Show that the function f x = sin 2 x + π 4 is decreasing function on the interval π 4 , π 2 .

Given that f x = sin 2 x + π 4 Differentiating with respect to x we get f ' x = 2 cos 2 x + π 4 Here given that x ∈ π 4 , π 2 ⇒ π 4 < x < π 2 ⇒ π 2 < 2 x < π ⇒ π 2 + π 4 < 2 x + π 4 < π + π 4 ⇒ 3 π 4 < 2 x + π 4 < 5 π 4 ⇒ cos 3 π 4 < cos 2 x + π 4 < cos 5 π 4 ⇒ cos 2 x + π 4 < - 1 2 < 0 ∴ f ' x < 0 . So function is decreasing on π 4 , π 2 .

Show that the function f x = cot - 1 sin x + cos x is decreasing on 0 , π 4 and increasing on π 4 , π 2 .

Given that f x = cot - 1 sin x + cos x Differentiate with respect to x f ' x = - 1 1 + sin x + cos x 2 cos x - sinx ⇒ sin x - cos x 1 + sin 2 x + cos 2 x + 2 sin x cos x ⇒ sin x - cos x 1 + 1 + 2 sin x cos x ⇒ sin x - cos x 2 + 2 sin x cos x ⇒ sin x - cos x 2 ( 1 + sin x cos x ) Now suppose that f ' x < 0 sin x - cos x 2 ( 1 + sin x cos x ) < 0 ⇒ sin x - cos x < 0 ⇒ sin x < cos x ⇒ tan x < 1 ⇒ x < π 4 So function is decreasing on 0 , π 4 . Now if f ' x > 0 sin x - cos x 2 ( 1 + sin x cos x ) > 0 ⇒ sin x - cos x > 0 ⇒ sin x > cos x ⇒ tan x > 1 ⇒ x > π 4 So function is increasing on π 4 , π 2 .

Show that f ( x ) = sin x − cos x is an increasing function on ( − π 4 , π 4 ) ?

Given that f x = sin x - cos x f ' x = cos x + sin x = cos x 1 + sin x cos x = cos x 1 + cot x Here, - π 4 < x < π 4 ⇒ cos x > 0 . . . 1 Also, - π 4 < x < π 4 ⇒ - 1 < cot x < 1 ⇒ 0 < 1 + cot x < 2 ⇒ 1 + cot x > 0 . . . 2 cos x 1 + cot x > 0 , ∀ x ∈ - π 4 , π 4 From eqs . ( 1 ) and ( 2 ) ⇒ f ' x > 0 , ∀ x ∈ - π 4 , π 4 So, f ( x ) is increasing interval - π 4 , π 4 .