$\def\frac{\dfrac}$ $\def\review{\begin{array}{|ll|}\hline\mbox{ Basic Trigonometric Identity (BTI) : }\\\qquad\begin{array}{rl}\sin \left(180^{\circ}-\theta\right) & =\sin \theta \\\cos \left(180^{\circ}-\theta\right) & =-\cos \theta\end{array}\\ \qquad\begin{array}{rl}\sin (90-\theta) & =\cos \theta \\\cos (90-\theta) & =\sin \theta \end{array}\\\qquad \begin{array}{rl}\sin (-\theta) & =-\sin \theta \\\cos (-\theta) & =\cos \theta \end{array}\\\text {Product to Sum Formula ( P2S):}\\ \qquad \begin{array}{rl}2 \sin \alpha \cos \beta & =\sin (\alpha+\beta)+\sin (\alpha-\beta) \\ 2 \cos \alpha \sin \beta & =\sin (\alpha+\beta)-\sin (\alpha-\beta) \\2 \cos \alpha \cos \beta &=\cos (\alpha+\beta)+\cos (\alpha-\beta) \\2 \sin \alpha \sin \beta &=\cos (\alpha-\beta)-\cos (\alpha+\beta)\end{array}\\\text{Triangular Note (TN):} \\\qquad \text{The three statements are equivalent}\\\qquad\begin{array}{ll}x+y+z=180^{\circ}\\x+y-z=180^{\circ}-2z\\x-y-z=2x-180^{\circ}\end{array}\\\text{Square Formula (SF):} \\\qquad\begin{array}{ll}\sin^2\alpha&=\dfrac{1-\cos2\alpha}{2}\\ \cos^2\alpha&=\dfrac{1+\cos2\alpha}{2}\end{array}\\\hline\end{array}}$
Problem 6: If $ A+B+C=180^{\circ}$, prove that $\sin ^{2} A+\sin ^{2} B-\sin ^{2} C=2 \sin A \sin B \cos C$ . $\review$ $\def\solvi{\begin{array}{lll}\text{R H S} &=(2 \sin A \sin B) \cos C \\&=(\cos (A-B)-\cos (A+B)) \cos C &\mbox{(P2S)}\\&=\cos (A-B) \cos C-\cos (A+B) \cos C \\&=\dfrac{1}{2}[\cos (A-B+C)+\cos (A-B-C)]\\&\quad-\frac{1}{2}[\cos (A+B+C)+\cos (A+B-C)]&\mbox{(P2S)} \\&=\dfrac{1}{2}\left[\cos \left(180^{\circ}-2 B\right)+\cos \left(2 A-180^{\circ}\right)\right.\\&\left.\quad-\cos 180^{\circ}-\cos \left(180^{\circ}-2 C\right)\right] &\mbox{(TN)}\\&=\dfrac{1}{2}[-\cos 2 B-\cos 2 A-(-1)+\cos 2 C] &\mbox{(BTI)}\\&=\dfrac 12\left(1-\cos2A-\cos2B +\cos2C\right)\\\text { LHS } &=\sin ^{2} A+\sin ^{2} B-\sin ^{2} C \\&=\dfrac{1-\cos 2 A}{2}+\dfrac{1-\cos 2 B}{2}-\dfrac{1-\cos 2 C}{2}&\mbox{(SF)} \\&=\dfrac{1}{2}[1-\cos 2 A-\cos 2 B+\cos 2 C]\end{array}}$
Problem 7: If $ A+B+C=180^{\circ} $, show that $ \cos ^{2} A+\cos ^{2} B+\cos ^{2} C=1-2 \cos A \cos B \cos C .$ $\review$ $\def\solvii{\begin{array}{lll}\mbox{RHS} &=1-(2 \cos A \cos B) \cos C \\&=1-(\cos (A+B)+\cos (A-B)) \cos C &\mbox{(P2S)} \\&=1-\cos (A+B) \cos C-\cos (A-B) \cos C \\&=1-\frac{1}{2}[\cos (A+B+C)+\cos (A+B-C)]\\&\quad-\frac{1}{2}[\cos (A-B+C)+\cos (A-B-C)] &\mbox{(P2S)}\\&=1-\frac{1}{2}\left(\cos 180^{\circ}+\cos (180-2 C)\right)\\&\quad-\frac{1}{2}(\cos (180-2 B)+\cos (2 A-180))&\mbox{(TN)} \\&=1-\frac{1}{2}(-1-\cos 2 C-\cos 2 B-\cos 2 A) &\mbox{(BTI)}\\&=\frac{1}{2}(3+\cos 2 A+\cos 2 B+\cos 2 C) \\\mbox{LHS} &=\cos ^{2} A+\cos ^{2} B+\cos ^{2} C \\&=\frac{1+\cos 2 A}{2}+\frac{1+\cos 2 B}{2}+\frac{1+\cos 2 C}{2}&\mbox{(SF)} \\&=\frac{1}{2}(3+\cos 2 A+\cos 2 B+\cos 2 C)\end{array}}$
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