Half-Angle Formulae - Derivation

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

$\displaystyle 2\alpha = \theta$ လို႔ ထားလိုက္မယ္။ ဒါဆိုရင္ $\displaystyle \theta =\frac{\alpha}{2}$ ေပါ့...။

မူလညီမွ်ျခင္းမွာ အစားသြင္းလိုက္ ရင္ ...

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

ဒါ့ေၾကာင့္ . . .

 $\displaystyle \sin \frac{\theta }{2}=\pm \sqrt{{\frac{{1-\cos \theta }}{2}}}$

$\displaystyle \cos 2\alpha =2{{\cos }^{2}}\alpha -1$ လို႔လည္း သိထားခဲ့ၿပီးသား မဟုတ္လား . . .။

အထက္မွာ ေျပာခဲ့တဲ့အတိုင္း $\displaystyle 2\alpha =\theta ,\alpha =\frac{\theta }{2}$ ကို အစားသြင္းလိုက္ရင္ ...

$\displaystyle \begin{array}{l}2{{\cos }^{2}}\displaystyle \frac{\theta }{2}=1+\cos \theta \\\\{{\cos }^{2}}\displaystyle \frac{\theta }{2}=\displaystyle \frac{{1+\cos \theta }}{2}\end{array}$

ဒါ့ေၾကာင့္ . . .

 $\displaystyle \cos \frac{\theta }{2}=\pm \sqrt{{\frac{{1+\cos \theta }}{2}}}$

$\displaystyle \sin \frac{\theta }{2}$ နဲ႕ $\displaystyle \cos \frac{\theta }{2}$ ကို သိၿပီဆိုေတာ့ ....

$\displaystyle \tan \displaystyle \frac{\theta }{2}= \displaystyle \frac{{\sin \displaystyle \frac{\theta }{2}}}{{\cos \displaystyle \frac{\theta }{2}}}$ ဆိုတဲ့ basic identity ကို သံုးၿပီး ဆက္ရွာလို႔ရၿပီေပါ့။

$\displaystyle \begin{array}{*{20}{l}} {\tan \displaystyle \frac{\theta }{2}=\displaystyle \frac{{\sin \displaystyle \frac{\theta }{2}}}{{\cos \displaystyle \frac{\theta }{2}}}} \\ {} \\ {\tan \displaystyle \frac{\theta }{2}=\pm \displaystyle \frac{{\sqrt{{\displaystyle \frac{{1-\cos \theta }}{2}}}}}{{\sqrt{{\displaystyle \frac{{1+\cos \theta }}{2}}}}}} \\ {} \\ {\tan \displaystyle \frac{\theta }{2}=\pm \sqrt{{\displaystyle \frac{{\displaystyle \frac{{1-\cos \theta }}{2}}}{{\displaystyle \frac{{1+\cos \theta }}{2}}}}}} \end{array}$

ဒါ့ေၾကာင့္ . . .

 $\displaystyle \tan \displaystyle \frac{\theta }{2}=\pm \sqrt{{\displaystyle \frac{{1-\cos \theta }}{{1+\cos \theta }}}}$

ဆက္ၿပီး derive လုပ္ၾကည့္မယ္ ...။

$\displaystyle \begin{array}{*{20}{l}} {\tan \displaystyle \frac{\theta }{2}=\sqrt{{\displaystyle \frac{{1-\cos \theta }}{{1+\cos \theta }}\times \displaystyle \frac{{1+\cos \theta }}{{1+\cos \theta }}}}} \\ {} \\ {\tan \displaystyle \frac{\theta }{2}=\sqrt{{\displaystyle \frac{{1-{{{\cos }}^{2}}\theta }}{{{{{(1+\cos \theta )}}^{2}}}}}}} \\ {} \\ {\tan \displaystyle \frac{\theta }{2}=\sqrt{{\displaystyle \frac{{{{{\sin }}^{2}}\theta }}{{{{{(1+\cos \theta )}}^{2}}}}}}} \end{array}$

ဒါ့ေၾကာင့္ . . .

 $\displaystyle \tan \frac{\theta }{2}=\frac{{\sin \theta }}{{1+\cos \theta }}$

တဖန္ . . .

$\displaystyle \begin{array}{*{20}{l}} {\tan \displaystyle \frac{\theta }{2}=\sqrt{{\displaystyle \frac{{1-\cos \theta }}{{1+\cos \theta }}\times \displaystyle \frac{{1-\cos \theta }}{{1-\cos \theta }}}}} \\ {} \\ {\tan \displaystyle \frac{\theta }{2}=\sqrt{{\displaystyle \frac{{{{{(1-\cos \theta )}}^{2}}}}{{1-{{{\cos }}^{2}}\theta }}}}} \\ {} \\ {\tan \displaystyle \frac{\theta }{2}=\sqrt{{\displaystyle \frac{{{{{(1-\cos \theta )}}^{2}}}}{{{{{\sin }}^{2}}\theta }}}}} \end{array}$

ဒါ့ေၾကာင့္ . . .

 $\displaystyle \tan \frac{\theta }{2}=\frac{{1-\cos \theta }}{{\sin \theta }}$