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I used the following way but got wrong answer
$$A.M. \ge G.M.$$ $$ \frac{\sin \theta + \csc \theta}{2} \ge \sqrt{\sin \theta \cdot \csc \theta}$$
Squaring both sides,
\begin{equation*} (\sin\theta + \csc\theta )^2 \ge 4 \tag{1} \end{equation*} Similarly \begin{equation*} (\cos\theta + \sec\theta )^2 \ge 4 \tag{2} \end{equation*}
Adding equation (1) and (2)

\begin{equation*} (\sin\theta + \csc\theta )^2+(\cos\theta + \sec\theta )^2 \ge 8 \end{equation*}
What is wrong?

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    $\begingroup$ The argument is correct, both terms are $\ge 4$. But there is interaction, the equality case of AM/GM cannot hold simultaneously for both parts. $\endgroup$ – André Nicolas Dec 13 '14 at 7:07
  • $\begingroup$ Why can't it simultaneously hold true? $\endgroup$ – maths_ms Dec 13 '14 at 7:19
  • $\begingroup$ For eq $(1)$, the equality holds when $\sin \theta = \csc\theta$, or $\sin \theta = \pm 1$. For eq $(2)$, the equality holds when $\cos \theta = \pm 1$. These conditions cannot occur simultaneously. $\endgroup$ – Dylan Dec 15 '14 at 0:48
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$$\sin^2\theta+\cos^2\theta+2+2+\csc^2\theta+\sec^2\theta$$

$$=5+\frac1{\sin^2\theta\cos^2\theta}=5+\frac4{(\sin2\theta)^2}=5+4\csc^22\theta$$

Now, $\csc^22\theta=1+\cot^22\theta\ge1$ for real $\theta$

The equality occurs if $\csc^22\theta=1\iff\sin^22\theta=1$

$\iff\cos2\theta=0\implies2\theta=(2n+1)\dfrac\pi2$ where $n$ is any integer

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  • $\begingroup$ Do you really need the last three lines? $$0 \leq \sin^2 2\theta \leq 1$$ therefore $\dfrac{4}{\sin^2 2\theta} \geq 4$ $\endgroup$ – Darth Geek Dec 13 '14 at 7:14
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    $\begingroup$ @DarthGeek, To make things more explicit $\endgroup$ – lab bhattacharjee Dec 13 '14 at 7:15
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Since $$\left(\sin{\theta}+\dfrac{1}{\sin{\theta}}\right)^2+\left(\cos{\theta}+\dfrac{1}{\cos{\theta}}\right)^2=5+\dfrac{1}{\sin^2{\theta}}+\dfrac{1}{\cos^2{\theta}}$$ Use Cauchy-Schwarz inequality we have $$\dfrac{1}{\sin^2{\theta}}+\dfrac{1}{\cos^2{\theta}}\ge\dfrac{(1+1)^2}{\sin^2{\theta}+\cos^2{\theta}}=4$$

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  • $\begingroup$ OP is asking for something more strict! $\endgroup$ – Mathronaut Dec 13 '14 at 7:28
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$\begin{align}(\sin\theta + \csc\theta)^2 + (\cos\theta +\sec\theta)^2 & =5+\sec^2 x+\csc^2 x\\&=5+1+\tan^2 x+1+\cot^2x\\&=7+\tan^2x+\cot^2x\\&\geq9\end{align}$

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Without AM-GM, you could expand the expression which, after simplifications, write $$f(x)=(\sin\theta + \csc\theta)^2 + (\cos\theta +\sec\theta)^2 =\csc ^2(x)+\sec ^2(x)+5$$ The derivative write $$f'(x)=2 \tan (x) \sec ^2(x)-2 \cot (x) \csc ^2(x)=-8 \sin (4 x) \csc ^4(2 x)$$ and cancels for $x=\frac{\pi}{4}, \frac{3\pi}{4}, \cdots$ and, for these values, $f(x)=9$. The second derivative test shows that this is a minimum value.

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Using Cauchy–Schwarz inequality:

$$(\sin x + \csc x)^2+(\cos x + \sec x)^2\\=[(\sin x + \csc x)^2+(\cos x + \sec x)^2][\sin^2 x+ \cos^2 x]\\\geq \mid (\sin x + \csc x)\sin x+(\cos x + \sec x)\cos x \mid^2\\=\mid (\sin^2 x +1)+(\cos^2 x +1) \mid^2\\=\mid 1+1+1 \mid^2=9$$

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