Question

Find the minimum value of $|sinx+cosx+\frac{cosx-sinx}{co{s}^{4}x-si{n}^{4}x}|$

Answer 1

Let $y=|\sin x+\cos x+\frac{\cos x-\sin x}{{\cos }^{4}x-{\sin }^{4}x}|....\left(1\right)$

we have to find minimum value of $y$

$\Rightarrow y=|\sin x+\cos x+\frac{\cos x-\sin x}{({\cos }^{2}x+{\sin }^{2}x)({\cos }^{2}x-{\sin }^{2}x)}|$

$\Rightarrow y=|\sin x+\cos x+\frac{\cos x-\sin x}{(\cos x-\sin x)(\cos x+\sin x)}|$

$\Rightarrow y=\left|\frac{(\sin x+\cos x{)}^2+1}{\cos x+\sin x}\right|$...... on expanding

$y=\left|\frac{2+2\sin x\cos x}{\cos x+\sin x}\right|......\left(2\right)$

$\Rightarrow$ In order to minimize $y=$ we have to maximize denominator Multyply & divide by $\sqrt{2}$

$\sqrt{2}(\frac{\cos x}{\sqrt{2}}+\frac{\sin x}{\sqrt{2}})\Rightarrow \sqrt{2}\left(\sin (x+\frac{\pi }{4})\right)$

now equation $\left(2\right)$ becomes

$y=\left|\frac{\sqrt{2}(1+\sin x\cos x)}{\sin (x+\frac{\pi }{4})}\right|.....\left(3\right)$

we know $\sin (x+\frac{\pi }{4})\le 1$

$x+\frac{\pi }{4}\le \frac{\pi }{2}\Rightarrow [x=\frac{\pi }{4}]$

$\Rightarrow$ Hence at $x=\frac{pi}{4}$ value of $y$ in minimum

$\Rightarrow$ now from equation $\left(3\right)$

$y=\left|\frac{\sqrt{2}(1+\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}})}{\sin \left(\frac{\pi }{2}\right)}\right|$

$y=\sqrt{2}\left(\frac{3}{2}\right)=\frac{3}{\sqrt{2}}$

$\Rightarrow (y=\frac{3}{\sqrt{2}})$