Consider a triangle ABC in which A+ B+ C= 3 (π)/(3) What is the value of A 2 (B)/(2) (C)/(2) ?

Correct answer: C. (1)/(8) aligned & Given A+ B+ C= 3 (π)/(3) & A+ B+ C & =2 (A+B)/(2) · (A-B)/(2)+(1-2 ^2 (C)/(2)) & =2 (C)/(2) (A-B)/(2)+1-2 ^2 (C)/(2) ( (A+B)/(2)=90^ ^ -(C)/(2)) & =1+2 (C)/(2)[ (A-B)/(2)- (C)/(2)] & =1+2 (C)/(2)[ (A-B)/(2)- (A+B)/(2)] & =1+4 (C)/(2) (A)/(2) (B)/(2) & A+ B+ C= 3 (π)/(3) & 1+4 (C)/(2) (A)/(2) (B)/(2)= 3 (π)/(3) aligned aligned & 1+4 (C)/(2) (A)/(2) (B)/(2)= 3 × 3 2 & 4 (A)/(2) (B)/(2) (C)/(2)=(3)/(2)-1 aligned (A)/(2) (B)/(2) (C)/(2)=(1)/(8)

NDA 2016

Exam: NDA
Year: 2016
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

NDA 2016 · Mathematics · Trigonometric Ratios & Identities

Consider a triangle ABC in which A+ B+ C= 3 (π)/(3) What is the...

Consider a triangle ABC in which

\cos A+\cos B+\cos C=\sqrt{3} \sin \frac{\pi}{3}

What is the value of

\sin \frac{\mathrm{A}}{2} \sin \frac{B}{2} \sin \frac{C}{2}

Options

A $\frac{1}{2}$
B $\frac{1}{4}$
C $\frac{1}{8}$
D $\frac{1}{16}$

Correct answer

C. $\frac{1}{8}$

Step-by-step solution

\begin{aligned} & \text{ Given } \cos A+\cos B+\cos C=\sqrt{3} \sin \frac{\pi}{3} \\ & \cos A+\cos B+\cos C \\ & =2 \cos \frac{A+B}{2} \cdot \cos \frac{A-B}{2}+\left(1-2 \sin ^2 \frac{C}{2}\right) \\ & =2 \sin \frac{C}{2} \cos \frac{A-B}{2}+1-2 \sin ^2 \frac{C}{2} \left(\because \frac{A+B}{2}=90^{{}^{\circ}}-\frac{C}{2}\right)\\ & =1+2 \sin \frac{C}{2}\left[\cos \frac{A-B}{2}-\sin \frac{C}{2}\right] \\ & =1+2 \sin \frac{C}{2}\left[\cos \frac{A-B}{2}-\cos \frac{A+B}{2}\right] \\ & =1+4 \sin \frac{C}{2} \sin \frac{A}{2} \sin \frac{B}{2} \\ & \cos A+\cos B+\cos C=\sqrt{3} \sin \frac{\pi}{3} \\ & 1+4 \sin \frac{C}{2} \sin \frac{A}{2} \sin \frac{B}{2}=\sqrt{3} \sin \frac{\pi}{3}\end{aligned}

\begin{aligned} & \Rightarrow 1+4 \sin \frac{C}{2} \sin \frac{A}{2} \sin \frac{B}{2}=\sqrt{3} \times \frac{\sqrt{3}}{2} \\ & \Rightarrow 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}=\frac{3}{2}-1\end{aligned}

\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}=\frac{1}{8}$

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