Double Angle Formulae - Derivation

$ \displaystyle \sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta $ ဆိုတာ သိခဲ့ၿပီး ျဖစ္မယ္ ထင္ပါတယ္။

ဒီ ပံုေသနည္းဟာ မည္သည့္ေထာင့္ $ \displaystyle \alpha$ နဲ႔ $ \displaystyle \beta$ အတြက္မဆို မွန္ပါတယ္။

ဒါဆိုရင္ $ \displaystyle \alpha=\beta$ အတြက္လည္း မွန္တာေပါ့။... ဒါေၾကာင့္

$ \displaystyle \sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta $

$ \displaystyle \alpha=\beta,$ ျဖစ္တဲ့အခါ

$ \displaystyle \sin (\alpha +\alpha )=\sin \alpha \cos \alpha +\cos \alpha \sin \alpha $

ဒါ့ေၾကာင့္

$ \displaystyle \sin 2\alpha =2\sin \alpha \cos \alpha $


အလားတူပါပဲ.....။

$ \displaystyle \cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta $

$ \displaystyle \alpha=\beta,$ ျဖစ္တဲ့အခါ

$ \displaystyle \cos (\alpha +\alpha )=\cos \alpha \cos \alpha -\sin \beta \sin \beta $

ဒါ့ေၾကာင့္...

$ \displaystyle \cos 2\alpha ={{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha $


$ \displaystyle {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$ ဆိုတဲ့ Pythagorean Identity ကို မွတ္မိမယ္ ထင္ပါတယ္။

$ \displaystyle {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$ ျဖစ္တာေၾကာင့္ $ \displaystyle {{\sin }^{2}}\alpha =1-{{\cos }^{2}}\alpha $ နဲ႔ $ \displaystyle {{\cos }^{2}}\alpha =1-{{\sin }^{2}}\alpha $ ျဖစ္ပါတယ္။

$ \displaystyle \cos 2\alpha ={{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha $ ဆိုတဲ့ equation မွာ သက္ဆိုင္ရာ တန္ဖိုးေတြကို အစားသြင္းလိုက္ရင္ ...

$ \displaystyle \begin{array}{l}\cos 2\alpha ={{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha \\\\\cos 2\alpha =1-{{\sin }^{2}}\alpha -{{\sin }^{2}}\alpha \end{array}$

ဒါ့ေၾကာင့္...

$ \displaystyle \cos 2\alpha =1-2{{\sin }^{2}}\alpha $


အလားတူပါပဲ...။

$ \displaystyle \begin{array}{l}\cos 2\alpha ={{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha \\\\\cos 2\alpha ={{\cos }^{2}}\alpha -(1-{{\cos }^{2}}\alpha )\end{array}$

ဒါ့ေၾကာင့္...

$ \displaystyle \cos 2\alpha =2{{\cos }^{2}}\alpha -1$


$ \displaystyle \tan 2\alpha $ အတြက္ ဆက္ရွာၾကည့္ပါမယ္။

$ \displaystyle \tan (\alpha +\beta )=\frac{{\tan \alpha +\tan \beta }}{{1-\tan \alpha \tan \beta }}$ လို႔ သိခဲ့ၿပီးပါၿပီ။..

$ \displaystyle \alpha=\beta,$ ျဖစ္တဲ့အခါ

$ \displaystyle \tan (\alpha +\alpha )=\frac{{\tan \alpha +\tan \alpha }}{{1-\tan \alpha \tan \alpha }}$ ..

ဒါ့ေၾကာင့္...

$ \displaystyle \tan 2\alpha =\frac{{2\tan \alpha }}{{1-{{{\tan }}^{2}}\alpha }}$