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manoj jangra Grade: 12
        

Prove that sinx.sin2x.sin3x <9/16

6 years ago

Answers : (1)

Sumit Majumdar
IIT Delhi
askIITians Faculty
132 Points
										
Dear student,
We have:
sin\left ( x \right )sin\left ( 2x \right )sin\left ( 3x \right )=2sin^{2}\left ( x \right )cos\left ( x \right )\left ( 3sin\left ( x \right ) \right-4sin^{3}\left ( x \right ) )=6sin^{3}\left ( x \right )cos\left ( x \right )-8sin^{5}\left ( x \right )cos\left ( x \right )On maximising this, we get the maximum value of sin(x) as equal to:
sin\left ( x \right )=\sqrt{\frac{8\pm \sqrt{10}}{12}}
This gives the required result.

Thanks & Regards


Sumit Majumdar,


askIITians Faculty


Ph.D,IIT Delhi

2 years ago
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