# Problem Study (Trigonometric Identity)

Prove that $\displaystyle 4\sin (x+30{}^\circ )\sin (x-30{}^\circ )=3-4{{\cos }^{2}}x.$

$\displaystyle \begin{array}{l}\ \ \ 4\sin (x+30{}^\circ )\sin (x-30{}^\circ )\\=4(\sin x\cos 30{}^\circ +\cos x\sin 30{}^\circ )(\sin x\cos 30{}^\circ -\cos x\sin 30{}^\circ )\ \\=4(\frac{{\sqrt{3}}}{2}\sin x+\frac{1}{2}\cos x)(\frac{{\sqrt{3}}}{2}\sin x-\frac{1}{2}\cos x)\\=4(\frac{3}{4}{{\sin }^{2}}x-\frac{{\sqrt{3}}}{4}\sin x\cos x+\frac{{\sqrt{3}}}{4}\sin x\cos x-\frac{1}{4}{{\cos }^{2}}x)\\=3{{\sin }^{2}}x-{{\cos }^{2}}x\\=3{{\sin }^{2}}x+3{{\cos }^{2}}x-3{{\cos }^{2}}x-{{\cos }^{2}}x\ \\=3({{\sin }^{2}}x+{{\cos }^{2}}x)-4{{\cos }^{2}}x\\=3(1)-4{{\cos }^{2}}x\\=3-4{{\cos }^{2}}x\end{array}$

ဒုတိယ အဆင့္ $\displaystyle 4(\sin x\cos 30{}^\circ +\cos x\sin 30{}^\circ )(\sin x\cos 30{}^\circ -\cos x\sin 30{}^\circ )\$
မွာ sum difference formula ကို သံုးလိုက္ပါတယ္။

တတိယအဆင့္မွာ special angle ရဲ့ trigonometric ratio ေတြျဖစ္တဲ့ $\displaystyle \cos 30{}^\circ =\frac{{\sqrt{3}}}{2}$ နဲ႔ $\displaystyle \sin30{}^\circ =\frac{1}{2}$  ကို သံုးပါတယ္။

အဆင့္ (5) မွာ $\displaystyle -a+a=0$ ဆိုတဲ့ identity ကို သံုးပါတယ္။

အဆင့္ (7) မွာ $\displaystyle {{\sin }^{2}}x+{{\cos }^{2}}x=1$ ဆိုတဲ့ identity ကို သံုးပါတယ္။

အဆင္ေျပပါေစ.....။