Trigonometrical Inequality?

Algebra Level 4

For x , y R x, y \in \mathbb R with 0 < x < π 2 0 < x < \frac { \pi }{ 2 } such that

( sin x ) 2 y ( cos x ) y 2 2 + ( cos x ) 2 y ( sin x ) y 2 2 = sin 2 x \large \frac {(\sin x)^{2y}}{(\cos x)^{\frac {y^2}2}} +\frac {(\cos x)^{2y}}{(\sin x)^{\frac {y^2}2}} = \sin 2x

then find the value of y y .


The answer is 2.

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1 solution

Jd Money
Dec 9, 2017

Let x = pi/4. Then we have sqrt(1/2)^(2y-.5y^2) + sqrt(1/2)^(2y-.5y^2) = sin(pi/2). So, sqrt(1/2)^(2y-.5y^2) = 1/2. So, 2y-.5y^2 = 2. So, y=2.

@JD Money ,

Can you please put that in latex form. That would look more decent.

Priyanshu Mishra - 3 years, 5 months ago

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