Q.11. If , then the value of is :

Q: What is the correct answer for JEE Main Mathematics Q11?

The correct answer is A. Sol. We use the identity $\sin \theta \sin(60^\circ - \theta) \sin(60^\circ + \theta) = \frac{1}{4} \sin(3\theta)$. Let $\theta = \frac{\pi}{18} = 10^\circ$. Then $60^\circ - \theta = 60^\circ - 10^\circ = 50^\circ = \frac{5\pi}{18}$. And $60^\circ + \theta = 60^\circ + 10^\circ = 70^\circ = \frac{7\pi}{18}$. So, $K = \sin \left( \frac{\pi}{18} \right) \sin \left( \frac{5\pi}{18} \right) \sin \lef

Q: Q.11. If , then the value of is :

The correct answer is A. Sol. We use the identity $\sin \theta \sin(60^\circ - \theta) \sin(60^\circ + \theta) = \frac{1}{4} \sin(3\theta)$. Let $\theta = \frac{\pi}{18} = 10^\circ$. Then $60^\circ - \theta = 60^\circ - 10^\circ = 50^\circ = \frac{5\pi}{18}$. And $60^\circ + \theta = 60^\circ + 10^\circ = 70^\circ = \frac{7\pi}{18}$. So, $K = \sin \left( \frac{\pi}{18} \right) \sin \left( \frac{5\pi}{18} \right) \sin \lef

Q: What are the options for this JEE Main Mathematics question?

A. \frac{\sqrt{3}+1}{2\sqrt{2}} · B. \frac{\sqrt{3}}{2\sqrt{2}} · C. \frac{\sqrt{3}}{2} · D. \frac{1}{2}

Question

Q.11. If , then the value of  is :

FirstInTest · Accepted Answer

Sol. We use the identity $\sin 	heta \sin(60^\circ - 	heta) \sin(60^\circ + 	heta) = \frac{1}{4} \sin(3	heta)$.
Let $	heta = \frac{\pi}{18} = 10^\circ$.
Then $60^\circ - 	heta = 60^\circ - 10^\circ = 50^\circ = \frac{5\pi}{18}$.
And $60^\circ + 	heta = 60^\circ + 10^\circ = 70^\circ = \frac{7\pi}{18}$.
So, $K = \sin \left( \frac{\pi}{18} ight) \sin \left( \frac{5\pi}{18} ight) \sin \left( \frac{7\pi}{18} ight) = \sin(10^\circ) \sin(50^\circ) \sin(70^\circ)$.
Using the identity, $K = \frac{1}{4} \sin(3 	imes 10^\circ) = \frac{1}{4} \sin(30^\circ) = \frac{1}{4} 	imes \frac{1}{2} = \frac{1}{8}$.

Now we need to find the value of $\sin \left( \frac{10K\pi}{3} ight)$.
Substitute $K = \frac{1}{8}$:
$\sin \left( \frac{10 	imes \frac{1}{8} 	imes \pi}{3} ight) = \sin \left( \frac{10\pi}{24} ight) = \sin \left( \frac{5\pi}{12} ight)$.

To calculate $\sin \left( \frac{5\pi}{12} ight)$:
$\frac{5\pi}{12} = \frac{5 	imes 180^\circ}{12} = 5 	imes 15^\circ = 75^\circ$.
$\sin(75^\circ) = \sin(45^\circ + 30^\circ) = \sin(45^\circ)\cos(30^\circ) + \cos(45^\circ)\sin(30^\circ)$
$= \frac{1}{\sqrt{2}} 	imes \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} 	imes \frac{1}{2}$
$= \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$.

Answer

\frac{\sqrt{3}+1}{2\sqrt{2}}

Answer

\frac{\sqrt{3}}{2\sqrt{2}}

Answer

\frac{\sqrt{3}}{2}

Answer

\frac{1}{2}

Related Questionsin Mathematics

More from JEE Main

Browse Engineering Questions by Subject

Practice More JEE Main Questions