$ \displaystyle ΔABC$ မွာ $ \displaystyle ∠CAB$ ကို $ \displaystyle \alpha$ လို႔ သတ္မွတ္ပါမယ္။
$ \displaystyle ΔABC\simΔCDF$ ျဖစ္တာေၾကာင့္ $ \displaystyle ∠CDF=\alpha$ ျဖစ္ပါတယ္။
$ \displaystyle ΔACD$ မွာေတာ့ $ \displaystyle ∠CAD$ ကို $ \displaystyle \beta$ လို႔ သတ္မွတ္ပါမယ္။
ဒါဆိုရင္ $ \displaystyle ΔABC$ မွာ...
$ \displaystyle \sin \alpha=\frac{BC}{AC}$ နဲ႕ $ \displaystyle \cos \alpha=\frac{AB}{AC}$ ျဖစ္ပါတယ္။
ဒါ့ေၾကာင့္ $ \displaystyle BC =AC \sin \alpha$ နဲ႕ $ \displaystyle AB =AC \cos \alpha$ လို႔ ဆိုႏိုင္ပါတယ္။
တဖန္ $ \displaystyle ΔCDF$ မွာ...
$ \displaystyle \sin \alpha=\frac{FC}{DC}$ နဲ႕ $ \displaystyle \cos \alpha=\frac{DF}{DC}$ ျဖစ္ပါတယ္။
ဒီမွာလည္း $ \displaystyle FC =DC \sin \alpha$ နဲ႕ $ \displaystyle DF =DC \cos \alpha$ လို႔ ဆိုႏိုင္ပါတယ္။
$ \displaystyle ΔACD$ မွာလည္း ...
$ \displaystyle \sin \beta=\frac{DC}{AD}$ နဲ႕ $ \displaystyle \cos \beta=\frac{AC}{AD}$ ျဖစ္ပါတယ္။
ဒါဆိုရင္ $ \displaystyle DC =AD \sin \alpha$ နဲ႕ $ \displaystyle AC =AD \cos \alpha$ လို႔ ဆိုႏိုင္ပါတယ္။
$ \displaystyle ΔADE$ အတြက္ ဆက္ၾကည့္ရေအာင္...
$ \displaystyle \sin (\alpha+\beta)=\frac{DE}{AD}$ ျဖစ္ပါတယ္။
ဒါ့ေၾကာင့္ $ \displaystyle \sin (\alpha +\beta )=\frac{{DF}}{{AD}}+\frac{{FE}}{{AD}}$ လို႔ ေျပာႏိုင္ပါတယ္။ ။
တဖန္ ့ $ \displaystyle BCFE$ က rectangle ျဖစ္တာေၾကာင့္ $ \displaystyle FE=BC$ လို႔ ေျပာႏိုင္ျပန္ပါတယ္။ ။
ဒါ့ေၾကာင့္ $ \displaystyle \sin (\alpha +\beta )=\frac{{DF}}{{AD}}+\frac{{BC}}{{AD}}$ လို႔ ေျပာႏိုင္ျပန္ပါတယ္။။
$ \displaystyle DF =DC \cos \alpha, BC =AC \sin \alpha$ လို႔ အထက္မွာ သိခဲ့ၿပီးပါၿပီ။ ဒါဆိုရင္ ။
$ \displaystyle \sin (\alpha +\beta )=\frac{{DC}}{{AD}}\cos \alpha +\frac{{AC}}{{AD}}\sin \alpha $ လို႔ ေျပာလို႔ရတာေပါ့။ ။
ဒီအခါမွာလည္း $ \displaystyle \sin \beta=\frac{DC}{AD}, \cos \beta=\frac{AC}{AD}$ လို႕သိခဲ့ၿပီးပါၿပီ။ ဒါေၾကာင့္။
$ \displaystyle \sin (\alpha +\beta )=\sin \beta \cos \alpha +\cos \beta \sin \alpha $ လို႔ ေျပာလို႔ရပါၿပီ။ ျပန္စီလိုက္ရင္ ...။
| $ \displaystyle \sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta $ |
အထက္ပါအတိုင္း ... $ \displaystyle \cos (\alpha +\beta )$ အတြက္ ပံုေသနည္းကို ဆက္ရွာႏိုင္ပါတယ္။ ..။
$ \displaystyle \begin{array}{l}\cos (\alpha +\beta )=\displaystyle \frac{{AE}}{{AD}}\\\\\cos (\alpha +\beta )=\displaystyle \frac{{AB-EB}}{{AD}}\\\\\cos (\alpha +\beta )=\displaystyle \frac{{AB-FC}}{{AD}}\ \ \ \ \left[ {\because EB=FC} \right]\\\\\cos (\alpha +\beta )=\displaystyle \frac{{AB}}{{AD}}-\displaystyle \frac{{FC}}{{AD}}\\\\\cos (\alpha +\beta )=\displaystyle \frac{{AC}}{{AD}}\cos \alpha -\displaystyle \frac{{DC}}{{AD}}\sin \alpha \\\\\text{Since}\ \displaystyle \frac{{AC}}{{AD}}=\cos \beta \ \operatorname{and}\ \displaystyle \frac{{DC}}{{AD}}=\sin \beta ,\\\\\text{Therefore,}\end{array}$
| $ \displaystyle \cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta $ |
$ \displaystyle \sin (\alpha +\beta )$ နဲ႔ $ \displaystyle \cos (\alpha +\beta )$ ကို သိၿပီဆိုေတာ့ $ \displaystyle \tan (\alpha +\beta )$ ကို ရွာႏိုင္ၿပီေပါ့။
$ \displaystyle \begin{array}{l}\tan (\alpha +\beta )=\displaystyle \frac{{\sin (\alpha +\beta )}}{{\cos (\alpha +\beta )}}\\\\\tan (\alpha +\beta )=\displaystyle \frac{{\sin \alpha \cos \beta +\cos \alpha \sin \beta }}{{\cos \alpha \cos \beta -\sin \alpha \sin \beta }}\\\\\text{Dividing the numerator and denominator }\\\text{with}\ \cos \alpha \cos \beta ,\\\\\tan (\alpha +\beta )=\displaystyle \frac{{\displaystyle \frac{{\sin \alpha \cos \beta }}{{\cos \alpha \cos \beta }}+\displaystyle \frac{{\cos \alpha \sin \beta }}{{\cos \alpha \cos \beta }}}}{{\displaystyle \frac{{\cos \alpha \cos \beta }}{{\cos \alpha \cos \beta }}-\displaystyle \frac{{\sin \alpha \sin \beta }}{{\cos \alpha \cos \beta }}}}\\\\\text{Therefore,}\end{array}$
| $ \displaystyle \tan (\alpha +\beta )=\frac{{\tan \alpha +\tan \beta }}{{1-\tan \alpha \tan \beta }}$ |
$ \displaystyle \begin{array}{l}\ \ \ \ \sin (-\alpha )=-\sin \alpha ,\\\\\ \ \ \ \cos (-\alpha )=\cos \alpha ,\\\\\ \ \ \ \tan (-\alpha )=-\tan \alpha \\\\\ \ \ \ \sin \left( {\alpha -\beta } \right)=\sin \left[ {\alpha +(-\beta )} \right]\\\\\ \ \ \ \sin \left( {\alpha -\beta } \right)=\sin \alpha \cos (-\beta )+\cos \alpha \sin (-\beta )\\\\\text{Therefore,}\end{array}$
| $ \displaystyle \sin \left( {\alpha -\beta } \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta $ |
$ \displaystyle \begin{array}{l}\ \ \ \ \cos \left( {\alpha -\beta } \right)=\cos \left[ {\alpha +(-\beta )} \right]\\\\\ \ \ \ \cos \left( {\alpha -\beta } \right)=\cos \alpha \cos (-\beta )-\sin \alpha \sin (-\beta )\\\\\text{Therefore,}\end{array}$
| $ \displaystyle \cos \left( {\alpha -\beta } \right)=\cos \alpha \cos \beta +\sin \alpha \sin \beta $ |
$ \displaystyle \begin{array}{l}\ \ \ \ \tan \left( {\alpha -\beta } \right)=\tan \left[ {\alpha +(-\beta )} \right]\\\\\ \ \ \ \tan \left( {\alpha -\beta } \right)=\displaystyle \frac{{\tan \alpha +\tan (-\beta )}}{{1-\tan \alpha \tan (-\beta )}}\end{array}$
| $ \displaystyle \tan \left( {\alpha -\beta } \right)=\frac{{\tan \alpha -\tan \beta }}{{1+\tan \alpha \tan \beta }}$ |
အားလံုးျပန္ေပါင္းရရင္....
| $ \displaystyle \begin{array}{l} \sin \left( {\alpha \pm \beta } \right)=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta \\\\ \cos \left( {\alpha \pm \beta } \right)=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta \\\\ \tan \left( {\alpha \pm \beta } \right)=\displaystyle \frac{{\tan \alpha \pm \tan (-\beta )}}{{1\mp \tan \alpha \tan (-\beta )}}\end{array}$ |
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