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If $\sin\theta_1+\sin\theta_2+\sin\theta_3=3$ then $\cos\theta_1+\cos\theta_2+\cos\theta_3$ is equal to

$(a)\;3\qquad(b)\;2\qquad(c)\;1\qquad(d)\;0$

1 Answer

$\sin\theta_1+\sin\theta_2+\sin\theta_3=3$
$\theta_1=\theta_2=\theta_3=\large\frac{\pi}{2}$
$\cos\theta_1+\cos\theta_2+\cos\theta_3=0$
Hence (d) is the correct answer.
answered Oct 8, 2013 by sreemathi.v
 

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