This-y-2cos-2-x-sin2x-2sin-2-x-find-the-smallest-value-of-the-function-

Question Number 12551 by @ANTARES_VY last updated on 25/Apr/17

This y = \frac{2 \cos^{2} x + \sin 2 x}{2 \sin^{2} x} find the

smallest value of the function .

Answered by mrW1 last updated on 25/Apr/17

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

= \frac{2 \cos^{2} x + 2 sinx \cos x}{2 \sin^{2} x}

= \frac{\cos^{2} x + sinx \cos x}{\sin^{2} x}

= \cot^{2} x + \cot x

= \cot^{2} x + 2 \times \frac{1}{2} \cot x + {(\frac{1}{2})}^{2} - \frac{1}{4}

= {(\cot x + \frac{1}{2})}^{2} - \frac{1}{4} ⩾ - \frac{1}{4}

\Rightarrow s m a l l e s v a l u e o f f u n c t i o n = - \frac{1}{4}

m i n i m u m w h e n \cot x + \frac{1}{2} = 0

o r \tan x = - 2

o r x = n π - \tan^{- 1} (2)

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