Bài tập TOÁN 11 (CT mới) – Chuyên đề CÔNG THỨC LƯỢNG GIÁC.

a) Tính các giá trị lượng giác của góc $75^o.$

+) $\sin 75^o$ $=\sin(30^o+45^o)$ $=\sin 30^o\cos 45^o+\cos 30^o\sin 45^o$ $=\dfrac{1}{2}\cdot\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{2}$ $=\dfrac{\sqrt{2}+\sqrt{6}}{4}.$

+) $\cos 75^o$ $=\cos(30^o+45^o)$ $=\cos 30^o\cos 45^o-\sin 30^o\sin 45^o$ $=\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}\cdot\dfrac{\sqrt{2}}{2}$ $=\dfrac{\sqrt{6}-\sqrt{2}}{4}.$

+) $\tan 75^o$ $=\dfrac{\sin 75^o}{\cos 75^o}$ $=\dfrac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}.$

+) $\cot 75^o$ $=\dfrac{\cos 75^o}{\sin 75^o}$ $=\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}.$

b) Tính các giá trị lượng giác của góc $795^o.$

Ta có: $795^o=75^o+2\cdot 360^o.$

Do đó:

+) $\sin 795^o$ $=\sin 75^o$ $=\dfrac{\sqrt{2}+\sqrt{6}}{4}.$

+) $\cos 795^o$ $=\cos 75^o$ $=\dfrac{\sqrt{6}-\sqrt{2}}{4}.$

+) $\tan 795^o$ $=\dfrac{\sin 795^o}{\cos 795^o}$ $=\dfrac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}.$

+) $\cot 795^o$ $=\dfrac{\cos 795^o}{\sin 795^o}$ $=\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}.$

a) Tính các giá trị lượng giác của góc $\dfrac{\pi}{12}.$

+) $\sin\dfrac{\pi}{12}$ $=\sin(\dfrac{\pi}{3}-\dfrac{\pi}{4})$ $=\sin\dfrac{\pi}{3}\cos\dfrac{\pi}{4}-\sin\dfrac{\pi}{4}\cos\dfrac{\pi}{3}$ $=\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2}$ $=\dfrac{\sqrt{6}-\sqrt{2}}{4}.$

+) $\cos\dfrac{\pi}{12}$ $=\cos(\dfrac{\pi}{3}-\dfrac{\pi}{4})$ $=\cos\dfrac{\pi}{3}\cos\dfrac{\pi}{4}+\sin\dfrac{\pi}{4}\sin\dfrac{\pi}{3}$ $=\dfrac{1}{2}\cdot\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{3}}{2}$ $=\dfrac{\sqrt{2}+\sqrt{6}}{4}.$

+) $\tan\dfrac{\pi}{12}$ $=\dfrac{\sin\dfrac{\pi}{12}}{\cos\dfrac{\pi}{12}}$ $=\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}.$

+) $\cot\dfrac{\pi}{12}$ $=\dfrac{\cos\dfrac{\pi}{12}}{\sin\dfrac{\pi}{12}}$ $=\dfrac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}.$

b) Tính các giá trị lượng giác của góc $\dfrac{73\pi}{12}.$

Ta có: $\dfrac{73\pi}{12}$ $=\dfrac{\pi}{12}+6\pi$ $=\dfrac{\pi}{12}+3\cdot 2\pi.$

Do đó:

+) $\sin\dfrac{73\pi}{12}$ $=\sin\dfrac{\pi}{12}$ $=\dfrac{\sqrt{6}-\sqrt{2}}{4}.$

+) $\cos\dfrac{73\pi}{12}$ $=\cos\dfrac{\pi}{12}$ $=\dfrac{\sqrt{2}+\sqrt{6}}{4}.$

+) $\tan\dfrac{73\pi}{12}$ $=\dfrac{\sin\dfrac{73\pi}{12}}{\cos\dfrac{73\pi}{12}}$ $=\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}.$

+) $\cot\dfrac{73\pi}{12}$ $=\dfrac{\cos\dfrac{73\pi}{12}}{\sin\dfrac{73\pi}{12}}$ $=\dfrac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}.$

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

Do đó: $\cos\alpha=-\sqrt{1-\sin^2\alpha}$ $=-\sqrt{1-\left(\dfrac{3}{\sqrt{10}}\right)^2}$ $=-\sqrt{\dfrac{1}{10}}$ $=\dfrac{-1}{\sqrt{10}}.$

Vậy:

+) $\cos\left(\alpha-\dfrac{\pi}{4}\right)$ $=\cos\alpha\cos\dfrac{\pi}{4}+\sin\alpha\sin\dfrac{\pi}{4}$ $=\dfrac{-1}{\sqrt{10}}\cdot\dfrac{\sqrt{2}}{2}+\dfrac{3}{\sqrt{10}}\cdot\dfrac{\sqrt{2}}{2}$ $=\dfrac{1}{\sqrt{5}}.$

+) $\sin\left(\dfrac{\pi}{6}+\alpha\right)$ $=\sin\dfrac{\pi}{6}\cos\alpha+\cos\dfrac{\pi}{6}\sin\alpha$ $=\dfrac{1}{2}\cdot\dfrac{-1}{\sqrt{10}}+\dfrac{\sqrt{3}}{2}\cdot\dfrac{3}{\sqrt{10}}$ $=\dfrac{3\sqrt{3}-1}{2\sqrt{10}}.$

Ta có: $\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$ $=-3.$

Do đó:

+) $\tan\left(\dfrac{\pi}{3}-\alpha\right)$ $=\dfrac{\tan\dfrac{\pi}{3}-\tan\alpha}{1+\tan\dfrac{\pi}{3}\tan\alpha}$ $=\dfrac{\sqrt{3}-(-3)}{1+\sqrt{3}\cdot(-3)}$ $=\dfrac{3+\sqrt{3}}{1-3\sqrt{3}}.$

a) Ta có:

VP = $\sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right)$ $=\sqrt{2}\left(\sin x\cos\dfrac{\pi}{4}+\cos x\sin\dfrac{\pi}{4}\right)$ $=\sqrt{2}\left(\sin x\cdot\dfrac{\sqrt{2}}{2}+\cos x\cdot\dfrac{\sqrt{2}}{2}\right)$ $=\sqrt{2}\cdot\dfrac{\sqrt{2}}{2}(\sin x+\cos x)$ $=\sin x+\cos x$ = VT.

Suy ra đpcm.

b) Ta có:

VP = $2\sin\left(x-\dfrac{\pi}{3}\right)$ $=2\left(\sin x\cos\dfrac{\pi}{3}-\cos x\sin\dfrac{\pi}{3}\right)$ $=2\left(\sin x\cdot \dfrac{1}{2}-\cos x\cdot\dfrac{\sqrt{3}}{2}\right)$ $=2\cdot\dfrac{1}{2}\left(\sin x-\sqrt{3}\cos x\right)$ $=\sin x-\sqrt{3}\cos x$ =VT.

Suy ra đpcm.

c) Ta có:

VP = $2\sin\left(x+\dfrac{\pi}{6}\right)$ $=2\left(\sin x\cos\dfrac{\pi}{6}+\cos x\sin\dfrac{\pi}{6}\right)$ $=2\left(\sin x\cdot\dfrac{\sqrt{3}}{2}+\cos x\cdot\dfrac{1}{2}\right)$ $=2\cdot\dfrac{1}{2}\left(\sqrt{3}\sin x+\cos x\right)$ $=\sqrt{3}\sin x+\cos x$ = VT.

Suy ra đpcm.

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

Do đó: $\cos\alpha = -\sqrt{1-\sin^2 x}$ $=-\sqrt{1-\left(\dfrac{1}{\sqrt{3}}\right)^2}$ $=-\sqrt{\dfrac{2}{3}}$ $=\dfrac{-\sqrt{2}}{\sqrt{3}}.$

Từ đó tính được:

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

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

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

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

a) $A=\dfrac{\sin\dfrac{\pi}{5}-\sin\dfrac{2\pi}{15}}{\cos\dfrac{\pi}{5}-\cos\dfrac{2\pi}{15}}.$

Ta có:

+) $\sin\dfrac{\pi}{5}-\sin\dfrac{2\pi}{15}$ $=2\cos\dfrac{\dfrac{\pi}{5}+\dfrac{2\pi}{15}}{2}\sin\dfrac{\dfrac{\pi}{5}-\dfrac{2\pi}{15}}{2}$ $=2\cos\dfrac{\pi}{6}\sin\dfrac{\pi}{30}.$

+) $\cos\dfrac{\pi}{5}-\cos\dfrac{2\pi}{15}$ $=-2\sin\dfrac{\dfrac{\pi}{5}+\dfrac{2\pi}{15}}{2}\sin\dfrac{\dfrac{\pi}{5}-\dfrac{2\pi}{15}}{2}$ $=-2\sin\dfrac{\pi}{6}\sin\dfrac{\pi}{30}.$

Do đó: $A=\dfrac{\sin\dfrac{\pi}{5}-\sin\dfrac{2\pi}{15}}{\cos\dfrac{\pi}{5}-\cos\dfrac{2\pi}{15}}$ $=\dfrac{2\cos\dfrac{\pi}{6}\sin\dfrac{\pi}{30}}{-2\sin\dfrac{\pi}{6}\sin\dfrac{\pi}{30}}$ $=-\cot\dfrac{\pi}{6}$ $=-\sqrt{3}.$

b) $B=\sin\dfrac{2\pi}{9}-\sin\dfrac{5\pi}{9}+\sin\dfrac{8\pi}{9}$

$=\sin\dfrac{2\pi}{9}+\sin\dfrac{8\pi}{9}-\sin\dfrac{5\pi}{9}$

$=2\sin\dfrac{\dfrac{2\pi}{9}+\dfrac{8\pi}{9}}{2}\cos\dfrac{\dfrac{2\pi}{9}-\dfrac{8\pi}{9}}{2}-\sin\dfrac{5\pi}{9}$

$=2\sin\dfrac{5\pi}{9}\cos\dfrac{-\pi}{3}-\sin\dfrac{5\pi}{9}$

$=2\sin\dfrac{5\pi}{9}\cdot\dfrac{1}{2}-\sin\dfrac{5\pi}{9}$

$=\sin\dfrac{5\pi}{9}-\sin\dfrac{5\pi}{9}$

$=0.$

c) $C=\cos\dfrac{2\pi}{9}+\cos\dfrac{4\pi}{9}+\cos\dfrac{10\pi}{9}$

$=\cos\dfrac{2\pi}{9}+\cos\dfrac{10\pi}{9}+\cos\dfrac{4\pi}{9}$

$=2\cos\dfrac{\dfrac{2\pi}{9}+\dfrac{10\pi}{9}}{2}\cos\dfrac{\dfrac{2\pi}{9}-\dfrac{10\pi}{9}}{2}+\cos\dfrac{4\pi}{9}$

$=2\cos\dfrac{2\pi}{3}\cos\dfrac{-4\pi}{9}+\cos\dfrac{4\pi}{9}$

$=2\cdot\dfrac{-1}{2}\cdot\cos\dfrac{4\pi}{9}+\cos\dfrac{4\pi}{9}$

$=-\cos\dfrac{4\pi}{9}+\cos\dfrac{4\pi}{9}$

$=0.$

a) $A=\sin 22^o30’\cdot\cos 202^o30’$

Ta có: $\cos 202^o30’$ $=\cos(22^o30’+180^o)$ $=-\cos 22^o30′.$

Do đó: $A=-\sin 22^o30’\cos 22^o30’$ $=-\dfrac{1}{2}\sin 2\cdot(22^o30′)$ $=-\dfrac{1}{2}\sin 45^o$ $=-\dfrac{1}{2}\cdot\dfrac{\sqrt{2}}{2}$ $=-\dfrac{\sqrt{2}}{4}.$

b) $B=4\sin^4\dfrac{\pi}{16}+2\cos\dfrac{\pi}{8}.$

Ta có: $4\sin^4\dfrac{\pi}{16}$ $=4\left(\sin^2\dfrac{\pi}{16}\right)^2$ $=4\left(\dfrac{1-\cos 2\cdot\dfrac{\pi}{16}}{2}\right)^2$ $=4\cdot\dfrac{\left(1-\cos\dfrac{\pi}{8}\right)^2}{4}$ $=\left(1-\cos\dfrac{\pi}{8}\right)^2$ $=1+\cos^2\dfrac{\pi}{8}-2\cos\dfrac{\pi}{8}.$

Do đó:

$B=4\sin^4\dfrac{\pi}{16}+2\cos\dfrac{\pi}{8}$

$=1+\cos^2\dfrac{\pi}{8}-2\cos\dfrac{\pi}{8}+2\cos\dfrac{\pi}{8}$

$=1+\cos^2\dfrac{\pi}{8}$

$=1+\dfrac{1+\cos 2\cdot\dfrac{\pi}{8}}{2}$

$=1+\dfrac{1+\cos\dfrac{\pi}{4}}{2}$

$=1+\dfrac{1+\dfrac{\sqrt{2}}{2}}{2}$

$=1+\dfrac{2+\sqrt{2}}{4}$

$=\dfrac{6+\sqrt{2}}{4}.$

a) $A=\dfrac{1}{\cos 290^o}+\dfrac{1}{\sqrt{3}\sin 250^o}.$

Ta có:

+) $\cos 290^o$ $=\cos(360^o-70^o)$ $=\cos(-70^o)$ $=\cos 70^o.$

+) $\sin 250^o$ $=\sin(70^o+180^o)$ $=-\sin 70^o.$

Do đó:

$A=\dfrac{1}{\cos 290^o}+\dfrac{1}{\sqrt{3}\sin 250^o}$

$=\dfrac{1}{\cos 70^o}+\dfrac{1}{\sqrt{3}(-\sin 70^o)}$

$=\dfrac{1}{\cos 70^o}-\dfrac{1}{\sqrt{3}\sin 70^o}$

$=\dfrac{\sqrt{3}\sin 70^o-\cos 70^o}{\sqrt{3}\sin 70^o\cos 70^o}$

$=4\cdot \dfrac{\dfrac{\sqrt{3}}{2}\sin 70^o-\dfrac{1}{2}\cos 70^o}{\sqrt{3}\cdot 2\sin 70^o\cos 70^o}$

$=4\cdot \dfrac{\cos 30^o\sin 70^o-\sin 30^o\cos 70^o}{\sqrt{3}\cdot \sin 140^o}$

$=4\cdot \dfrac{\sin 40^o}{\sqrt{3}\cdot \sin 140^o}$

$=4\cdot\dfrac{\sin 40^o}{\sqrt{3}\cdot\sin 40^o}$

$=\dfrac{4}{\sqrt{3}}.$

b) $B=(1+\tan 20^o)(1+\tan 25^o)$

$=\left(1+\dfrac{\sin 20^o}{\cos 20^o}\right)\left(1+\dfrac{\sin 25^o}{\cos 25^o}\right)$

$=\dfrac{(\cos 20^o+\sin 20^o)\cdot (\cos 25^o+\sin 25^o)}{\cos 20^o\cos 25^o}$

$=\dfrac{\sqrt{2}\cos(20^o-45^o)\cdot \sqrt{2}\cos(25^o-45^o)}{\cos 20^o\cos 25^o}$

$=\dfrac{2\cos(-25^o)\cos(-20^o)}{\cos 20^o\cos 25^o}$

$=\dfrac{2\cos 25^o\cos 20^o}{\cos 20^o\cos 25^o}$

$=2.$

c) $C=\tan 9^o-\tan 27^o-\tan 63^o+\tan 81^o.$

$=(\tan 9^o+\tan 81^o)-(\tan 27^o+\tan 63^o)$

$=\dfrac{\sin 9^o\cos 81^o+\cos 9^o\sin 81^o}{\cos 9^o\cos 81^o}-\dfrac{\sin 27^o\cos 63^o+\cos 27^o\sin 63^o}{\cos 27^o\cos 63^o}$

$=\dfrac{\sin 90^o}{\cos 9^o\sin 9^o}-\dfrac{\sin 90^o}{\cos 27^o\sin 27^o}$

$=\dfrac{2}{\sin 18^o}-\dfrac{2}{\sin 54^o}$

$=\dfrac{2(\sin 54^o-\sin 18^o)}{\sin 18^o\sin 54^o}$

$=\dfrac{4\cos 36^o\sin 18^o}{\sin 18^o\sin 54^o}$

$=\dfrac{4\sin 54^o\sin 18^o}{\sin 18^o\sin 54^o}$

$=4.$

d) $D=\sin^2\dfrac{\pi}{9}+\sin^2\dfrac{2\pi}{9}+\sin\dfrac{\pi}{9}\sin\dfrac{2\pi}{9}$

Đặt $x=\sin\dfrac{\pi}{9}$ và $y=\sin\dfrac{2\pi}{9}$ thì $A=x^2+y^2+xy=(x+y)^2-xy.$

Trong đó:

+) $x+y$ $=\sin\dfrac{\pi}{9}+\sin\dfrac{2\pi}{9}$ $=2\sin\dfrac{\pi}{6}\cos\dfrac{\pi}{18}$ $=\cos\dfrac{\pi}{18}.$

+) $xy$ $=\sin\dfrac{\pi}{9}\cdot\sin\dfrac{2\pi}{9}$ $=-\dfrac{1}{2}\left(\cos\dfrac{\pi}{3}-\cos\dfrac{\pi}{9}\right)$ $=-\dfrac{1}{2}\left(\dfrac{1}{2}-\cos\dfrac{\pi}{9}\right).$

Vậy:

$A=\cos^2\dfrac{\pi}{18}+\dfrac{1}{2}\left(\dfrac{1}{2}-\cos\dfrac{\pi}{9}\right)$

$=\dfrac{1+\cos\dfrac{\pi}{9}}{2}+\dfrac{1}{2}\left(\dfrac{1}{2}-\cos\dfrac{\pi}{9}\right)$

$\dfrac{3}{4}.$

e) $E=\sin\dfrac{\pi}{32}\cos\dfrac{\pi}{32}\cos\dfrac{\pi}{16}\cos\dfrac{\pi}{8}.$

Đặt $\dfrac{\pi}{32}=x$ thì $\dfrac{\pi}{16}=2x$ và $\dfrac{\pi}{8}=4x.$

Ta có:

$E=\sin x\cos x\cos 2x\cos 4x$

$=\dfrac{1}{2}\sin 2x\cos 2x\cos 4x$

$=\dfrac{1}{2}\cdot \dfrac{1}{2}\sin 4x\cos 4x$

$=\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot\dfrac{1}{2}\sin 8x$

$=\dfrac{1}{8}\sin 8x$

$=\dfrac{1}{8}\sin\left(8\cdot\dfrac{\pi}{32}\right)$

$=\dfrac{1}{8}\sin\dfrac{\pi}{4}$

$=\dfrac{\sqrt{2}}{16}.$

Ta có:

+) $\sin\left(\alpha+\dfrac{\pi}{6}\right)$ $=\sin\alpha\cos\dfrac{\pi}{6}+\cos\alpha\sin\dfrac{\pi}{6}$ $=\sin\alpha\cdot\dfrac{\sqrt{3}}{2}+\cos\alpha\cdot\dfrac{1}{2}$

+) $\sin\left(\alpha-\dfrac{\pi}{6}\right)$ $=\sin\alpha\cos\dfrac{\pi}{6}-\cos\alpha\sin\dfrac{\pi}{6}$ $=\sin\alpha\cdot\dfrac{\sqrt{3}}{2}-\cos\alpha\cdot\dfrac{1}{2}$

Do đó:

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

$=\dfrac{3}{4}\sin^2\alpha-\dfrac{1}{4}\cos^2\alpha$

$=\dfrac{3}{4}\sin^2\alpha-\dfrac{1}{4}(1-\sin^2\alpha)$

$=\sin^2\alpha-\dfrac{1}{4}$

$=\dfrac{9}{25}-\dfrac{1}{4}$

$=\dfrac{11}{100}.$

$P=\cos 4\alpha$ $=2\cos^2 2\alpha-1$ $=2\left(1-2\sin^2\alpha\right)^2-1$ $=2\left(1-4\sin^2\alpha+4\sin^4\alpha\right)-1$ $=1-8\sin^2\alpha+8\sin^4\alpha$ $=1-8\sin^2\alpha(1+\sin^2\alpha)$ $=1-8\cdot\dfrac{16}{25}\left(1+\dfrac{16}{25}\right)$ $=1-8\cdot\dfrac{16}{25}\cdot\dfrac{41}{25}$ $=1-\dfrac{5248}{625}$ $=\dfrac{-4623}{625}.$

$P=\dfrac{\sin 2\alpha}{\cos 4\alpha+1}$

$=\dfrac{\sin 2\alpha}{1-2\sin^2 2\alpha+1}$

$=\dfrac{\sin 2\alpha}{2-2\sin^2 2\alpha}$

Mà $\sin 2\alpha$ $=\dfrac{2\tan\alpha}{1+\tan^2\alpha}$ $=\dfrac{2\cdot(-2)}{1+(-2)^2}$ $=\dfrac{-4}{5}.$

Do đó:

$P=\dfrac{\dfrac{-4}{5}}{2-2\cdot\dfrac{16}{25}}$

$=\dfrac{\dfrac{-4}{5}}{\dfrac{18}{25}}=\dfrac{-10}{9}.$

a) $1-\sin 2\alpha=(\sin\alpha-\cos\alpha)^2.$

Đặt $\sin\alpha=x$ và $\cos\alpha=y$ thì $x^2+y^2=\sin^2\alpha+\cos^2\alpha=1.$

Ta có:

+) $VT=1-\sin 2\alpha$ $=1-2\sin\alpha\cos\alpha$ $=1-2xy.$

+) $VP=(\sin\alpha-\cos\alpha)^2$ $=(x-y)^2$ $=x^2+y^2-2xy$ $=1-2xy.$

Vậy $VT=VP.$ Suy ra đpcm.

b) $1+\sin 2\alpha=(\sin\alpha+\cos\alpha)^2.$

Tương tự câu a).

c) $\dfrac{1-\sin 2\alpha}{1+\sin 2\alpha}=\cot^2\left(\dfrac{\pi}{4}+\alpha\right).$

Áp dụng câu a) và câu b) ta được: $VT=\dfrac{1-\sin 2\alpha}{1+\sin 2\alpha}$ $=\left(\dfrac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}\right)^2$

Mặt khác: $VP=\cot^2\left(\dfrac{\pi}{4}+\alpha\right)$ $=\left(\dfrac{\cos\left(\dfrac{\pi}{4}+\alpha\right)}{\sin\left(\dfrac{\pi}{4}+\alpha\right)}\right)^2$ $=\left(\dfrac{\cos\dfrac{\pi}{4}\cos\alpha-\sin\dfrac{\pi}{4}\sin\alpha}{\sin\dfrac{\pi}{4}\cos\alpha+\cos\dfrac{\pi}{4}\sin\alpha}\right)^2$ $=\left(\dfrac{\dfrac{\sqrt{2}}{2}\cdot\cos\alpha-\dfrac{\sqrt{2}}{2}\cdot\sin\alpha}{\dfrac{\sqrt{2}}{2}\cdot\cos\alpha+\dfrac{\sqrt{2}}{2}\cdot\sin\alpha}\right)^2$ $=\left(\dfrac{\cos\alpha-\sin\alpha}{\cos\alpha+\sin\alpha}\right)^2.$

Vậy $VT=VP.$ Suy ra đpcm.

a) $\cos 3a=\cos(a+2a)$ $=\cos a\cos 2a-\sin a\sin2a$ $=\cos a(2\cos^2a-1)-\sin a\cdot 2\sin a\cos a$ $=2\cos^3a-\cos a-2\sin^2a\cos a$ $=2\cos^3a-\cos a-2(1-\cos^2a)\cos a$ $=2\cos^3a-\cos a-2\cos a+2\cos^3a$ $=4\cos^3a-3\cos a.$

b) Tương tự câu a).

a) $A=\dfrac{\cos a+2\cos 2a+\cos 3a}{\sin a+2\sin 2a+\sin 3a}.$

Ta có:

+) $\cos a+2\cos 2a+\cos 3a$ $=\cos a+\cos 3a+2\cos 2a$ $=2\cos 2a\cos a+2\cos 2a$ $=2\cos 2a(\cos a+1).$

+) $\sin a+\sin 2a+2\sin 3a$ $=\sin a+\sin 3a+2\sin 2a$ $=2\sin 2a\cos a+2\sin 2a$ $=2\sin 2a(\cos a+1).$

Vậy $A=\dfrac{2\cos 2a(\cos a+1)}{2\sin 2a(\cos a+1)}$ $=\dfrac{\cos 2a}{\sin 2a}$ $=\cot 2a.$

b) $B=\dfrac{\cos\left(a+\dfrac{\pi}{3}\right)+\cos\left(a-\dfrac{\pi}{3}\right)}{\cot a-\cot\dfrac{a}{2}}.$

Ta có:

+) $\cos\left(a+\dfrac{\pi}{3}\right)+\cos\left(a-\dfrac{\pi}{3}\right)$ $=2\cos a\cos\dfrac{\pi}{3}$ $=2\cos a\cdot\dfrac{1}{2}$ $=\cos a.$

+) $\cot a-\cot\dfrac{a}{2}$ $=\dfrac{\cos a}{\sin a}-\dfrac{\cos\dfrac{a}{2}}{\sin\dfrac{a}{2}}$ $=\dfrac{\sin\dfrac{a}{2}\cos a-\cos\dfrac{a}{2}\sin a}{\sin a\sin\dfrac{a}{2}}$ $=\dfrac{\sin\dfrac{-a}{2}}{\sin a\sin\dfrac{a}{2}}$ $=\dfrac{-\sin\dfrac{a}{2}}{\sin a\sin\dfrac{a}{2}}$ $=\dfrac{-1}{\sin a}.$

Vậy $B=\dfrac{\cos a}{\dfrac{-1}{\sin a}}$ $=-\sin a\cos a$ $=-\dfrac{\sin 2a}{2}.$

a) Vì $(A+B)+C=\pi$ (tổng ba góc của một tam giác) nên $\cos C=\cos [\pi-(A+B)]=-\cos(A+B).$

b) Vì $0 < A, B < \pi$ nên $\sin A, \sin B > 0.$ Do đó: $\sin A=\sqrt{1-\cos^2A}=\dfrac{3}{5}$ và $\sin B=\sqrt{1-\cos^2 B}=\dfrac{12}{13}.$

Ta có: $\cos(A+B)=\cos A\cos B-\sin A\sin B$ $=\dfrac{4}{5}\cdot\dfrac{5}{13}-\dfrac{3}{5}\cdot\dfrac{12}{13}$ $=\dfrac{-16}{65}.$

Suy ra: $\cos C=-\cos(A+B)=\dfrac{16}{65}.$

a) Ta có: $\sin^2 18^o=\dfrac{1-\cos 36^o}{2}.$

Do đó $2\sin^2 18^o=1-\cos 36^o$ $=1-\sin 54^o$ $=1-\sin(3\cdot 18^o)$ $=1-(3\sin 18^o-4\sin^3 18^o)$ $=1-3\sin 18^o+4\sin^3 18^o.$

Suy ra $4\sin^3 18^o-2\sin^2 18^o-3\sin 18^o+1=0$ $\Leftrightarrow (\sin 18^o-1)(4\sin^2 18^o+2\sin 18^o-1)=0$ $\Leftrightarrow \left[\begin{matrix} \sin 18^o=1 \\ \sin 18^o=\dfrac{\sqrt{5}-1}{2} \\ \sin 18^o=\dfrac{\sqrt{5}+1}{2} \end{matrix}\right.$

Vì $0 < \sin 18^o < 1$ nên ta chọn $\sin 18^o=\dfrac{\sqrt{5}-1}{2}.$

b) Ta có: $\cot\dfrac{5\pi}{8}=\cot\left(\dfrac{\pi}{8}+\dfrac{\pi}{2}\right)=-\tan\dfrac{\pi}{8}.$

Mặt khác: $\tan\left(2\cdot\dfrac{\pi}{8}\right)=\dfrac{2\tan\dfrac{\pi}{8}}{1-\tan^2\dfrac{\pi}{8}}$ $\Rightarrow 1=\dfrac{2\tan\dfrac{\pi}{8}}{1-\tan^2\dfrac{\pi}{8}}$ $\Rightarrow \tan^2\dfrac{\pi}{8}+2\tan\dfrac{\pi}{8}-1=0$ $\Rightarrow \tan\dfrac{\pi}{8}=-1-\sqrt{2}\;hoặc\; \tan\dfrac{\pi}{8}=-1+\sqrt{2}.$

Do $\tan\dfrac{\pi}{8} > 0$ nên ta chọn $\tan\dfrac{\pi}{8}=-1+\sqrt{2}.$

Suy ra $\cot\dfrac{5\pi}{8}=-\tan\dfrac{\pi}{8}=1-\sqrt{2}.$

a) $A=\sin 10^o\sin 30^o\sin 50^o\sin 70^o$ $=\cos 80^o\dfrac{1}{2}\cos 40^o\cos 20^o$ $=\dfrac{1}{2}\cos 20^o\cos 40^o\cos 80^o.$

Suy ra: $16\sin 20^o\cdot A=8\sin 20^o\cos 20^o\cos 40^o\cos 80^o$ $=4\sin 40^o\cos 40^o\cos 80^o$ $=2\sin 80^o\cos 80^o$ $=\sin 160^o$ $=\sin 20^o.$

Do đó, vì $\sin 20^o\neq 0$ nên $A=\dfrac{\sin 20^o}{16\sin 20^o}=\dfrac{1}{16}.$

b) $B=\cos\dfrac{\pi}{5}+\cos\dfrac{3\pi}{5}$ $=2\cos\dfrac{\pi}{5}\cos\dfrac{2\pi}{5}.$

Suy ra: $2\sin\dfrac{\pi}{5}\cdot B=4\sin\dfrac{\pi}{5}\cos\dfrac{\pi}{5}\cos\dfrac{2\pi}{5}$ $=2\sin\dfrac{2\pi}{5}\cos\dfrac{2\pi}{5}$ $=\sin\dfrac{4\pi}{5}$ $=\sin\dfrac{\pi}{5}.$

Do đó, vì $\sin\dfrac{\pi}{5}\neq 0$ nên $B=\dfrac{\sin\dfrac{\pi}{5}}{2\sin\dfrac{\pi}{5}}=\dfrac{1}{2}.$

c) $C=\cos^2\dfrac{\pi}{7}+\cos^2\dfrac{2\pi}{7}+\cos^2\dfrac{3\pi}{7}$ $=\dfrac{1+\cos\dfrac{2\pi}{7}}{2}+\dfrac{1+\cos\dfrac{4\pi}{7}}{2}+\dfrac{1+\cos\dfrac{6\pi}{7}}{2}$ $=\dfrac{1}{2}\left(3+\cos\dfrac{2\pi}{7}+\cos\dfrac{4\pi}{7}+\cos\dfrac{6\pi}{7}\right).$

Ta tính riêng $T=\cos\dfrac{2\pi}{7}+\cos\dfrac{4\pi}{7}+\cos\dfrac{6\pi}{7}.$

Ta có: $2\sin\dfrac{\pi}{7}\cdot T=2\sin\dfrac{\pi}{7}\cos\dfrac{2\pi}{7}+2\sin\dfrac{\pi}{7}\cos\dfrac{4\pi}{7}+2\sin\dfrac{\pi}{7}\cos\dfrac{6\pi}{7}$ $=\left(\sin\dfrac{3\pi}{7}-\sin\dfrac{\pi}{7}\right)+\left(\sin\dfrac{5\pi}{7}-\sin\dfrac{3\pi}{7}\right)+\left(\sin\pi-\sin\dfrac{5\pi}{7}\right)$ $=-\sin\dfrac{\pi}{7}.$

Do đó, vì $\sin\dfrac{\pi}{7}\neq 0$ nên $T=\dfrac{-\sin\dfrac{\pi}{7}}{2\sin\dfrac{\pi}{7}}=\dfrac{-1}{2}.$

Vậy $C=\dfrac{1}{2}\left(3+\dfrac{-1}{2}\right)=\dfrac{5}{4}.$

a) $A=\sin x+\cos x$ $=\sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right).$

Ta có: $-1\leq \sin\left(x+\dfrac{\pi}{4}\right)\leq 1$

Suy ra $-\sqrt{2} \leq \sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right) \leq \sqrt{2},$ hay $-\sqrt{2}\leq A\leq \sqrt{2}.$

Khi $x=\dfrac{3\pi}{4}$ thì $A=-\sqrt{2}.$ Khi $x=\dfrac{\pi}{4}$ thì $A=-\sqrt{2}.$

Vậy $max\; A=\sqrt{2}$ và $min\; A=-\sqrt{2}.$

b) $B=\sin^4 x+\cos^4 x$ $=\left(\sin^2x+\cos^2x\right)^2-2\sin^2x\cos^2x$ $=1-\dfrac{1}{2}\left(2\sin x\cos x\right)^2$ $=1-\dfrac{1}{2}\sin^2 2x$ $=1-\dfrac{1}{2}\cdot \dfrac{1-\cos 4x}{2}$ $=\dfrac{3}{4}+\dfrac{1}{4}\cos 4x.$

Vì $-1\leq \cos 4x\leq 1$ nên $\dfrac{1}{2}\leq \dfrac{3}{4}+\dfrac{1}{4}\cos 4x\leq 1,$ hay $\dfrac{1}{2}\leq B\leq 1.$

Khi $x=\dfrac{\pi}{4}$ thì $B=\dfrac{1}{2}.$ Khi $x=0$ thì $B=1.$

Vậy $max\;B=1$ và $min\;B=\dfrac{1}{2}.$

$\dfrac{\sin B}{\sin C}=2\cos A$ $\Leftrightarrow \sin B=2\sin C\cos A$ $\Leftrightarrow \sin B=\sin(C+A)+\sin(C-A)\;\;\;(1)$

Mặt khác: $A+B+C=\pi$ nên $\sin B=\sin(C+A)\;\;\;(2)$

Từ $(1)$ và $(2)$ suy ra $\sin(C-A)=0.$

Do đó, $C=A,$ hay tam giác $ABC$ cân tại $B.$