Exercise 3.7-Additional Questions - Chapter 3 Trigonometry 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Additional Questions
Question 1.
$A+B+C=\pi$, prove that $\sin 2 A-\sin 2 B+\sin 2 C=4 \cos A \sin B \cos C$
Solution:
LHS $=\sin 2 \mathrm{~A}-\sin 2 \mathrm{~B}+\sin 2 \mathrm{C}$
$=\sin 2 \mathrm{~A}+\sin 2 \mathrm{C}-\sin 2 \mathrm{~B}$
$=2 \sin \frac{2 \mathrm{~A}+2 \mathrm{C}}{2} \cos \frac{2 \mathrm{~A}-2 \mathrm{C}}{2}-\sin 2 \mathrm{~B}$
$=2 \sin (\mathrm{A}+\mathrm{C}) \cos (\mathrm{A}-\mathrm{C})-2 \sin \mathrm{B} \cos \mathrm{B}$
$=2 \sin \left(180^{\circ}-\mathrm{B}\right) \cos (\mathrm{A}-\mathrm{C})-2 \sin \mathrm{B} \cos \mathrm{B}$
$=2 \sin \mathrm{B} \cos (\mathrm{A}-\mathrm{C})-2 \sin \mathrm{B} \cos \mathrm{B}$
$=2 \sin \mathrm{B}[\cos (\mathrm{A}-\mathrm{C})-\cos \mathrm{B}]$
$=2 \sin \mathrm{B}\left[\cos (\mathrm{A}-\mathrm{C})-\cos \left(180^{\circ}-(\mathrm{A}+\mathrm{C})\right)\right]$
$=2 \sin B[\cos (A-C)+\cos (A+C)]$
$=2 \sin B[2 \cos A \cos C]$
$=4 \cos \mathrm{A} \sin \mathrm{B} \cos \mathrm{C}=$ RHS

Question 2
Prove that $A+B+C=\pi$, prove that $\sin ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}+\sin ^2 \frac{C}{2}=1-2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$
Solution:

$\begin{aligned}
& \text { LHS }=\sin ^2 \frac{\mathrm{A}}{2}+\sin ^2 \frac{\mathrm{B}}{2}+\sin ^2 \frac{\mathrm{C}}{2} \\
& =\frac{1}{2}-\frac{1}{2} \cos \mathrm{A}+\frac{1}{2}-\frac{1}{2} \cos \mathrm{B}+\frac{1}{2}-\frac{1}{2} \cos \mathrm{C} \\
& =\frac{3}{2}-\frac{1}{2}(\cos \mathrm{A}+\cos \mathrm{B}+\cos \mathrm{C}) \\
& \text { consider } \cos \mathrm{A}+\cos \mathrm{B}+\cos \mathrm{C} \\
& =2 \cos \frac{\mathrm{A}+\mathrm{B}}{2} \cos \frac{\mathrm{A}-\mathrm{B}}{2}+1-2 \sin ^2 \frac{\mathrm{C}}{2} \\
& =2 \cos \left(90^{\circ}-\frac{\mathrm{C}}{2}\right) \cos \frac{\mathrm{A}-\mathrm{B}}{2}+1-2 \sin ^2 \frac{\mathrm{C}}{2} \\
& =2 \sin \frac{\mathrm{C}}{2} \cos \frac{\mathrm{A}-\mathrm{B}}{2}-2 \sin ^2 \frac{\mathrm{C}}{2}+1 \\
& =2 \sin \frac{\mathrm{C}}{2}\left(\cos \frac{\mathrm{A}-\mathrm{B}}{2}-\sin \frac{\mathrm{C}}{2}\right)+1 \\
&
\end{aligned}$

$
\begin{aligned}
& =2 \sin \frac{\mathrm{C}}{2}\left(\cos \frac{\mathrm{A}-\mathrm{B}}{2}-\sin \left(90^{\circ}-\frac{\mathrm{A}+\mathrm{B}}{2}\right)\right)+1 \\
& =2 \sin \frac{\mathrm{C}}{2}\left(\cos \frac{\mathrm{A}-\mathrm{B}}{2}-\cos \frac{\mathrm{A}+\mathrm{B}}{2}\right)+1 \\
& =2 \sin \frac{\mathrm{C}}{2}\left(2 \sin \frac{\mathrm{A}}{2} \sin \frac{\mathrm{B}}{2}\right)+1=4 \sin \frac{\mathrm{A}}{2} \sin \frac{\mathrm{B}}{2} \sin \frac{\mathrm{C}}{2}+1=\mathrm{RHS} \\
& =1+4 \sin \frac{\mathrm{A}}{2} \sin \frac{\mathrm{B}}{2} \sin \frac{\mathrm{C}}{2}
\end{aligned}
$
substitute in (1) we get,
$
\begin{aligned}
\text { LHS } & =\frac{3}{2}-\frac{1}{2}\left(1+4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\right) \\
& =\frac{3}{2}-\frac{1}{2}-2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \\
& =1-2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}
\end{aligned}
$