$\begin{array}{|ll|}\hline\mbox{ Basic Trigonometric Identity (BTI) : }\\\qquad\begin{array}{rl}\sin \left(180^{\circ}-\theta\right) & =\sin \theta \\\cos \left(180^{\circ}-\theta\right) & =-\cos \theta\end{array}\\ \qquad\begin{array}{rl}\sin (90-\theta) & =\cos \theta \\\cos (90-\theta) & =\sin \theta \end{array}\\\qquad \begin{array}{rl}\sin (-\theta) & =-\sin \theta \\\cos (-\theta) & =\cos \theta \end{array}\\\text {Product to Sum Formula ( P2S):}\\ \qquad \begin{array}{rl}2 \sin \alpha \cos \beta & =\sin (\alpha+\beta)+\sin (\alpha-\beta) \\ 2 \cos \alpha \sin \beta & =\sin (\alpha+\beta)-\sin (\alpha-\beta) \\2 \cos \alpha \cos \beta &=\cos (\alpha+\beta)+\cos (\alpha-\beta) \\2 \sin \alpha \sin \beta &=\cos (\alpha-\beta)-\cos (\alpha+\beta)\end{array}\\\text{Triangular Note (TN):} \\\qquad \text{The three statements are equivalent}\\\qquad\begin{array}{ll}x+y+z=180^{\circ}\\x+y-z=180^{\circ}-2z\\x-y-z=2x-180^{\circ}\end{array}\\\hline\end{array}$
Problem 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 $\cos 2 A+\cos 2 B+\cos 2C =-1-4 \cos A \cos B \cos C$
ReplyDeleteAQA A- Level Mathematics Past Papers
Post a Comment