Bài tập TOÁN 11 (CT mới) – Chuyên đề GIÁ TRỊ LƯỢNG GIÁC CỦA MỘT GÓC LƯỢNG GIÁC.

Điểm $N$ là điểm chính giữa của cung tròn $AB’.$

$\cos\left(-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}.$

$\sin\left(-\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}.$

$\tan\left(-\dfrac{\pi}{4}\right)=\dfrac{\sin\left(-\dfrac{\pi}{4}\right)}{\cos\left(-\dfrac{\pi}{4}\right)}=-1.$

$\cot\left(-\dfrac{\pi}{4}\right)=\dfrac{\cos\left(-\dfrac{\pi}{4}\right)}{\sin\left(-\dfrac{\pi}{4}\right)}=-1.$

a) $\sin\alpha=\dfrac{1}{3}$ và $90^o < \alpha < 180^o.$

+) Ta có: $\sin^2\alpha + \cos^2\alpha=1.$ Suy ra $\cos^2\alpha = 1-\sin^2\alpha = 1-\left(\dfrac{1}{3}\right)^2=\dfrac{8}{9}.$

Vì $90^o < \alpha < 180^o$ (góc phần tư thứ II) nên $\cos\alpha < 0.$

Do đó, $\cos\alpha = -\sqrt{\dfrac{8}{9}}=-\dfrac{2\sqrt{2}}{3}.$

+) $\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha}=\dfrac{\dfrac{1}{3}}{-\dfrac{2\sqrt{2}}{3}}=-\dfrac{1}{2\sqrt{2}}.$

+) $\cot\alpha = \dfrac{1}{\tan\alpha}=-2\sqrt{2}.$

b) $\cos\alpha=-\dfrac{2}{3}$ và $\pi < \alpha < \dfrac{3\pi}{2}.$

+) Ta có: $\sin^2\alpha+\cos^2\alpha=1.$ Suy ra $\sin^2\alpha = 1-\cos^2\alpha=1-\left(-\dfrac{2}{3}\right)^2=\dfrac{5}{9}.$

Vì $\pi < \alpha < \dfrac{3\pi}{2}$ (góc phần tư thứ III) nên $\sin\alpha < 0.$

Do đó $\sin\alpha = -\sqrt{\dfrac{5}{9}}=-\dfrac{\sqrt{5}}{3}.$

+) $\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{-\dfrac{\sqrt{5}}{3}}{-\dfrac{2}{3}}=\dfrac{\sqrt{5}}{2}.$

+) $\cot\alpha=\dfrac{1}{\tan\alpha}=\dfrac{2}{\sqrt{5}}.$

c) $\tan\alpha=-2\sqrt{2}$ và $0 < \alpha < \pi.$

+) $\cot\alpha=\dfrac{1}{\tan\alpha}=-\dfrac{1}{2\sqrt{2}}.$

+) Ta có: $1+\tan^2\alpha=\dfrac{1}{\cos^2\alpha}.$ Suy ra $\cos^2\alpha = \dfrac{1}{1+\tan^2\alpha}=\dfrac{1}{1+\left(-2\sqrt{2}\right)^2}=\dfrac{1}{9}.$

Vì $0 < \alpha < \pi$ (nửa đường tròn phía trên) nên $\sin\alpha > 0.$ Mà $\tan\alpha =-2\sqrt{2} < 0$ nên $\cos\alpha < 0.$

Suy ra $\cos\alpha=-\sqrt{\dfrac{1}{9}}=-\dfrac{1}{3}.$

+) Ta có: $\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}.$ Suy ra $\sin\alpha=\tan\alpha\cdot \cos\alpha=(-2\sqrt{2})\cdot \left(-\dfrac{1}{3}\right)=\dfrac{2\sqrt{2}}{3}.$

d) $\cot\alpha=-\sqrt{2}$ và $\dfrac{\pi}{2} < \alpha < \dfrac{3\pi}{2}.$

+) $\tan\alpha=\dfrac{1}{\cot\alpha}=-\dfrac{1}{\sqrt{2}}.$

+) Ta có: $1+\cot^2\alpha=\dfrac{1}{\sin^2\alpha}.$ Suy ra $\sin^2\alpha=\dfrac{1}{1+\cot^2\alpha}=\dfrac{1}{1+\left(-\sqrt{2}\right)^2}=\dfrac{1}{3}.$

Vì $\dfrac{\pi}{2} < \alpha < \dfrac{3\pi}{2}$ (nửa đường tròn bên trái) nên $\cos\alpha < 0.$ Mà $\cot\alpha=-\sqrt{2} < 0$ nên $\sin\alpha > 0.$

Suy ra $\sin\alpha=\sqrt{\dfrac{1}{3}}=\dfrac{1}{\sqrt{3}}.$

+) Ta có: $\cot\alpha=\dfrac{\cos\alpha}{\sin\alpha}.$ Suy ra $\cos\alpha=\cot\alpha\cdot \sin\alpha=(-\sqrt{2})\cdot \dfrac{1}{\sqrt{3}}=-\sqrt{\dfrac{2}{3}}.$

a) Do $\dfrac{\pi}{2} < \alpha < \pi$ nên $\pi < \dfrac{\pi}{2}+\alpha < \dfrac{3\pi}{2}$ (góc phần tư thứ III).

Do đó $\sin\left(\dfrac{\pi}{2}+\alpha\right) < 0.$

b) Do $\dfrac{\pi}{2} < \alpha < \pi$ nên $\pi > \dfrac{3\pi}{2}-\alpha > \dfrac{\pi}{2}$ (góc phần tư thứ II).

Do đó, $\sin\left(\dfrac{3\pi}{2}-\alpha\right) > 0$ và $\cos\left(\dfrac{3\pi}{2}-\alpha\right) < 0.$

Suy ra $\tan\left(\dfrac{3\pi}{2}-\alpha\right) < 0.$

c) Do $\dfrac{\pi}{2} < \alpha < \pi$ nên $0 < -\dfrac{\pi}{2}+\alpha < \dfrac{\pi}{2}$ (góc phần tư thứ I) và $\dfrac{\pi}{2} > \pi-\alpha > 0$ (góc phần tư thứ I).

Do đó, $\cos\left(-\dfrac{\pi}{2}+\alpha\right) > 0$ và $\tan(\pi-\alpha) > 0.$

Suy ra $\cos\left(-\dfrac{\pi}{2}+\alpha\right)\cdot \tan(\pi-\alpha) > 0.$

b) Do $\dfrac{\pi}{2} < \alpha < \pi$ nên $\dfrac{3\pi}{2} < \pi+\alpha < 2\pi$ (góc phần tư thứ IV). Suy ra $\cot(\pi+\alpha) < 0.$

Mặt khác, $\dfrac{3\pi}{2} < \dfrac{14\pi}{9} < 2\pi$ (góc phần tư thứ IV). Suy ra $\sin\dfrac{14\pi}{9} < 0.$

Vậy $\sin\dfrac{14\pi}{9}\cdot \cot(\pi+\alpha) > 0.$

a) Ta có:

+) $\tan 368^o=\tan (8^o+2\cdot 180^o)=\tan 8^o.$

+) $\sin 2550^o=\sin (30^o+7\cdot 360^o) = \sin 30^o=\dfrac{1}{2}.$

+) $\cos (-188^o)=\cos 188^o=\cos (8^o+180^o)=-\cos 8^o.$

+) $\cos 638^o=\cos (-82^o+2\cdot 360^o)=\cos (-82^o)=\cos 82^o=\sin 8^o.$

+) $\cos 98^o=-\cos 82^o=-sin 8^o.$

Do đó:

$A=\dfrac{1}{\tan 8^o}+\dfrac{2\cdot \dfrac{1}{2} (-\cos 8^o)}{2\sin 8^o-\sin 8^o}$ $=\cot 8^o-\dfrac{\cos 8^o}{\sin 8^o}$ $=\cot 8^o- \cot 8^o$ $=0.$

b) $B=\sin^2 25^o + \sin^2 45^o + \sin^2 60^o + \sin^2 65^o$ $=(\sin^2 25^o+\sin^2 65^o)+\sin^2 45^o+\sin^2 60^o$ $=(\cos^2 25^o+\sin^2 65^o)+\left(\dfrac{\sqrt{2}}{2}\right)^2+\left(\dfrac{\sqrt{3}}{2}\right)^2$ $=1+\dfrac{1}{2}+\dfrac{3}{4}$ $=\dfrac{9}{4}.$

c) $C=\tan^2\dfrac{\pi}{8}\cdot \tan\dfrac{3\pi}{8}\cdot \tan\dfrac{5\pi}{8}.$

Vì $\dfrac{3\pi}{8}+\dfrac{5\pi}{8}=\pi$ nên $\tan\dfrac{5\pi}{8}=-\tan\dfrac{3\pi}{8}.$

Vì $\dfrac{\pi}{8}+\dfrac{3\pi}{8}=\dfrac{\pi}{2}$ nên $\tan\dfrac{\pi}{8}=\cot\dfrac{3\pi}{8}.$

Do đó $C=\left(\cot\dfrac{3\pi}{8}\right)^2\cdot\tan\dfrac{3\pi}{8}\cdot\left(-\tan\dfrac{3\pi}{8}\right)$ $=\cot^2\dfrac{3\pi}{8}\cdot \left(-\tan^2\dfrac{3\pi}{8}\right)=-1.$

a) Ta có: $\tan x+\cot x=2$ $\Leftrightarrow \tan x+\dfrac{1}{\tan x}=2$ $\Leftrightarrow \tan x=1.$

Suy ra: $\cot x=\dfrac{1}{\tan x}=1.$

Vì $0 < x < \dfrac{\pi}{2}$ nên $\sin x,\cos x > 0.$ Do đó:

+) $\sin x=\dfrac{1}{\sqrt{1+\cot^2 x}}=\dfrac{1}{\sqrt{2}}.$

+) $\cos x=\sqrt{1-\sin^2x}=\dfrac{1}{\sqrt{2}}.$

b) Ta có: $\tan x-\cot x=-\dfrac{2}{\sqrt{3}}$ $\Leftrightarrow \tan x-\dfrac{1}{\tan x}=-\dfrac{2}{\sqrt{3}}$ $\Leftrightarrow \left\{\begin{matrix}\sqrt{3}\tan^2x+2\tan x-\sqrt{3}=0 \\ \tan x\neq 0 \end{matrix}\right.$ $\Leftrightarrow \tan x=\dfrac{1}{\sqrt{3}} \vee \tan x=-\sqrt{3}$

Vì $\pi < x < \dfrac{3\pi}{2}$ nên $\sin x < 0, \cos x < 0, \tan x > 0.$

Do đó, ta chọn $\tan x=\dfrac{1}{\sqrt{3}}.$

Suy ra:

+) $\cot x=\dfrac{1}{\tan x}=\sqrt{3}.$

+) $\sin x=-\dfrac{1}{\sqrt{1+\cot^2 x}}=-\dfrac{1}{2}.$

+) $\cos x=-\sqrt{1-\sin^2x}=-\dfrac{\sqrt{3}}{2}.$

Vì $\tan\alpha$ và $\cot\alpha$ luôn cùng dấu và $\tan\alpha+\cot\alpha < 0$ nên $\tan\alpha < 0$ và $\cot\alpha < 0.$

Ta có: $1+\cot^2\alpha=\dfrac{1}{\sin^2\alpha}$

Suy ra $\cot^2\alpha=\dfrac{1}{\sin^2\alpha}-1=\dfrac{1}{\dfrac{1}{25}}-1=25-1=24.$

Mà $\cot\alpha < 0$ (cmt)

Nên $\cot\alpha=-\sqrt{24}=-2\sqrt{6}.$

Suy ra $\tan\alpha=\dfrac{1}{\cot\alpha}=\dfrac{-1}{2\sqrt{6}}$

Vì $\cot\alpha=\dfrac{\cos\alpha}{\sin\alpha}$ nên $\cos\alpha=\cot\alpha\cdot\sin\alpha=(-2\sqrt{6})\cdot \dfrac{1}{5}=\dfrac{-2\sqrt{6}}{5}.$

$A=\tan\left(\dfrac{7\pi}{2}-\alpha\right)$ $=\tan\left(3\pi+\dfrac{\pi}{2}-\alpha\right)$ $=\tan\left(\dfrac{\pi}{2}-\alpha\right)$ $=\cot\alpha.$

Ta có: $\sin(\pi+\alpha)=-\sin\alpha.$ Đề cho: $\sin(\pi+\alpha)=-\dfrac{1}{3}.$ Suy ra $-\sin\alpha=-\dfrac{1}{3},$ hay $\sin\alpha=\dfrac{1}{3}.$

Do đó: $\cot^2\alpha=\dfrac{1}{\sin^2\alpha}-1=8.$

Vì $\dfrac{\pi}{2} < \alpha < \pi$ nên $\cot\alpha < 0.$

Suy ra $\cot\alpha = -\sqrt{8}=-2\sqrt{2}.$

Vậy $A=-2\sqrt{2}.$

Ta có: $1+\tan^2\alpha=\dfrac{1}{\cos^2\alpha}.$

Suy ra: $\cos^2\alpha=\dfrac{1}{1+\tan^2\alpha}=\dfrac{1}{1+(-3)^2}=\dfrac{1}{10}.$

Vì $0\leq\alpha\leq\pi$ nên $\sin\alpha \leq 0.$ Mà $\tan\alpha=-3 < 0$ nên $\cos\alpha < 0.$

Do đó, $\cos\alpha=-\dfrac{1}{\sqrt{10}}.$

Suy ra $\sin\alpha=\tan\alpha\cdot\cos\alpha=(-3)\cdot \left(-\dfrac{1}{\sqrt{10}}\right)=\dfrac{3}{\sqrt{10}}.$

Vậy $\dfrac{\cos\alpha-2\sin\alpha}{2\cos\alpha+\sin\alpha}$ $=\dfrac{-\dfrac{1}{\sqrt{10}}-2\cdot\dfrac{3}{\sqrt{10}}}{2\left(-\dfrac{1}{\sqrt{10}}\right)+\dfrac{3}{\sqrt{10}}}$ $=\dfrac{\dfrac{-7}{\sqrt{10}}}{\dfrac{1}{\sqrt{10}}}$ $=-7.$

Cách khác: Chia cả tử và mẫu của $\dfrac{\cos\alpha-2\sin\alpha}{2\cos\alpha+\sin\alpha}$ cho $\cos\alpha,$ và lưu ý $\dfrac{\sin\alpha}{\cos\alpha}=\tan\alpha,$ ta được:

$\dfrac{\cos\alpha-2\sin\alpha}{2\cos\alpha+\sin\alpha}$ $=\dfrac{1-2\tan\alpha}{2+\tan\alpha}$ $=\dfrac{1-2\cdot (-3)}{2+(-3)}$ $=\dfrac{7}{-1}$ $=-7.$

a) Chứng minh $\cos^4 x +2\sin^2 x = 1 + \sin^4 x.$

Ta có: $VT=\cos^4 x+2\sin^2 x$ $=\cos^2 x(1-\sin^2 x)+2\sin^2 x$ $=\cos^2 x-cos^2 x\sin^2 x +\sin^2 x + \sin^2 x$ $=(\cos^2 x+\sin^2 x)+\sin^2 x(1-\cos^2 x)$ $=1+\sin^2 x \sin^2 x$ $=1+\sin^4 x=VP.$ $\Rightarrow đpcm.$

b) Chứng minh $\dfrac{\sin x + \cos x}{\sin^3 x}=\cot^3 x + \cot^2 x + \cot x +1.$

Cách 1 (đi từ vế phải):

Ta có: $\cot^3 x + \cot^2 x + \cot x +1$ $=\dfrac{\cos^3 x}{\sin^3 x}+\dfrac{\cos^2 x}{\sin^2 x}+\dfrac{\cos x}{\sin x}+1$ $=\dfrac{\cos^3 x+\cos^2 x\sin x+\cos x\sin^2 x+\sin^3 x}{\sin^3 x}.$

Vậy ta cần chứng minh $\cos^3 x+\cos^2 x\sin x+\cos x\sin^2 x+\sin^3 x=\sin x+\cos x.$

Ta thấy: $(\sin x+\cos x)-(\cos^3 x+\cos^2 x\sin x+\cos x\sin^2 x+\sin^3 x)$ $=\sin x(1-\cos^2 x)+\cos x(1-\sin^2 x)-\cos^3 x-\sin^3 x$ $=\sin x\sin^2 x+\cos x\cos^2 x-\cos^3 x-\sin^3 x$ $=\sin^3 x+\cos^3 x-\cos^3 x-\sin^3 x=0.$

Suy ra $\cos^3 x+\cos^2 x\sin x+\cos x\sin^2 x+\sin^3 x=\sin x+\cos x.$

Suy ra điều phải chứng minh.

Cách 2 (đi từ vế trái):

Ta có: $VT=\dfrac{\sin x + \cos x}{\sin^3 x}$ $=\dfrac{\sin x+\cos x}{\sin x}\cdot \dfrac{1}{\sin^2 x}$ $=(1+\cot x)\cdot (1+\cot^2 x)$ $=\cot^3 x+\cot^2 x+\cot x+1=VP.$

Suy ra điều phải chứng minh.

c) Chứng minh $\dfrac{\cot^2 x – \cot^2 y}{\cot^2 x\cdot \cot^2 y}=\dfrac{\cos^2 x – \cos^2 y}{\cos^2 x\cdot \cos^2 y}.$

Ta có: $VP=\dfrac{\cos^2 x – \cos^2 y}{\cos^2 x\cdot \cos^2 y}$ $=\dfrac{1}{\cos^2 y}-\dfrac{1}{\cos^2 x}$ $=(1+\tan^2 y)-(1+\tan^2 x)$ $=\tan^2 y – \tan^2 x$ $=\dfrac{1}{\cot^2 y}-\dfrac{1}{\cot^2 x}$ $=\dfrac{\cot^2 x-\cot^2}{\cot^2 x\cdot \cot^2 y}=VT.$

Suy ra điều phải chứng minh.

d) Chứng minh $\sqrt{\sin^4 x + 4\cos^2 x} + \sqrt{\cos^4 x + 4\sin^2 x} = 3\tan\left(x+\dfrac{\pi}{3}\right)\tan\left(\dfrac{\pi}{6}-x\right).$

Ta có:

+) $\sin^4 x + 4\cos^2 x$ $=\sin^4 x+4(1-\sin^2 x)$ $=\sin^4 x-4\sin^2 x+4$ $=\left(\sin^2 x-2\right)^2.$ Do đó $\sqrt{\sin^4 x + 4\cos^2 x}=2-\sin^2 x$ (vì $\sin^2 x < 2).$

+) $\cos^4 x + 4\sin^2 x$ $=\cos^4 x+4(1-\cos^2 x)$ $=\cos^4 x-4\cos^2 x+4$ $=\left(\cos^2 x-2\right)^2.$ Do đó $\sqrt{\cos^4 x + 4\sin^2 x}=2-\cos^2 x$ (vì $\sin^2 x < 2).$

Do đó: $VT=(2-\sin^2 x)+(2-\cos^2 x)$ $=4-(\sin^2 x+\cos^2 x)$ $=4-1=3.$

Mặt khác, do $\left(x+\dfrac{\pi}{3}\right)+\left(\dfrac{\pi}{6}-x\right)=\dfrac{\pi}{2}$ nên $\tan\left(\dfrac{\pi}{6}-x\right)=\cot\left(x+\dfrac{\pi}{3}\right).$

Do đó: $VP=3\tan\left(x+\dfrac{\pi}{3}\right)\tan\left(\dfrac{\pi}{6}-x\right)$ $=3\tan\left(x+\dfrac{\pi}{3}\right)\cot\left(x+\dfrac{\pi}{3}\right)$ $=3.$

Vậy $VT=VP(=3).$

Suy ra điều phải chứng minh.

Ta có: $A+B+C=\pi$ (tổng ba góc của một tam giác)

Do đó: $\dfrac{A+2B+C}{2}=\dfrac{(A+B+C)+B}{2}=\dfrac{\pi+B}{2}=\dfrac{\pi}{2}+\dfrac{B}{2}.$

Suy ra:

+) $\cos\left(\dfrac{A+2B+C}{2}\right)=\cos\left(\dfrac{\pi}{2}+\dfrac{B}{2}\right)=\sin\left(-\dfrac{B}{2}\right)=-\sin\dfrac{B}{2}.$

+) $\sin\left(\dfrac{A+2B+C}{2}\right)=\sin\left(\dfrac{\pi}{2}+\dfrac{B}{2}\right)=\cos\left(-\dfrac{B}{2}\right)=\cos\dfrac{B}{2}.$

Do đó:

$VT=\dfrac{\sin^3 \dfrac{B}{2}}{\cos\left(\dfrac{A+2B+C}{2}\right)} – \dfrac{\cos^3 \dfrac{B}{2}}{\sin\left(\dfrac{A+2B+C}{2}\right)}$ $=\dfrac{\sin^3 \dfrac{B}{2}}{-\sin\dfrac{B}{2}} – \dfrac{\cos^3 \dfrac{B}{2}}{\cos\dfrac{B}{2}}$ $=-\sin^2\dfrac{B}{2}-\cos^2\dfrac{B}{2}$ $=-\left(\sin^2\dfrac{B}{2}+\cos^2\dfrac{B}{2}\right)$ $=-1.$

$VP=\tan A\cdot\cot(B+C)$ $=\tan A\cdot\cot(\pi-A)$ $=\tan A\cdot \left(-\cot A\right)$ $=-1.$

Vậy $VT=VP(=-1).$

Suy ra điều phải chứng minh.

a) $A=\cos(5\pi-x)-\sin\left(\dfrac{3\pi}{2}+x\right)+\tan\left(\dfrac{3\pi}{2}-x\right)+\cot(3\pi-x).$

Ta có:

+) $\cos(5\pi-x)=\cos(\pi-x+2\cdot 2\pi)$ $=\cos(\pi-x)$ $=-\cos x.$

+) $\sin\left(\dfrac{3\pi}{2}+x\right)=\sin\left(2\pi-\dfrac{\pi}{2}+x\right)$ $=\sin\left(-\dfrac{\pi}{2}+x\right)$ $=\sin\left[-\left(\dfrac{\pi}{2}-x\right)\right)$ $=-\sin\left(\dfrac{\pi}{2}-x\right)$ $=-\cos x.$

+) $\tan\left(\dfrac{3\pi}{2}-x\right)=\tan\left(\pi+\dfrac{\pi}{2}-x\right)$ $=\tan\left(\dfrac{\pi}{2}-x\right)$ $=\cot x.$

+) $\cot(3\pi-x)=\cot(-x)=-\cot x.$

Do đó:

$A=-\cos x-(-\cos x)+\cot x+(-\cot x)=0.$

b) $B=\dfrac{\sin(900^o+x)-\cos(450^o-x)+\cot(1080^o-x)+\tan(630^o-x)}{\cos(450^o-x)+\sin(x-630^o)-\tan(810^o+x)-\tan(810^o-x)}.$

Ta có:

+) $\sin(900^o+x)=\sin(180^o+x+2\cdot 360^o)$ $=\sin(180^o+x)$ $=-\sin x.$

+) $\cos(450^o-x)=\cos(90^o-x+360^o)$ $=\cos(90^o-x)$ $=\sin x.$

+) $\cot(1080^o-x)=\cot(-x+6\cdot 180^o)$ $=\cot(-x)$ $=-\cot x.$

+) $\tan(630^o-x)=\tan(90^o-x+3\cdot 180^o)$ $=\tan(90^o-x)$ $=\cot x.$

+) $\sin(x-630^o)=\sin(x+90^o-2\cdot 360^o)$ $=\sin(x+90^o)$ $=\cos(-x)$ $=\cos x.$

+) $\tan(810^o+x)=\tan(90^o+x+4\cdot 180^o)$ $=\tan(90^o+x)$ $=\cot(-x)$ $=-\cot x.$

+) $\tan(810^o-x)=\tan(90^o-x+4\cdot 180^o)$ $=\tan(90^o-x)$ $=\cot x.$

Do đó:

$B=\dfrac{-\sin x-\sin x+(-\cot x)+\cot x}{\sin x+\cos x-(-\cot x)-\cot x}$ $=\dfrac{-2\sin x}{\sin x+\cos x}.$

+) Tính $\sqrt{\dfrac{1}{1+\cos x}+\dfrac{1}{1-\cos x}}$

Ta có: $\sqrt{\dfrac{1}{1+\cos x}+\dfrac{1}{1-\cos x}}$ $=\sqrt{\dfrac{1-\cos x+1+\cos x}{(1+\cos x)(1-\cos x)}}$ $=\sqrt{\dfrac{2}{1-\cos^2 x}}$ $=\sqrt{\dfrac{2}{\sin^2 x}}$ $=\dfrac{\sqrt{2}}{|\sin x|}.$

Vì $\pi < x < 2\pi$ nên $\sin x < 0.$ Suy ra $|\sin x|=-\sin x.$

Vậy $\sqrt{\dfrac{1}{1+\cos x}+\dfrac{1}{1-\cos x}}$ $=\dfrac{\sqrt{2}}{-\sin x}$ $=\dfrac{-\sqrt{2}}{\sin x}.$

+) Tính $\sin(x+2013\pi)$

Ta có: $\sin(x+2013\pi)$ $=\sin(x+\pi+1006\cdot 2\pi)$ $=\sin(x+\pi)$ $=-\sin x.$

+) Tính $A$

Từ những tính toán phía trên, ta có:

$A=\sqrt{2}-\dfrac{1}{\sin(x+2013\pi)}\cdot \sqrt{\dfrac{1}{1+\cos x}+\dfrac{1}{1-\cos x}}$

$=\sqrt{2}-\dfrac{1}{-\sin x}\cdot \dfrac{-\sqrt{2}}{\sin x}$

$=\sqrt{2}-\dfrac{\sqrt{2}}{\sin^2 x}$

$=\sqrt{2}\left(1-\dfrac{1}{\sin^2 x}\right)$

$=-\sqrt{2}\cot^2 x.$

a) Cho $\cos\alpha=\dfrac{2}{3}.$ Tính $A=\dfrac{\tan\alpha+3\cot\alpha}{\tan\alpha+\cot\alpha}.$

Ta có: $A=\dfrac{\tan\alpha+3\cot\alpha}{\tan\alpha+\cot\alpha}$ $=\dfrac{\tan\alpha+\dfrac{3}{\tan\alpha}}{\tan\alpha+\dfrac{1}{\tan\alpha}}$ $=\dfrac{\dfrac{\tan^2\alpha+3}{\tan\alpha}}{\dfrac{\tan^2\alpha+1}{\tan\alpha}}$ $=\dfrac{\tan^2\alpha+3}{\tan^2\alpha+1}.$

Mặt khác, $1+\tan^2\alpha=\dfrac{1}{\cos^2\alpha}=\dfrac{1}{\left(\dfrac{2}{3}\right)^2}=\dfrac{9}{4}.$

Suy ra $A=\dfrac{\tan^2\alpha+3}{\tan^2\alpha+1}$ $=\dfrac{\dfrac{9}{4}+2}{\dfrac{9}{4}}$ $=\dfrac{17}{9}.$

b) Cho $\tan\alpha=3.$ Tính $B=\dfrac{\sin\alpha-\cos\alpha}{\sin^3\alpha+3\cos^3\alpha+2\sin\alpha}.$

Ta có:

$B=\dfrac{\sin\alpha-\cos\alpha}{\sin^3\alpha+3\cos^3\alpha+2\sin\alpha}$ $=\dfrac{\dfrac{\sin\alpha}{\cos^3\alpha}-\dfrac{\cos\alpha}{\cos^3\alpha}}{\dfrac{\sin^3\alpha}{\cos^3\alpha}+\dfrac{3\cos^3\alpha}{\cos^3\alpha}+\dfrac{2\sin\alpha}{\cos^3\alpha}}$ $=\dfrac{\tan\alpha(1+\tan^2\alpha)-(1+\tan^2\alpha)}{\tan^3\alpha+3+2\tan\alpha(1+\tan^2\alpha)}$ $=\dfrac{3\cdot(1+3^2)-(1+3^2)}{3^3+3+2\cdot 3\cdot (1+3^2)}$ $=\dfrac{20}{90}$ $=\dfrac{2}{9}.$

c) Cho $\cot\alpha=\sqrt{5}.$ Tính $C=\sin^2\alpha-\sin\alpha\cos\alpha+\cos^2\alpha.$

Ta có: $(1+\cot^2\alpha)C$ $\dfrac{1}{\sin^2\alpha}C$ $=\dfrac{\sin^2\alpha-\sin\alpha\cos\alpha+\cos^2\alpha}{\sin^2\alpha}$ $=1-\dfrac{\cos\alpha}{\sin\alpha}+\dfrac{\cos^2\alpha}{\sin^2\alpha}$ $=1-\cot\alpha+\cot^2\alpha$ $=1-\sqrt{5}+5$ $=6-\sqrt{5}.$

Suy ra $C=\dfrac{6-\sqrt{5}}{1+\cot^2\alpha}$ $=\dfrac{6-\sqrt{5}}{6}.$

a)

+) Tìm $\sin\alpha\cos\alpha$

Ta có: $m^2=(\sin\alpha+\cos\alpha)^2$ $=\sin^2\alpha+\cos^2\alpha+2\sin\alpha\cos\alpha$ $=1+2\sin\alpha\cos\alpha.$

Suy ra $\sin\alpha\cos\alpha=\dfrac{m^2-1}{2}.$

+) Tìm $|\sin^4\alpha-\cos^4\alpha|$

Ta có: $\sin^4\alpha-\cos^4\alpha$ $=(\sin^2\alpha+\cos^2\alpha)(\sin^2\alpha-\cos^2\alpha)$ $=1\cdot(\sin\alpha+\cos\alpha)(\sin\alpha-\cos\alpha)$ $=m(\sin\alpha-\cos\alpha).$

Vậy $|\sin^4\alpha-\cos^4\alpha|=|m|\cdot |\sin\alpha-\cos\alpha|$ $=|m|\sqrt{(\sin\alpha-\cos\alpha)^2}$ $=|m|\sqrt{\sin^2\alpha+\cos^2\alpha-2\sin\alpha\cos\alpha}$ $=|m|\sqrt{1-2\cdot \dfrac{m^2-1}{2}}$ $=|m|\sqrt{2-m^2}.$

b) Ta có: $\sin\alpha\cos\alpha\leq\dfrac{\sin^2\alpha+\cos^2\alpha}{2}=\dfrac{1}{2}.$

Suy ra $\dfrac{m^2-1}{2}\leq\dfrac{1}{2},$ hay $m^2\leq 2.$

Do đó $|m|\leq\sqrt{2}.$

Đặt $u=\sin^2 x$ và $v=\cos^2 x,$ thì $u+v=1.$

Ta có:

+) $\sin^6 x+\cos^6 x$ $=u^3+v^3$ $=(u+v)(u^2+v^2-uv)$ $=1\cdot \left[(u+v)^2-3uv\right]$ $=1-3uv.$

+) $\sin^4 x+\cos^4 x$ $=u^2+v^2$ $=(u+v)^2-2uv$ $=1-2uv.$

Suy ra:

$A=\dfrac{\sin^6 x + \cos^6 x + 2}{\sin^4 x + \cos^4 x + 1}$

$=\dfrac{1-3uv+2}{1-2uv+1}$

$=\dfrac{3-3uv}{2-2uv}$

$=\dfrac{3(1-uv)}{2(1-uv)}$

$=\dfrac{3}{2}.$

Vậy giá trị của $A$ luôn bằng $\dfrac{3}{2}$ và không phụ thuộc vào $x.$

Đặt $u=\sin^2\alpha$ và $v=\cos^2\alpha$ thì $u+v=1.$

Ta có: $3\sin^4\alpha-\cos^4\alpha=\dfrac{1}{2}$ $\Leftrightarrow 3u^2-v^2=\dfrac{1}{2}$ $\Leftrightarrow 3u^2-(1-u)^2=\dfrac{1}{2}$ $\Leftrightarrow 2u^2+2u-1=\dfrac{1}{2}$ $\Leftrightarrow 4u^2+4u-3=0$ $\Leftrightarrow u=\dfrac{1}{2}\;hoặc\;u=\dfrac{-3}{2}.$

Mặt khác: $0\leq u=\sin^2\alpha\leq 1.$

Do đó $u=\dfrac{1}{2}.$ Suy ra $v=1-u=\dfrac{1}{2}.$

Ta có: $A=2\sin^4\alpha-\cos^4\alpha$ $=2u^2-v^2$ $=2\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^2$ $=2\cdot \dfrac{1}{4}-\dfrac{1}{4}$ $=\dfrac{1}{4}.$

Đặt $\cos^2\alpha=t.$

Ta có: $\dfrac{\sin^4\alpha}{a}+\dfrac{\cos^4\alpha}{b}=\dfrac{1}{a+b}$ $\Leftrightarrow \dfrac{(1-t)^2}{a}+\dfrac{t^2}{b}=\dfrac{1}{a+b}$ $\Leftrightarrow b(1-t)^2+at^2=\dfrac{ab}{a+b}$ $\Leftrightarrow (a+b)t^2-2bt+b=\dfrac{ab}{a+b}$ $\Leftrightarrow (a+b)^2t^2-2b(a+b)t+b^2=0$ $\Leftrightarrow \left((a+b)t-b\right)^2=0$ $\Leftrightarrow t=\dfrac{b}{a+b}.$

Suy ra $\cos^2\alpha=\dfrac{b}{a+b}$ và $\sin^2\alpha=\dfrac{a}{a+b}$ (vì $\sin^2\alpha+\cos^2\alpha=1).$

Do đó: $A=\dfrac{\sin^8\alpha}{a^3}+\dfrac{\cos^8\alpha}{b^3}$ $=\dfrac{a}{(a+b)^4}+\dfrac{b}{(a+b)^4}$ $=\dfrac{1}{(a+b)^3}.$