If sin, θ = sin, α then sin, frac{θ }{3} =

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Trigonometrical Equations

normal

If $sin\, \theta = sin\, \alpha$ then $sin\, \frac{\theta }{3}$ =

$sin\, \frac{\alpha }{3}$

$sin \, \left( {\frac{\pi }{3} - \frac{\alpha }{3}} \right)$

$- sin \, \left( {\frac{\pi }{3} + \frac{\alpha }{3}} \right)$

All of the above

$sin \theta = sin \alpha$

$\theta = n\pi +(-1)^n\alpha$

$n = 0$ $\theta = \alpha$ $sin\theta /3 = sin\alpha /3$

$n = 1$ $\theta = \pi -\alpha$ $sin\theta /3 = sin(\pi /3-\alpha /3)$

$n = -1$ $\theta = -\pi -\alpha$ $sin\theta /3 = sin(-\pi /3-\alpha /3)= -sin(\pi /3+\alpha /3)$

Standard 11

Mathematics

hard

normal

normal

(KVPY-2009)

medium

hard

(KVPY-2019)