Question

sinx and cos x are decreasing functions at interval, if 0≤x≤2π

Answer 1

Answer: YES THEY ARE

Step-by-step explanation:

Answer 2

Points $x=\frac{\pi}{4},\frac{5\pi}{4}$ divides interval $[0,2\pi]$ into $3$ disjoints intervals,

$[(0,\frac{\pi}{4}),( \frac{\pi}{4},\frac{5\pi}{4}),(\frac{5\pi}{4},2\pi)]$

Step-by-step explanation:

Given,

$sinx$ and $cosx$ are decreasing functions at interval, if $0\leq x\leq 2\pi$

Let,

$f(x)=sinx+cosx$  for $0\leq x\leq 2\pi$

Differentiate above equation with respect to $x$,

$f'(x)=\frac{d}{dx} (sinx+cosx)$

⇒$f'(x)=(cosx-sinx)$

Plug $f'(x)=0$,

Then,

$cosx-sinx=0$

⇒$cosx=sinx$

∴ $x=\frac{\pi}{4},\frac{5\pi}{4}$  as $0\leq x\leq 2\pi$

Points $x=\frac{\pi}{4},\frac{5\pi}{4}$ divides interval $[0,2\pi]$ into $3$ disjoints intervals,

$[(0,\frac{\pi}{4}),( \frac{\pi}{4},\frac{5\pi}{4}),(\frac{5\pi}{4},2\pi)]$