BÀI 3. CÔNG THỨC LƯỢNG GIÁC
$\cos \alpha = x = \overline {OH}$
$\sin \alpha = y = \overline {OK}$
$\tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }} = \overline {AT}$ với $\alpha \ne \dfrac{\pi }{2} + k\pi$
$\cot \alpha = \dfrac{{\cos \alpha }}{{\sin \alpha }} = \overline {BS}$ với $\alpha \ne k\pi$
Tính chất
Với mọi góc $\alpha$ ta có:
$- 1 \leqslant \cos \alpha \leqslant 1; - 1 \leqslant \sin \alpha \leqslant 1$
$\sin (\alpha + k2\pi ) = \sin \alpha $
$\cos (\alpha + k2\pi ) = \cos \alpha $
$\tan (\alpha + k\pi ) = \tan \alpha $
$\cot (\alpha + k\pi ) = \cot \alpha $
Dấu của các giá trị lượng giác
Giá trị lượng giác của các góc đặc biệt
Hai góc đối nhau
$\sin \left( { - \alpha } \right) = - \sin \alpha $
$\cos \left( { - \alpha } \right) = \cos \alpha $
$\tan \left( { - \alpha } \right) = - \tan \alpha $
$\cot \left( { - \alpha } \right) = - \cot \alpha $
Hai góc bù nhau
$\sin \left( {\pi - \alpha } \right) = \sin \alpha $
$\cos \left( {\pi - \alpha } \right) = - \cos \alpha $
$\tan \left( {\pi - \alpha } \right) = - \tan \alpha $
$\cot \left( {\pi - \alpha } \right) = - \cot \alpha $
Hai góc phụ nhau
$\sin \left( {\dfrac{\pi }{2} - \alpha } \right) = \cos \alpha $
$\cos \left( {\dfrac{\pi }{2} - \alpha } \right) = \sin \alpha $
$\tan \left( {\dfrac{\pi }{2} - \alpha } \right) = \cot \alpha $
$\cot \left( {\dfrac{\pi }{2} - \alpha } \right) = \tan \alpha $
Hai góc hơn kém $\pi $
$\sin \left( {\alpha + \pi } \right) = - \sin \alpha $
$\cos \left( {\alpha + \pi } \right) = - \cos \alpha $
$\tan \left( {\alpha + \pi } \right) = \tan \alpha $
$\cot \left( {\alpha + \pi } \right) = \cot \alpha $
Hai góc hơn kém $\dfrac{\pi }{2}$
$\sin \left( {\alpha + \dfrac{\pi }{2}} \right) = \cos \alpha $
$\cos \left( {\alpha + \dfrac{\pi }{2}} \right) = - \sin \alpha $
$\tan \left( {\alpha + \dfrac{\pi }{2}} \right) = - \cot \alpha $
$\cot \left( {\alpha + \dfrac{\pi }{2}} \right) = - \tan \alpha $
Công thức lượng giác cơ bản
${\sin ^2}\alpha + {\cos ^2}\alpha = 1$
$\tan \alpha .\cot \alpha = 1$
$\tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }}$
$\cot \alpha = \dfrac{{\cos \alpha }}{{\sin \alpha }}$
$\dfrac{1}{{{{\cos }^2}\alpha }} = 1 + {\tan ^2}\alpha $
$\dfrac{1}{{{{\sin }^2}\alpha }} = 1 + {\cot ^2}\alpha $
Công thức cộng
$\sin \left( {\alpha + \beta } \right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta $
$\sin \left( {\alpha - \beta } \right) = \sin \alpha \cos \beta - \cos \alpha \sin \beta {\text{ }}$
$\cos \left( {\alpha + \beta } \right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta $
$\cos \left( {\alpha - \beta } \right) = \cos \alpha \cos \beta + \sin \alpha \sin \beta $
$\tan \left( {\alpha + \beta } \right) = \dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }}$
$\tan \left( {\alpha - \beta } \right) = \dfrac{{\tan \alpha - \tan \beta }}{{1 + \tan \alpha \tan \beta }}$
Công thức nhân đôi
$\sin 2\alpha = 2\sin \alpha \cos \alpha {\text{ }}$
$\cos 2\alpha = {\cos ^2}\alpha - {\sin ^2}\alpha $
$\cos 2\alpha = 1 - 2{\sin ^2}\alpha $
$\cos 2\alpha = 2{\cos ^2}\alpha - 1{\text{ }}$
${\text{tan2}}\alpha {\text{ = }}\dfrac{{{\text{2tan}}\alpha }}{{1 - {{\tan }^2}\alpha }}$
$\cot 2\alpha = \dfrac{{{{\cot }^2}\alpha - 1}}{{2\cot \alpha }}$
Công thức hạ bậc
${\cos ^2}\alpha = \dfrac{{1 + \cos 2\alpha }}{2}$
${\sin ^2}\alpha = \dfrac{{1 - \cos 2\alpha }}{2}$
${\tan ^2}\alpha = \dfrac{{1 - \cos 2\alpha }}{{1 + \cos 2\alpha }}$
Công thức nhân ba
$\cos 3\alpha = 4{\cos ^3}\alpha - 3\cos \alpha $
$\sin 3\alpha = 3\sin \alpha - 4{\sin ^3}\alpha $
$\tan 3\alpha = \dfrac{{3\tan \alpha - {{\tan }^3}\alpha }}{{1 - 3{{\tan }^2}\alpha }}$
${\cos ^3}\alpha = \dfrac{{3\cos \alpha + \cos 3\alpha }}{4}$
${\sin ^3}\alpha = \dfrac{{3\sin \alpha - \sin 3\alpha }}{4}$
Công thức biến đổi tích thành tổng
$\cos \alpha \cos \beta = \dfrac{1}{2}\left[ {\cos \left( {\alpha + \beta } \right) + \cos \left( {\alpha - \beta } \right)} \right]$
$\sin \alpha \sin \beta = - \dfrac{1}{2}\left[ {\cos \left( {\alpha + \beta } \right) - \cos \left( {\alpha - \beta } \right)} \right]{\text{ }}$
$\sin \alpha \cos \beta = \dfrac{1}{2}\left[ {\sin \left( {\alpha + \beta } \right) + \sin \left( {\alpha - \beta } \right)} \right]$
Công thức biến đổi tổng thành tích
$\cos \alpha + \cos \beta = 2\cos \dfrac{{\alpha + \beta }}{2}\cos \dfrac{{\alpha - \beta }}{2}$
$\cos \alpha - \cos \beta = - 2\sin \dfrac{{\alpha + \beta }}{2}\sin \dfrac{{\alpha - \beta }}{2}$
$\sin \alpha + \sin \beta = 2\sin \dfrac{{\alpha + \beta }}{2}\cos \dfrac{{\alpha - \beta }}{2}$
$\sin \alpha - \sin \beta = 2\cos \dfrac{{\alpha + \beta }}{2}\sin \dfrac{{\alpha - \beta }}{2}$
Một số công thức khác
$\tan \alpha + \tan \beta = \dfrac{{\sin (\alpha + \beta )}}{{\cos \alpha .\cos \beta }}$
$\tan \alpha - \tan \beta = \dfrac{{\sin (\alpha - \beta )}}{{\cos \alpha .\cos \beta }}$
$\cot \alpha + \cot \beta = \dfrac{{\sin (\alpha + \beta )}}{{\sin \alpha .\sin \beta }}$
$\cot \alpha - \cot \beta = \dfrac{{\sin (\beta - \alpha )}}{{\sin \alpha .\sin \beta }}$
$\sin \alpha + \cos \alpha = \sqrt 2 .\sin \left( {\alpha + \dfrac{\pi }{4}} \right) = \sqrt 2 .\cos \left( {\alpha - \dfrac{\pi }{4}} \right)$
$\sin \alpha - \cos \alpha = \sqrt 2 \sin \left( {\alpha - \dfrac{\pi }{4}} \right) = - \sqrt 2 \cos \left( {\alpha + \dfrac{\pi }{4}} \right)$
$\tan \left( {\dfrac{\pi }{4} + \alpha } \right) = \dfrac{{1 + \tan \alpha }}{{1 - \tan \alpha }}$
$\tan \left( {\dfrac{\pi }{4} - \alpha } \right) = \dfrac{{1 - \tan \alpha }}{{1 + \tan \alpha }}$
Công thức tính theo $\tan \dfrac{x}{2}$
Đặt $t = \tan \dfrac{x}{2}$. Khi đó:
$\cos x = \dfrac{{1 - {t^2}}}{{1 + {t^2}}}$
$\sin x = \dfrac{{2t}}{{1 + {t^2}}}$
$\tan x = \dfrac{{2t}}{{1 - {t^2}}}$
$\cot x = \dfrac{{1 - {t^2}}}{{2t}}$
Hằng đẳng thức hay dùng
${A^2} + {B^2} = {(A + B)^2} - 2AB$
${A^2} - {B^2} = (A - B)(A + B)$
${A^4} + {B^4} = {({A^2} + {B^2})^2} - 2{A^2}{B^2}$
${A^3} + {B^3} = (A + B)({A^2} - AB + {B^2})$
${A^3} - {B^3} = (A - B)({A^2} + AB + {B^2})$
Ví dụ 1: Chứng minh rằng:
a) ${\sin ^4}\alpha + {\cos ^4}\alpha = \dfrac{3}{4} + \dfrac{{\cos 4\alpha }}{4}$.
b) ${\sin ^6}\alpha + {\cos ^6}\alpha = \dfrac{5}{8} + \dfrac{3}{8}\cos 4\alpha $.
Lời giải
a) Ta có ${\sin ^4}\alpha + {\cos ^4}\alpha$$ = {\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)^2} - 2{\sin ^2}\alpha {\cos ^2}\alpha$$ = 1 - \dfrac{1}{2}{\sin ^2}2\alpha $$ = 1 - \dfrac{{1 - \cos 4\alpha }}{4}$$ = \dfrac{3}{4} + \dfrac{{\cos 4\alpha }}{4}$.
b) Ta có ${\sin ^6}\alpha + {\cos ^6}\alpha$$ = {\left( {{{\sin }^2}\alpha } \right)^3} + {\left( {{{\cos }^2}\alpha } \right)^3} + 3{\sin ^2}\alpha {\cos ^2}\alpha \left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) - 3{\sin ^2}\alpha {\cos ^2}\alpha \left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)$$ = {\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)^3} - 3{\sin ^2}\alpha {\cos ^2}\alpha$$ = 1 - \dfrac{3}{4}{\left( {2\sin \alpha \cos \alpha } \right)^2}$$ = 1 - \dfrac{3}{4}{\sin ^2}2\alpha$$ = 1 - \dfrac{3}{8}\left( {1 - \cos 4\alpha } \right)$$ = \dfrac{5}{8} + \dfrac{3}{8}\cos 4\alpha $.
Ví dụ 2: Chứng minh rằng:
a) $\sin (\alpha + \beta ).\sin (\alpha - \beta )$$ = {\sin ^2}\alpha - {\sin ^2}\beta $.
b) $\cot \dfrac{\alpha }{2}\cot \dfrac{\beta }{2} = 2$ với $\sin \alpha + \sin \beta = 3\sin \left( {\alpha + \beta } \right),\alpha + b \ne k2\pi $.
c) $\dfrac{{\sin \alpha + \sin \beta \cos \left( {\alpha + \beta } \right)}}{{\cos \alpha - \sin \beta \sin \left( {\alpha + \beta } \right)}}$$ = \tan \left( {\alpha + \beta } \right)$.
Lời giải
a) Ta có $\sin (\alpha + \beta ).\sin (\alpha - \beta )$$ = - \dfrac{1}{2}\left[ {\cos 2\alpha - \cos 2\beta } \right]$ $ = - \dfrac{1}{2}\left[ {\left( {1 - 2{{\sin }^2}\alpha } \right) - \left( {1 - 2{{\sin }^2}\beta } \right)} \right]$$ = {\sin ^2}\alpha - {\sin ^2}\beta $.
b) Từ giả thiết ta có $2\sin \dfrac{{\alpha + \beta }}{2}\cos \dfrac{{\alpha - \beta }}{2}$$ = 6\sin \dfrac{{\alpha + \beta }}{2}\cos \dfrac{{\alpha + \beta }}{2}$.
Do $\alpha + \beta \ne k2\pi \Rightarrow \sin \dfrac{{\alpha + \beta }}{2} \ne 0$ suy ra $\cos \dfrac{{\alpha - \beta }}{2}$$ = 3\cos \dfrac{{\alpha + \beta }}{2}$
$ \Leftrightarrow \cos \dfrac{\alpha }{2}\cos \dfrac{\beta }{2} + \sin \dfrac{\alpha }{2}\sin \dfrac{\beta }{2}$$ = 3\left( {\cos \dfrac{\alpha }{2}\cos \dfrac{\beta }{2} - \sin \dfrac{\alpha }{2}\sin \dfrac{\beta }{2}} \right)$
$ \Leftrightarrow 2\sin \dfrac{\alpha }{2}\sin \dfrac{\beta }{2}$$ = \cos \dfrac{\alpha }{2}\cos \dfrac{\beta }{2}$
$ \Leftrightarrow \cot \dfrac{\alpha }{2}\cot \dfrac{\beta }{2} = 2$ (ĐPCM).
c) Ta có $VT = \dfrac{{\sin \alpha + \dfrac{1}{2}\left[ {\sin \left( {\alpha + 2\beta } \right) + \sin \left( { - \alpha } \right)} \right]}}{{\cos \alpha - \left( { - \dfrac{1}{2}} \right)\left[ {\cos \left( {\alpha + 2\beta } \right) - \cos \left( { - \alpha } \right)} \right]}}$$ = \dfrac{{\sin \alpha + \sin \left( {\alpha + 2\beta } \right)}}{{\cos \alpha + \cos \left( {\alpha + 2\beta } \right)}}$
$ = \dfrac{{2\sin \left( {\alpha + \beta } \right)\cos \left( { - \beta } \right)}}{{2\cos \left( {\alpha + \beta } \right)\cos \left( { - \beta } \right)}}$$ = \tan \left( {\alpha + \beta } \right) = VP$ (ĐPCM).
Ví dụ 3: Rút gọn biểu thức:
a) $A = {\cos ^2}\alpha + {\cos ^2}\left( {\dfrac{{2\pi }}{3} + \alpha } \right) + {\cos ^2}\left( {\dfrac{{2\pi }}{3} - \alpha } \right)$.
b) $B = \cos \left( {\alpha - \dfrac{\pi }{3}} \right).\cos \left( {\alpha + \dfrac{\pi }{4}} \right) + \cos \left( {\alpha + \dfrac{\pi }{6}} \right).\cos \left( {\alpha + \dfrac{{3\pi }}{4}} \right)$.
Lời giải
a) Ta có: $A = {\cos ^2}\alpha + {\cos ^2}\left( {\dfrac{{2\pi }}{3} + \alpha } \right) + {\cos ^2}\left( {\dfrac{{2\pi }}{3} - \alpha } \right) $
$ = \dfrac{1}{2}\left[ {3 + \cos 2\alpha + \cos \left( {\dfrac{{4\pi }}{3} + 2\alpha } \right) + \cos \left( {\dfrac{{4\pi }}{3} - 2\alpha } \right)} \right]$
$ = \dfrac{1}{2}\left[ {3 + \cos 2\alpha + 2\cos \dfrac{{4\pi }}{3}\cos 2\alpha } \right] = \dfrac{3}{2}$.
b) Vì $\alpha + \dfrac{\pi }{6}$$ = \left( {\alpha - \dfrac{\pi }{3}} \right) + \dfrac{\pi }{2}$$ \Rightarrow \cos \left( {\alpha + \dfrac{\pi }{6}} \right)$$ = - \sin \left( {\alpha - \dfrac{\pi }{3}} \right)$ và $\cos \left( {\alpha + \dfrac{{3\pi }}{4}} \right)$$ = - \sin \left( {\alpha + \dfrac{\pi }{4}} \right)$ nên
$B = \cos \left( {\alpha - \dfrac{\pi }{3}} \right).\cos \left( {\alpha + \dfrac{\pi }{4}} \right) + \sin \left( {\alpha - \dfrac{\pi }{3}} \right).\sin \left( {\alpha + \dfrac{\pi }{4}} \right)$
$= \cos \left[ {\left( {\alpha - \dfrac{\pi }{3}} \right) - \left( {\alpha + \dfrac{\pi }{4}} \right)} \right]$$ = \cos \left( { - \dfrac{\pi }{3} - \dfrac{\pi }{4}} \right)$$ = \cos \left( {\dfrac{\pi }{3} + \dfrac{\pi }{4}} \right) $$= \cos \dfrac{\pi }{3}\cos \dfrac{\pi }{4} - \sin \dfrac{\pi }{3}\sin \dfrac{\pi }{4}$$ = \dfrac{1}{2}.\dfrac{{\sqrt 2 }}{2} - \dfrac{{\sqrt 3 }}{2}.\dfrac{{\sqrt 2 }}{2}$$ = \dfrac{{\sqrt 2 - \sqrt 6 }}{4}$.
Ví dụ 4: Đơn giản biểu thức sau:
a) $A = \dfrac{{\cos a + 2\cos 2a + \cos 3a}}{{\sin a + \sin 2a + \sin 3a}}$.
b) $B = \dfrac{{\cos \left( {a + \dfrac{\pi }{3}} \right) + \cos \left( {a - \dfrac{\pi }{3}} \right)}}{{\cot a - \cot \dfrac{a}{2}}}$.
Lời giải
a) $A = \dfrac{{\left( {\cos a + \cos 3a} \right) + 2\cos 2a}}{{\left( {\sin a + \sin 3a} \right) + 2\sin 2a}}$$ = \dfrac{{2\cos 2a\cos a + 2\cos 2a}}{{2\sin 2a\cos a + 2\sin 2a}}$$ = \dfrac{{2\cos 2a\left( {\cos a + 1} \right)}}{{2\sin 2a\left( {\cos a + 1} \right)}}$$ = \cot 2a$.
b) Ta có $\cos \left( {a + \dfrac{\pi }{3}} \right) + \cos \left( {a - \dfrac{\pi }{3}} \right)$$ = 2\cos a\cos \dfrac{\pi }{3}$$ = \cos a$ và $\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}}}$$ = \dfrac{{\sin \left( {\dfrac{a}{2} - a} \right)}}{{\sin a\sin \dfrac{a}{2}}}$$ = \dfrac{{ - \sin \dfrac{a}{2}}}{{\sin a\sin \dfrac{a}{2}}}$$ = - \dfrac{1}{{\sin a}}$.
Suy ra $B = \dfrac{{\cos a}}{{ - \dfrac{1}{{\sin a}}}}$$ = - \sin a\cos a$$ = - \dfrac{{\sin 2a}}{2}$.
Ví dụ 5: Chứng minh rằng: $\sin 3\alpha$$ = 3\sin \alpha - 4{\sin ^3}\alpha$$ = 4\sin \alpha .\sin \left( {\dfrac{\pi }{3} - \alpha } \right).\sin \left( {\dfrac{\pi }{3} + \alpha } \right)$.
Lời giải
Ta có $\sin 3\alpha = \sin \left( {2\alpha + \alpha } \right)$$ = \sin 2\alpha \cos \alpha + \cos 2\alpha \sin \alpha $$= 2\sin \alpha {\cos ^2}\alpha + \cos 2\alpha \sin \alpha $$= 2\sin \alpha \left( {1 - {{\sin }^2}\alpha } \right) + \left( {1 - 2{{\sin }^2}\alpha } \right)\sin \alpha$$= 3\sin \alpha - 4{\sin ^3}\alpha (1) $.
Mặt khác $4\sin \alpha .\sin \left( {\dfrac{\pi }{3} - \alpha } \right).\sin \left( {\dfrac{\pi }{3} + \alpha } \right)$$ = - 4\sin \alpha .\dfrac{1}{2}\left( {\cos \dfrac{{2\pi }}{3} - \cos \left( { - 2\alpha } \right)} \right)$$= - 2\sin \alpha .\left( { - \dfrac{1}{2} - \cos 2\alpha } \right) = 2\sin \alpha \left( {\dfrac{1}{2} + 1 - 2{{\sin }^2}\alpha } \right)$$= 3\sin \alpha - 4{\sin ^3}\alpha (2)$.
Từ (1) và (2) ta có ĐPCM.
Ví dụ 6: Chứng minh rằng: ${\sin ^4}\alpha$$ = \dfrac{3}{8} - \dfrac{1}{2}\cos 2\alpha + \dfrac{1}{8}\cos 4\alpha $.
Lời giải
${\sin ^4}\alpha$$ = {\left( {\dfrac{{1 - \cos 2\alpha }}{2}} \right)^2}$$ = \dfrac{{1 - 2\cos 2\alpha + {{\cos }^2}2\alpha }}{4}$$ = \dfrac{{1 - 2\cos 2\alpha + \dfrac{{1 + \cos 4\alpha }}{2}}}{4}$$ = \dfrac{3}{8} - \dfrac{1}{2}\cos 2\alpha + \dfrac{1}{8}\cos 4\alpha $.
Ví dụ 7: Cho $\sin x = 2\sin \left( {x + y} \right),$ $x + y \ne \dfrac{\pi }{2} + k\pi $. Chứng minh $\tan \left( {x + y} \right)$$ = \dfrac{{\sin y}}{{\cos y - 2}}$.
Lời giải
$\sin x = \sin \left[ {\left( {x + y} \right) - y} \right]$$ = \sin \left( {x + y} \right)\cos y - \cos \left( {x + y} \right)\sin y$
$\Rightarrow \sin \left( {x + y} \right)\cos y - \cos \left( {x + y} \right)\sin y$$ = 2\sin \left( {x + y} \right) $$ \Rightarrow \left( {\cos y - 2} \right)\sin \left( {x + y} \right)$$ = \cos \left( {x + y} \right)\sin y $$ \Rightarrow \tan \left( {x + y} \right)$$ = \dfrac{{\sin y}}{{\cos y - 2}}$.
Ví dụ 8: Chứng minh các hệ thức sau:
a) $4\left( {{{\cos }^3}\alpha \sin \alpha - {{\sin }^3}\alpha \cos \alpha } \right) = \sin 4\alpha $.
b) $\tan x + \tan y = \dfrac{{2\sin (x + y)}}{{\cos (x + y) + \cos (x - y)}}$.
c) $\tan x.\tan 3x = \dfrac{{{{\tan }^2}2x - {{\tan }^2}x}}{{1 - {{\tan }^2}2x.{{\tan }^2}x}}$.
Lời giải
a) $VT = 4\sin \alpha \cos \alpha \left( {{{\cos }^2}\alpha - {{\sin }^2}\alpha } \right)$$ = 2\sin 2\alpha \cos 2\alpha = \sin 4\alpha = VP$.
b) $VP = \dfrac{{2\left( {\sin x\cos y + \sin y\cos x} \right)}}{{2\cos x\cos y}}$$ = \tan x + \tan y = VT$.
c) $VP = \dfrac{{\left( {\tan 2x + \tan x} \right)\left( {\tan 2x - \tan x} \right)}}{{\left( {1 - \tan 2x.\tan x} \right)\left( {1 + \tan 2x.\tan x} \right)}}$$ = \tan \left( {2x + x} \right)\tan \left( {2x - x} \right) = VT$.
Ví dụ 9: Đơn giản biểu thức sau:
a) $A = \dfrac{{1 - \cos \alpha + \cos 2\alpha }}{{\sin 2\alpha - \sin \alpha }}$.
b) $B = \dfrac{{\cos a - \cos 3a + \cos 5a - \cos 7a}}{{\sin a + \sin 3a + \sin 5a + \sin 7a}}$.
c) $C = \cos a - \dfrac{{\cos \left( {2a - \dfrac{\pi }{6}} \right) - \cos \left( {2a + \dfrac{\pi }{6}} \right)}}{{2\cos a}}$.
Lời giải
a) $A = \dfrac{{1 - \cos \alpha + 2{{\cos }^2}\alpha - 1}}{{\sin \alpha \left( {2\cos \alpha - 1} \right)}} = \cot \alpha $.
b) $B = \dfrac{{ - 2\sin 4a\sin \left( { - 3a} \right) - 2\sin 4a\sin 2a}}{{2\sin 4a\cos 3a + 2\sin 4a\cos 2a}}$$ = \dfrac{{\sin 3a - \sin 2a}}{{\cos 3a - \cos 2a}}$$ = \dfrac{{2\cos \dfrac{{5a}}{2}\sin \dfrac{a}{2}}}{{ - 2\sin \dfrac{{5a}}{2}\sin \dfrac{a}{2}}}$$ = - \cot \dfrac{{5a}}{2}$.
c) $C = \dfrac{{2{{\cos }^2}a - 2\sin 2a\sin \left( { - \dfrac{\pi }{6}} \right)}}{{2\cos a}}$$ = \dfrac{{2{{\cos }^2}a + 4\sin a\cos a}}{{2\cos a}}$$ = \cos a + 2\sin a$.
Ví dụ 10: Chứng minh các hệ thức sau:
a) Nếu $2\tan a = \tan (a + b)$ thì $\sin b = \sin a.\cos (a + b)$.
b) Nếu $2\tan a = \tan (a + b)$ thì $3\sin b = \sin (2a + b)$.
c) Nếu $\tan (a + b).\tan b = - 3$ thì $\cos (a + 2b) + 2\cos a = 0$.
d) Nếu $3\sin \left( {a + b} \right) = \cos \left( {a - b} \right)$ thì $8{\sin ^2}\left( {a + b} \right) = \cos 2a\cos 2b$.
Lời giải
a) $2\tan a = \tan (a + b)$$ \Rightarrow \tan a = \tan (a + b) - \tan a$
$ \Rightarrow \tan a = \dfrac{{\sin b}}{{\cos (a + b)\cos a}}$
$ \Rightarrow \sin a\cos (a + b) = \sin b$.
b) $2\tan a = \tan (a + b)$$ \Rightarrow 3\tan a = \tan (a + b) + \tan a$$ = \dfrac{{\sin \left( {2a + b} \right)}}{{\cos (a + b)\cos a}}$
$ \Rightarrow 3\sin a\cos \left( {a + b} \right)$$ = \sin \left( {2a + b} \right)$.
Theo câu a) ta có $\sin b = \sin a.\cos (a + b)$ suy ra $3\sin b = \sin (2a + b)$.
c) $\tan (a + b).\tan b = - 3$$ \Rightarrow \sin \left( {a + b} \right)\sin b = - 3\cos \left( {a + b} \right)\cos b$
$ \Rightarrow \cos \left( {a + b} \right)\cos b + \sin \left( {a + b} \right)\sin b = - 2\cos \left( {a + b} \right)\cos b$$\Rightarrow \cos a = - \left[ {\cos \left( {2a + b} \right) + \cos a} \right]$$ \Rightarrow \cos (a + 2b) + 2\cos a = 0$.
d) Từ giả thiết ta có $9{\sin ^2}\left( {a + b} \right)$$ = {\cos ^2}\left( {a - b} \right)$
$ \Rightarrow 9.\dfrac{{1 - \cos 2\left( {a + b} \right)}}{2}$$ = \dfrac{{1 + \cos 2\left( {a - b} \right)}}{2}$$ \Rightarrow 8\left[ {1 - \cos 2\left( {a + b} \right)} \right]$$ = \cos 2\left( {a + b} \right) + \cos 2\left( {a - b} \right) $$ \Rightarrow 16{\sin ^2}\left( {a + b} \right) = 2\cos 2a\cos 2b$.
Hay $8{\sin ^2}\left( {a + b} \right) = \cos 2a\cos 2b$ (ĐPCM).
Ví dụ 11: Chứng minh trong mọi tam giác $ABC$ ta đều có:
a) $\sin A + \sin B + \sin C$$ = 4\cos \dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2}$.
b) ${\sin ^2}A + {\sin ^2}B + {\sin ^2}C$$ = 2(1 + \cos A\cos B\cos C)$.
c) $\sin 2A + \sin 2B + \sin 2C$$ = 4\sin A\sin B\sin C$.
Lời giải
a) $VT = 2\sin \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2} + 2\sin \dfrac{C}{2}\cos \dfrac{C}{2}$.
Mặt khác trong tam giác $ABC$ ta có $A + B + C = \pi \Rightarrow $$\dfrac{{A + B}}{2} = \dfrac{\pi }{2} - \dfrac{C}{2}$.
Suy ra $\sin \dfrac{{A + B}}{2}$$ = \cos \dfrac{C}{2},\sin \dfrac{C}{2}$$ = \cos \dfrac{{A + B}}{2}$.
Vậy $VT = 2\cos \dfrac{C}{2}\cos \dfrac{{A - B}}{2} + 2\cos \dfrac{{A + B}}{2}\cos \dfrac{C}{2} $$= 2\cos \dfrac{C}{2}\left( {\cos \dfrac{{A - B}}{2} + \cos \dfrac{{A + B}}{2}} \right)$
$ = 4\cos \dfrac{C}{2}\cos \dfrac{A}{2}\cos \dfrac{B}{2} = VP$ (ĐPCM).
b) $VT = \dfrac{{1 - \cos 2A}}{2} + \dfrac{{1 - \cos 2B}}{2} + 1 - {\cos ^2}C$$ = 2 - \dfrac{{\cos 2A + \cos 2B}}{2} - {\cos ^2}C$
$ = 2 - \cos \left( {A + B} \right)\cos \left( {A - B} \right) - {\cos ^2}C$.
Vì $A + B + C = \pi \Rightarrow \cos \left( {A + B} \right) = - \cos C$ nên
$VT = 2 + \cos C\cos \left( {A - B} \right) + \cos C\cos \left( {A + B} \right)$$ = 2 + \cos C\left[ {\cos \left( {A - B} \right) + \cos \left( {A + B} \right)} \right]$
$ = 2 + \cos C.2\cos A\cos B = 2(1 + \cos A\cos B\cos C) = VP$ (ĐPCM).
c) $VT = 2\sin \left( {A + B} \right)\cos \left( {A - B} \right) + 2\sin C\cos C$.
Vì $A + B + C = \pi \Rightarrow \cos C$$ = - \cos \left( {A + B} \right),\sin \left( {A + B} \right) = \sin C$ nên $VT = 2\sin C\cos \left( {A - B} \right) - 2\sin C\cos \left( {A + B} \right)$$ = 2\sin C\left[ {\cos \left( {A - B} \right) - \cos \left( {A + B} \right)} \right]$
$ = 2\sin C.\left[ { - 2\sin A\sin \left( { - B} \right)} \right]$$ = 4\sin A\sin B\sin C = VP$ (ĐPCM).
Ví dụ 12: Chứng minh trong mọi tam giác $ABC$ không vuông ta có:
a) $\tan A + \tan B + \tan C = \tan A.\tan B.\tan C$.
b) $\cot A.\cot B + \cot B.\cot C + \cot C.\cot A = 1$.
Lời giải
a) Đẳng thức tương đương với $\tan A + \tan B = \tan A.\tan B.\tan C - \tan C$
$ \Leftrightarrow \tan A + \tan B$$ = \tan C\left( {\tan A\tan B - 1} \right)\left( * \right)$.
Do tam giác $ABC$ không vuông nên $A + B \ne \dfrac{\pi }{2}$.
$ \Rightarrow \tan A\tan B - 1$$ = \dfrac{{\sin A\sin B}}{{\cos A\cos B}} - 1$$ = \dfrac{{\sin A\sin B - \cos A\cos B}}{{\cos A\cos B}}$$ = - \dfrac{{\cos \left( {A + B} \right)}}{{\cos A\cos B}} \ne 0$.
Suy ra $\left( * \right) \Leftrightarrow \dfrac{{\tan A + \tan B}}{{\tan A\tan B - 1}}$$ = \tan C $$\Leftrightarrow \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$$ = - \tan C$$ \Leftrightarrow \tan \left( {A + B} \right) = - \tan C$.
Đẳng thức cuối đúng vì $A + B + C = \pi $ (ĐPCM).
b) Vì $A + B + C = \pi$$ \Rightarrow \cot \left( {A + B} \right) = - \cot C$.
Theo công thức cộng ta có:
$\cot \left( {A + B} \right)$$ = \dfrac{1}{{\tan \left( {A + B} \right)}} = \dfrac{{1 - \tan A\tan B}}{{\tan A + \tan B}}$$ = \dfrac{{1 - \dfrac{1}{{\cot A\cot B}}}}{{\dfrac{1}{{\cot A}} + \dfrac{1}{{\cot B}}}}$$ = \dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}}$.
Suy ra $\dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}}$$ = - \cot C \Rightarrow \cot A\cot B - 1$$ = - \cot C\left( {\cot A + \cot B} \right)$.
Hay $\cot A.\cot B + \cot B.\cot C + \cot C.\cot A = 1$ (ĐPCM).
Ví dụ 13: Chứng minh trong mọi tam giác $ABC$ ta có: $\cos A + \cos B + \cos C \leqslant \dfrac{3}{2}$.
Lời giải
a) Ta có $\cos A + \cos B + \cos C$$ = 2\cos \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2} + \cos C$.
Vì $\dfrac{{A + B}}{2}$$ = \dfrac{\pi }{2} - \dfrac{C}{2}$ nên $\cos \dfrac{{A + B}}{2}$$ = \sin \dfrac{C}{2}$.
Mặt khác $\cos C = 1 - 2{\sin ^2}\dfrac{C}{2}$ do đó
$\cos A + \cos B + \cos C$$ = 2\sin \dfrac{C}{2}\cos \dfrac{{A - B}}{2} + 1 - 2{\sin ^2}\dfrac{C}{2} $$= - 2\left( {{{\sin }^2}\dfrac{C}{2} - \sin \dfrac{C}{2}\cos \dfrac{{A - B}}{2} - \dfrac{1}{2}} \right)$
$ = - 2\left( {{{\sin }^2}\dfrac{C}{2} - 2\sin \dfrac{C}{2}.\dfrac{1}{2}\cos \dfrac{{A - B}}{2} + \dfrac{1}{4}{{\cos }^2}\dfrac{{A - B}}{2}} \right)$$ + 1 + \dfrac{1}{2}{\cos ^2}\dfrac{{A - B}}{2}$
$ = - 2{\left( {\sin \dfrac{C}{2} + \dfrac{1}{2}\cos \dfrac{{A - B}}{2}} \right)^2} + 1$$ + \dfrac{1}{2}{\cos ^2}\dfrac{{A - B}}{2}$.
Vì $\left| {\cos \dfrac{{A - B}}{2}} \right| \leqslant 1$$ \Rightarrow {\cos ^2}\dfrac{{A - B}}{2} \leqslant 1$ nên
$\cos A + \cos B + \cos C \leqslant 1 + \dfrac{1}{2} = \dfrac{3}{2}$ (ĐPCM).
Ví dụ 14: Cho $\cos 2x = \dfrac{3}{5}$(với $\dfrac{{3\pi }}{4} < x < \pi $). Tính $\sin x$, $\cos x$.
Lời giải
Vì $\dfrac{{3\pi }}{4} < x < \pi $ nên $\sin x > 0,\cos x < 0$.
Áp dụng công thức hạ bậc, ta có :
${\sin ^2}x = \dfrac{{1 - \cos 2x}}{2}$$ = \dfrac{1}{5} \Rightarrow \sin x$$ = \dfrac{1}{{\sqrt 5 }}$ ;
${\cos ^2}x = \dfrac{{1 + \cos 2x}}{2}$$ = \dfrac{4}{5} \Rightarrow \cos x$$ = - \dfrac{2}{{\sqrt 5 }}$.
Ví dụ 15: Cho $\cos 2x = - \dfrac{4}{5}$, với $\dfrac{\pi }{4} < x < \dfrac{\pi }{2}$.
a) Tính $\sin x,$$\cos x$, $\sin \left( {x + \dfrac{\pi }{3}} \right)$,$\cos \left( {2x - \dfrac{\pi }{4}} \right)$.
Lời giải
Vì $\dfrac{\pi }{4} < x < \dfrac{\pi }{2}$ nên $\sin x > 0,\cos x > 0$.
Áp dụng công thức hạ bậc, ta có:
${\sin ^2}x = \dfrac{{1 - \cos 2x}}{2} = \dfrac{9}{{10}}$$ \Rightarrow \sin x = \dfrac{3}{{\sqrt {10} }}$ ;
${\cos ^2}x = \dfrac{{1 + \cos 2x}}{2} = \dfrac{1}{{10}}$$ \Rightarrow \cos x = \dfrac{1}{{\sqrt {10} }}$.
Theo công thức cộng, ta có :
$\sin \left( {x + \dfrac{\pi }{3}} \right)$$ = \sin x\cos \dfrac{\pi }{3} + \cos x\sin \dfrac{\pi }{3}$$ = \dfrac{3}{{\sqrt {10} }}.\dfrac{1}{2} + \dfrac{1}{{\sqrt {10} }}.\dfrac{{\sqrt 3 }}{2}$$ = \dfrac{{3 + \sqrt 3 }}{{2\sqrt {10} }}$ ;
$\cos \left( {2x - \dfrac{\pi }{4}} \right)$$ = \cos 2x\sin \dfrac{\pi }{4} + \cos \dfrac{\pi }{4}\sin 2x$$ = - \dfrac{4}{5}.\dfrac{{\sqrt 2 }}{2} + \dfrac{{\sqrt 2 }}{2}.2.\dfrac{3}{{\sqrt {10} }}.\dfrac{1}{{\sqrt {10} }}$$ = - \dfrac{{\sqrt 2 }}{{10}}$.
Ví dụ 16: Cho $\cos 4\alpha + 2 = 6{\sin ^2}\alpha $ với $\dfrac{\pi }{2} < \alpha < \pi $. Tính $\tan 2\alpha $.
Lời giải
Ta có $\cos 4\alpha + 2 = 6{\sin ^2}\alpha$$ \Leftrightarrow 2{\cos ^2}2\alpha - 1 + 2 = 3\left( {1 - \cos 2\alpha } \right)$
$ \Leftrightarrow 2{\cos ^2}2\alpha + 3\cos 2\alpha - 2 = 0$$ \Leftrightarrow \left( {2\cos 2\alpha - 1} \right)\left( {\cos 2\alpha + 2} \right) = 0$$ \Leftrightarrow \cos 2\alpha = \dfrac{1}{2}$ (Vì $\cos 2\alpha + 2 > 0$).
Ta có $1 + {\tan ^2}2\alpha = \dfrac{1}{{{{\cos }^2}2\alpha }}$$ \Rightarrow {\tan ^2}2\alpha = \dfrac{1}{{{{\cos }^2}2\alpha }} - 1 = 3$.
Vì $\dfrac{\pi }{2} < \alpha < \pi$$ \Rightarrow \pi < \alpha < 2\pi $ nên $\sin 2\alpha < 0$. Mặt khác $\cos 2\alpha > 0$ do đó $\tan 2\alpha < 0$.
Vậy $\tan 2\alpha = - \sqrt 3 $.
Ví dụ 17: Tính giá trị của các biểu thức sau:
a) $A = {\cos ^2}{73^0} + {\cos ^2}{47^0} + \cos {73^0}\cos {47^0}$.
b) $B = \sin {6^0}\sin {42^0}\sin {66^0}\sin {78^0}$.
c) $C = \cos \dfrac{\pi }{7}\cos \dfrac{{4\pi }}{7}\cos \dfrac{{5\pi }}{7}$.
d) $D = \dfrac{1}{{\sin {{10}^0}}} - 4\sin {70^0}$.
Lời giải
a) $A = {\left( {\cos {{73}^0} + \cos {{47}^0}} \right)^2} - \cos {73^0}\cos {47^0}$$ = {\left( {2\cos {{60}^0}\cos {{18}^0}} \right)^2} - \dfrac{1}{2}\left( {\cos {{120}^0} + \cos {{36}^0}} \right)$
$ = \dfrac{{1 + \cos {{36}^0}}}{2} + \dfrac{1}{4} - \dfrac{{\cos {{36}^0}}}{2} = \dfrac{3}{4}$.
b) $B = \sin {6^0}\cos {48^0}\cos {24^0}\cos {12^0}$$ = \dfrac{{\sin {{12}^0}}}{{2\cos {6^0}}}.\dfrac{{\sin {{24}^0}}}{{2\sin {{12}^0}}}.\dfrac{{\sin {{48}^0}}}{{2\sin {{24}^0}}}.\dfrac{{\sin {{96}^0}}}{{2\cos {{48}^0}}} = \dfrac{1}{{16}}$.
c) $C = - \cos \dfrac{\pi }{7}\cos \dfrac{{4\pi }}{7}\cos \dfrac{{2\pi }}{7}$$ = - \dfrac{{\sin \dfrac{{2\pi }}{7}}}{{2\sin \dfrac{\pi }{7}}}.\dfrac{{\sin \dfrac{{4\pi }}{7}}}{{2\sin \dfrac{{2\pi }}{7}}}.\dfrac{{\sin \dfrac{{8\pi }}{7}}}{{2\sin \dfrac{{4\pi }}{7}}} = \dfrac{1}{8}$.
d) $D = \dfrac{{1 - 4\sin {{70}^0}\sin {{10}^0}}}{{\sin {{10}^0}}}$$ = \dfrac{{1 + 2\left( {\cos {{80}^0} - \cos {{60}^0}} \right)}}{{\sin {{10}^0}}} = 2$.
