Question

Let $F\left(x\right)=si{n}^{4}x+co{s}^{4}x$. Then $F$is an increasing function in the interval?

Answer 1

Determine derivative:

$F\left(x\right)=si{n}^{4}x+co{s}^{4}x$

$$\begin{gathered} \Rightarrow F\text{'}\left(x\right)=4{\sin }^{3}x\cos x+4{\cos }^{3}x(-\sin x) \\ \Rightarrow F\text{'}\left(x\right)=4\sin x\cos x({\sin }^{2}x-{\cos }^{2}x) \\ \Rightarrow F\text{'}\left(x\right)=2\sin 2x({\sin }^{2}x-{\cos }^{2}x)\left[2sin\theta cos\theta =sin2\theta \right] \\ \Rightarrow F\text{'}\left(x\right)=-2\sin 2x\left(\cos 2x\right)\left[si{n}^2\theta -co{s}^2\theta =-cos2\theta \right] \\ \Rightarrow F\text{'}\left(x\right)=-\sin 4x \end{gathered}$$

Thus, $F\left(x\right)$ is increasing when $F\text{'}\left(x\right)>0$

$$\begin{gathered} \Rightarrow -\sin 4x>0 \\ \Rightarrow \sin 4x<0 \\ \Rightarrow 4x\in \left(\mathrm{\pi },2\mathrm{\pi }\right) \\ \Rightarrow x\in \left(\frac{\mathrm{\pi },}{4}\frac{\mathrm{\pi }}{2}\right) \end{gathered}$$

In genereal $x\in \left(\frac{\mathrm{n\pi }}{2}+\frac{\mathrm{\pi }}{4},\frac{n\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{2}\right)$

Hence, $F$is increasing function in $\mathit{x}\mathbf{\in }\left(\frac{n\pi }{2}+\frac{\pi }{4},\frac{n\pi }{2}+\frac{\pi }{2}\right)$