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 }}$ |
စာဖတ်သူ၏ အမြင်ကို လေးစားစွာစောင့်မျှော်လျက်!