a) Ta có sin4(x + kπ/2) = sin(4x + k2π) = sin4x với k ∈ Z.

Từ đó suy ra hàm số y = sin4x là hàm số tuần hoàn với chu kì π/2.

Vì hàm số y = sin4x là hàm số lẻ nên đồ thị của nó có tâm đối xứng là gốc tọa độ O.

Các hàm số y = sin4x (C1) và y = sin4x + 1 (C2) có đồ thị như trên hình 1 và hình 2.

b) Vì sin4x + 1 = m ⇔ sin4x = m – 1

và -1 ≤ sin4x ≤ 1

nên -1 ≤ m – 1 ≤ 1

⇔ 0 ≤ m ≤ 2.

Từ đó, phương trình (1) có nghiệm khi 0 ≤ m ≤ 2 và vô nghiệm khi m > 2 hoặc m < 0.

c) Phương trình tiếp tuyến của (C2) có dạng

y   -   y o   =   y ’ ( x o ) ( x   -   x o ) .

b) Đồ thị hàm số \(y=\left|\cos2x\right|\)

ta có : \(VT=\dfrac{2cos2x-sin4x}{2cos2x+sin4x}=\dfrac{2cos2x-2sin2x.cos2x}{2cos2x+2sin2x.cos2x}\)

\(=\dfrac{2cos2x\left(1-sin2x\right)}{2cos2x\left(1+sin2x\right)}=\dfrac{1-sin2x}{1+sin2x}=\dfrac{sin^2x-2sinx.cosx+cos^2x}{sin^2x+2sinx.cosx+cos^2x}\)

\(=\left(\dfrac{sinx-cosx}{sinx+cosx}\right)^2=\left(\dfrac{\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)}{\sqrt{2}cos\left(x-\dfrac{\pi}{4}\right)}\right)=tan^2\left(x-\dfrac{\pi}{4}\right)\)

\(=tan^2\left(\dfrac{\pi}{4}-x\right)=VP\left(đpcm\right)\)

ta có : \(VT=\dfrac{2cos2x-sin4x}{2cos2x+sin4x}=\dfrac{2cos2x-2sin2x.cos2x}{2cos2x+2sin2x.cos2x}\)

\(=\dfrac{2cos2x\left(1-sin2x\right)}{2cos2x\left(1+sin2x\right)}=\dfrac{1-sin2x}{1+sin2x}=\dfrac{sin^2x-2sinx.cosx+cos^2x}{sin^2x+2sinx.cosx+cos^2x}\)

\(=\left(\dfrac{sinx-cosx}{sinx+cosx}\right)^2=\left(\dfrac{\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)}{\sqrt{2}cos\left(x-\dfrac{\pi}{4}\right)}\right)=tan^2\left(x-\dfrac{\pi}{4}\right)\)

\(=tan^2\left(\dfrac{\pi}{4}-x\right)=VP\left(đpcm\right)\)

ta có : \(VT=\dfrac{2cos2x-sin4x}{2cos2x+sin4x}=\dfrac{2cos2x-2sin2x.cos2x}{2cos2x+2sin2x.cos2x}\)

\(=\dfrac{2cos2x\left(1-sin2x\right)}{2cos2x\left(1+sin2x\right)}=\dfrac{1-sin2x}{1+sin2x}=\dfrac{sin^2x-2sinx.cosx+cos^2x}{sin^2x+2sinx.cosx+cos^2x}\)

\(=\left(\dfrac{sinx-cosx}{sinx+cosx}\right)^2=\left(\dfrac{\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)}{\sqrt{2}cos\left(x-\dfrac{\pi}{4}\right)}\right)=tan^2\left(x-\dfrac{\pi}{4}\right)\)

\(=tan^2\left(\dfrac{\pi}{4}-x\right)=VP\left(đpcm\right)\)

a) y=\(sin^4x+cos^4x-3=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x-3=-2-\dfrac{1}{2}.sin^22x\)

Có \(0\le sin^22x\le1\)

\(\Leftrightarrow-2\ge y\ge-\dfrac{5}{2}\)

Min xảy ra \(\Leftrightarrow sin^22x=1\Leftrightarrow sin2x=1\Leftrightarrow2x=\dfrac{\Pi}{2}+k2\Pi\left(k\in Z\right)\)

\(\Leftrightarrow x=\dfrac{\Pi}{4}+k\Pi\left(k\in Z\right)\)

Max xảy ra \(\Leftrightarrow sin2x=0\Leftrightarrow2x=k\Pi\Leftrightarrow x=\dfrac{k\Pi}{2}\)

b, \(x\in\left[0;\pi\right]\)

=>\(sin\left(x-\dfrac{\pi}{4}\right)\in\left[-\dfrac{\sqrt{2}}{2};1\right]\)

\(\Leftrightarrow2sin\left(x-\dfrac{\pi}{4}\right)\in\left[-\sqrt{2};2\right]\)

\(\Rightarrow\left\{{}\begin{matrix}Miny=-\sqrt{2}\\Maxy=2\end{matrix}\right.\)

Min xảy ra \(\Leftrightarrow x=0\)

Max xảy ra \(\Leftrightarrow x=\dfrac{\pi}{2}\)

1) \(\dfrac{1-cosx+cos2x}{sin2x-sinx}=cotx\)

\(VT=\dfrac{1-cosx+2cos^2x-1}{2sinx.cosx-sinx}\)

\(VT=\dfrac{cosx\left(2cos-1\right)}{sinx\left(2cosx-1\right)}\)

\(VT=\dfrac{cosx}{sinx}=cotx=VP\) ( đpcm )

b) \(\dfrac{sinx+sin\dfrac{x}{2}}{1+cosx+cos\dfrac{x}{2}}=tan\dfrac{x}{2}\)

\(VT=\dfrac{sin\left(2.\dfrac{x}{2}\right)+sin\dfrac{x}{2}}{1+cos\left(2.\dfrac{x}{2}\right)+cos\dfrac{x}{2}}\)

\(VT=\dfrac{2sin\dfrac{x}{2}.cos\dfrac{x}{2}+sin\dfrac{x}{2}}{1+2cos^2\dfrac{x}{2}-1+cos\dfrac{x}{2}}\)

\(VT=\dfrac{2sin\dfrac{x}{2}.cos\dfrac{x}{2}+sin\dfrac{x}{2}}{2cos^2\dfrac{x}{2}+cos\dfrac{x}{2}}\)

\(VT=\dfrac{sin\dfrac{x}{2}\left(2cos\dfrac{x}{2}+1\right)}{cos\dfrac{x}{2}\left(2cos\dfrac{x}{2}+1\right)}\)

\(VT=\dfrac{sin\dfrac{x}{2}}{cos\dfrac{x}{2}}=tan\dfrac{x}{2}=VP\) ( đpcm )

c) \(\dfrac{2cos2x-sin4x}{2cos2x+sin4x}=tan^2\left(\dfrac{\pi}{4}-x\right)\)

\(VT=\dfrac{2cos2x-sin\left(2.2x\right)}{2cos2x+sin\left(2.2x\right)}\)

\(VT=\dfrac{2cos2x-2sin2x.cos2x}{2cos2x+2sin2x.cos2x}\)

\(VT=\dfrac{2cos2x\left(1-sin2x\right)}{2cos2x\left(1+sin2x\right)}\)

\(VT=\dfrac{1-sin2x}{1+sin2x}\)

\(VP=tan^2\left(\dfrac{\pi}{4}-x\right)=\dfrac{1-cos2\left(\dfrac{\pi}{4}-x\right)}{1+cos2\left(\dfrac{\pi}{4}-x\right)}\)

\(VP=\dfrac{1-cos\left(\dfrac{\pi}{2}-2x\right)}{1+cos\left(\dfrac{\pi}{2}-2x\right)}\)

\(VP=\dfrac{1-sin2x}{1+cos2x}=VT\) ( đpcm )

d) \(tanx-tany=\dfrac{sin\left(x-y\right)}{cosx.cosy}\)

\(VP=\dfrac{sin\left(x-y\right)}{cosx.cosy}=\dfrac{sinx.cosy-cosx.siny}{cosx.cosy}\)

\(VP=\dfrac{sinx.cosy}{cosx.cosy}-\dfrac{cosx.siny}{cosx.cosy}\)

\(VP=\dfrac{sinx}{cosx}-\dfrac{siny}{cosy}=tanx-tany=VT\) ( đpcm )

Hàm số này có tập xác định là R \ {0}