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Trignometry

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17 Jan 2012 14:11:06 IST

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finding max. and min. values

Mathematics

how to find the maxinmum and minimum values of t-functions?e.g.- what is the min. value of (sin^4x+cos^2x)

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Gaurav

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17 Jan 2012 16:05:02 IST

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easy...u must remembermin. value of sin x = 0max. value of sin x = 1min. value of cos x = -1max. value of cos x = 1so min value of (sin^4x+cos^2x) is 0+0=0

arvindpandit

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17 Jan 2012 21:01:13 IST

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but at the same time ...i.e the same value of theta u cant hav sinx and cosx both equal to

arvindpandit

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17 Jan 2012 21:01:40 IST

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equal to zero

jagdish singh

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18 Jan 2012 13:21:50 IST

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$\hspace{-16}$Let $\mathbf{A=\sin^4x+\cos^2 x=(1-\cos^2x)^2+\cos^2 x}$\\\\ $\mathbf{A=1+\cos^4x-\cos^2x}$\\\\ $\mathbf{A=\sin^2 x+\cos^2 x+\cos^4 x-\cos^2 x}$\\\\ $\mathbf{A=\sin^2 x+\cos^4 x}$$

$\textbf{Let $\mathbf{A=sin^2\;x+cos^4\;x=sin^2x+cos^2x.\left(1-sin^2\;x\right)$}}$\\\\\\ $\textbf{So $\mathbf{A=1-cos^2\;x.sin^2\;x=1-\frac{1}{4}.\left(2sinx.cosx\right)^2}$}$\\\\\\ \mathbf{A=1-\frac{1}{4}.sin^2\;2x\Leftrightarrow {\frac{1}{4}sin^2\2x=1-A.....................................(1)}}$\\\\\\ \textbf{Now $\mathbf{0 \leq sin^2\;2x\leq 1\Leftrightarrow 0\leq \frac{1}{4}.sin^2\;2x\leq \frac{1}{4}}$}$\\\\\\ $\textbf{Now Put value from equation..(1)}$\\\\ \mathbf{0\leq 1-A\leq \frac{1}{4}\Leftrightarrow -\frac{1}{4}\leq A-1\leq 0}$\\\\ \textbf{So $\boxed{\boxed{\mathbf{\frac{3}{4}\leq A\leq 1}}}$}$

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