Question

Let A and B denote the statements
A: $cos{\theta }_1+cos{\theta }_2+cos{\theta }_3=0$
B: $sin{\theta }_1+sin{\theta }_2+sin{\theta }_3=0.$

If $cos({\theta }_1-{\theta }_2)+cos({\theta }_2-{\theta }_3)+cos({\theta }_3-{\theta }_1)=\frac{-3}{2}$

Then which one of following is correct ?

• A. A is true and B is false
• B. A is false and B is true
• C. both A and B are true
• D. both A and Bare false

Answer 1

Answer: (C)

both A and B are true

Given

$\cos ({\theta }_1-{\theta }_2)+\cos ({\theta }_2-{\theta }_3)+\cos ({\theta }_3-{\theta }_1)=\frac{-3}{2}$

$2[\cos {\theta }_1\cos {\theta }_2+\sin {\theta }_1\sin {\theta }_2]+2[\cos {\theta }_2\cos {\theta }_3+\sin {\theta }_2\sin {\theta }_3]+2[\cos {\theta }_1\cos {\theta }_3+\sin {\theta }_1\sin {\theta }_3]=-3$

$(\cos (A-B)=\cos A\cos B+\sin A\sin B)$

$3+2\cos {\theta }_1\cos {\theta }_2+2\sin {\theta }_1\sin {\theta }_2+2\cos {\theta }_2\cos {\theta }_3+2\sin {\theta }_2\sin {\theta }_3+2\cos {\theta }_1\cos {\theta }_3+2\sin {\theta }_1\sin {\theta }_3=0$

$({\cos }^2{\theta }_1+{\sin }^1{\theta }_1)+({\cos }^2{\theta }_2+{\sin }^2{\theta }_2)+({\cos }^2{\theta }_3+{\sin }^2{\theta }_3)+2\cos {\theta }_1\cos {\theta }_2$

$+2\sin {\theta }_1\sin {\theta }_2+2\cos {\theta }_2\cos {\theta }_3+2\sin {\theta }_2\sin {\theta }_3+2\cos {\theta }_1\cos {\theta }_3+2\sin {\theta }_1\sin {\theta }_3=0$

$({\sin }^2\theta +{\cos }^2\theta =1)$

$({\cos }^2{\theta }_1+{\cos }^2{\theta }_2+{\cos }^2{\theta }_3+2\cos {\theta }_1\cos {\theta }_2+2\cos {\theta }_2\cos {\theta }_3+2\cos {\theta }_1\cos {\theta }_3)$

$+({\sin }^2{\theta }_1+{\sin }^2{\theta }_2+{\sin }^2{\theta }_3+2\sin {\theta }_1\sin {\theta }_2+2\sin {\theta }_2\sin {\theta }_3+2\sin {\theta }_1\sin {\theta }_3)=0$

$(\cos {\theta }_1+\cos {\theta }_2+\cos {\theta }_3{)}^2+(\sin {\theta }_1+\sin {\theta }_2+\sin {\theta }_3{)}^2=0$

$((a+b+c{)}^2=a^2+b^2+c^2+2ab+2bc+2ac)$

$⟹\cos {\theta }_1+\cos {\theta }_2+\cos {\theta }_3=0$

$⟹\sin {\theta }_1+\sin {\theta }_2+\sin {\theta }_3=0$