\theta _{1})^{2}(1-cos(\theta _{2})^{2})-2cos(\theta _{1})cos(\theta _{2})sin(\theta _{1})sin(\theta _{2}))\,$
                     -- Carefully selected Pythagoras again on the left hand side

                     $=1-(cos(\theta _{1})^{2}-cos(\theta _{1})^{2}sin(\theta _{2})^{2}+sin(\theta _{1})^{2}-sin(\theta _{1})^{2}cos(\theta _{2})^{2}-2cos(\theta _{1})cos(\theta _{2})sin(\theta _{1})sin(\theta _{2}))\,$
                     -- Multiplied out

                     $=1-(1-cos(\theta _{1})^{2}sin(\theta _{2})^{2}-sin(\theta _{1})^{2}cos(\theta _{2})^{2}-2cos(\theta _{1})cos(\theta _{2})sin(\theta _{1})sin(\theta _{2}))\,$
                     -- Carefully selected Pythagoras 

                     $=cos(\theta _{1})^{2}sin(\theta _{2})^{2}+sin(\theta _{1})^{2}cos(\theta _{2})^{2}+2cos(\theta _{1})cos(\theta _{2})sin(\theta _{1})sin(\theta _{2})\,$
                     -- Algebraic simplification

                     $=(cos(\theta _{1})sin(\theta _{2})+sin(\theta _{1})cos(\theta _{2}))^{2}\,$

兩邊開平方得到“正弦角和公式”

$\Rightarrow sin(\theta _{1}+\theta _{2})=cos(\theta _{1})sin(\theta _{2})+sin(\theta _{1})cos(\theta _{2})$

我們可以使用類似的技術從“餘弦半形公式”中找到“正弦半形公式”

$cos\left({\frac {\theta }{2}}\right)={\sqrt {\frac {1+cos(\theta )}{2}}}$

我們知道 $1-cos\left({\frac {\theta }{2}}\right)^{2}=sin\left({\frac {\theta }{2}}\right)^{2}$，所以“餘弦半形公式”兩邊平方並從 1 中減去

 $\Rightarrow 1-cos\left({\frac {\theta }{2}}\right)^{2}=1-{\frac {1+cos(\theta )}{2}}$
 $\Rightarrow sin\left({\frac {\theta }{2}}\right)^{2}={\frac {1-cos(\theta )}{2}}$
 $\Rightarrow sin\left({\frac {\theta }{2}}\right)={\sqrt {\frac {1-cos(\theta )}{2}}}$

到目前為止一切順利，但我們仍然有一個 ${\frac {cos(\theta )}{2}}$ 需要去掉。再次使用畢達哥拉斯定理得到“正弦半形公式”

         $sin\left({\frac {\theta }{2}}\right)={\sqrt {\frac {1-{\sqrt {1-sin(\theta )^{2}}}}{2}}}$  

或者更清晰一點

         $sin\left({\frac {\theta }{2}}\right)={\sqrt {\frac {1}{2}}}{\sqrt {1-{\sqrt {1-sin(\theta )^{2}}}}}$