Trigonometric Problem (P2S Method)

 $\begin{array}{|ll|}\hline\mbox{ Basic Trigonometric Identity  (BTI) : }\\\qquad\begin{array}{rl}\sin \left(180^{\circ}-\theta\right) & =\sin \theta \\\cos \left(180^{\circ}-\theta\right) & =-\cos \theta\end{array}\\ \qquad\begin{array}{rl}\sin (90-\theta) & =\cos \theta \\\cos (90-\theta) & =\sin \theta \end{array}\\\qquad \begin{array}{rl}\sin (-\theta) & =-\sin \theta \\\cos (-\theta) & =\cos \theta \end{array}\\\text {Product to Sum Formula ( P2S):}\\ \qquad \begin{array}{rl}2 \sin \alpha \cos \beta & =\sin (\alpha+\beta)+\sin (\alpha-\beta) \\ 2 \cos \alpha \sin \beta & =\sin (\alpha+\beta)-\sin (\alpha-\beta) \\2  \cos \alpha \cos \beta &=\cos (\alpha+\beta)+\cos (\alpha-\beta) \\2  \sin \alpha \sin \beta &=\cos (\alpha-\beta)-\cos (\alpha+\beta)\end{array}\\\text{Triangular Note (TN):} \\\qquad \text{The three statements are equivalent}\\\qquad\begin{array}{ll}x+y+z=180^{\circ}\\x+y-z=180^{\circ}-2z\\x-y-z=2x-180^{\circ}\end{array}\\\hline\end{array}$

Problem 1:   If $ \alpha+\beta+\gamma=180^{\circ},$  prove that $ \sin 2 \alpha$ $+\sin 2 \beta$ $+\sin 2 \gamma$ $=4 \sin \alpha \sin \beta \sin \gamma .$ 

Proof:
$\begin{array}{lll}\quad& 4 \sin \alpha \sin \beta \sin \gamma \\&=(2 \sin \alpha \sin \beta)(2 \sin r) \\&=(\cos (\alpha-\beta)-\cos (\alpha+\beta))(2 \sin \gamma) &\mbox{(P2S)}\\&=2 \cos (\alpha-\beta) \sin \gamma-2 \cos (\alpha+\beta) \sin \gamma \\&=\sin (\alpha-\beta+\gamma)-\sin (\alpha-\beta-\gamma)\\&\quad -(\sin (\alpha+\beta+\gamma)-\sin (\alpha+\beta-\gamma)) &\mbox{(P2S)}\\&=\sin \left(180^{\circ}-2 \beta\right)-\sin \left(2 \alpha-180^{\circ}\right)\\&\quad -\sin \left(180^{\circ}\right)+\sin (180-2 \gamma) &\mbox{ (TN)}\\&=\sin 2 \beta+\sin 2 \alpha+\sin 2 \gamma &\mbox{(BTI)}\end{array}$

Problem 2:  If  $A+B+C=180^{\circ}$, prove that $ \cos A+\cos B+\cos C=1+4 \sin \dfrac{A}{2} \sin \dfrac{B}{2} \sin \dfrac{C}{2} \cdot$ 

Click here for solution from 00:00 to 01:21

Problem 3: If $ \alpha+\beta+\gamma=180^{\circ}$  show that $ \sin \alpha+\sin \beta+\sin\gamma=4 \cos \displaystyle\frac{\alpha}{2} \cos \displaystyle\frac{\beta}{2} \cos \displaystyle\frac{\gamma}{2}.$

Click here for solution from 01:21 to 2:53

Problem 4: If  $A+B+C=180^{\circ},$  prove that  $\sin 2 A-\sin 2 B+\sin 2 C=4 \cos A \sin B \cos C.$ 

Proof:

$\begin{array}{lll}\quad& 4 \cos A \sin B \cos C \\=&(2 \cos A \sin B)(2 \cos C) \\=&[\sin (A+B)-\sin (A-B)](2 \cos C) &\mbox{(P2S)}\\=& 2 \sin (A+B) \cos C-2 \sin(A-B) \cos C \\=&[\sin(A+B+C)+\sin (A+B-C)] \\&-[\sin (A-B+C)+\sin (A-B-C)]&\mbox{(P2S)} \\=& \sin 180^{\circ}+\sin \left(180^{\circ}-2 C\right)\\&-\sin (180-2 B)-\sin (2 A-180)&\mbox{(TN)} \\=& 0+\sin 2 C-\sin 2 B+\sin 2 A &\mbox{(BTI)}\\=& \sin 2 A-\sin 2 B+\sin 2 C .\end{array}$


Problem 5: If $ A+B+C=180,$ then prove that 

$ \cos 2 A+\cos 2 B+\cos 2 C $ $= -1 -4 \cos A \cos B \cos C $ 

Solution:

$\begin{array}{lll}\quad&4 \cos A \cos B \cos C\\ &=(2 \cos A \cos B)(2 \cos C) \\&=[\cos (A+B)+\cos (A-B)](2 \cos C)&\mbox{(P2S)} \\&=2 \cos (A+B) \cos C+2 \cos (A-B) \cos C \\&=[\cos (A+B+C)+\cos (A+B-C)]\\&\quad+[\cos (A-B+C)+\cos (A-B-C)]&\mbox{(P2S)} \\&=\cos 180^{\circ}+\cos \left(180^{\circ}-2 c\right)\\&\quad+\cos (180-2 B)+\cos (2 A-180)&\mbox{(TN)} \\&=-1-\cos 2 C-\cos 2 B-\cos 2 A &\mbox{(BTI)}\\\end{array}$

Hence $\cos 2 A+\cos 2 B+\cos 2C =-1-4 \cos A \cos B \cos C$



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