Bài 1: Giá trị lượng giác của một góc từ 0° đến 180°. Định lí sin và côsin trong tam giác

Hướng dẫn giải:

a) $A=a^{2} \cdot 1+b^{2} \cdot 0+c^{2} \cdot(-1)=a^{2}-c^{2}$

b) $B=3-(1)^{2}+2\left(\dfrac{1}{2}\right)^{2}-3\left(\dfrac{\sqrt{2}}{2}\right)^{2}=1$

c) $C=\sin ^{2} 45^{\circ}+3 \cos ^{2} 45^{\circ}-2\left(\sin ^{2} 50^{\circ}+\sin ^{2} 40^{\circ}\right)+4 \tan 55^{\circ} \cdot \cot 55^{\circ}$
$C=\left(\dfrac{\sqrt{2}}{2}\right)^{2}+3\left(\dfrac{\sqrt{2}}{2}\right)^{2}-2\left(\sin ^{2} 50^{\circ}+\cos ^{2} 40^{\circ}\right)+4=\dfrac{1}{2}+\dfrac{3}{2}-2+4=4$

Hướng dẫn giải:

a) $A=\left(\sin ^{2} 3^{\circ}+\sin ^{2} 87^{\circ}\right)+\left(\sin ^{2} 15^{\circ}+\sin ^{2} 75^{\circ}\right)$

$ \begin{aligned} &=\left(\sin ^{2} 3^{\circ}+\cos ^{2} 3^{\circ}\right)+\left(\sin ^{2} 15^{\circ}+\cos ^{2} 15^{\circ}\right) \\ &=1+1=2. \end{aligned} $

b) $B=\left(\cos 0^{\circ}+\cos 180^{\circ}\right)+\left(\cos 20^{\circ}+\cos 160^{\circ}\right)+\ldots+\left(\cos 80^{\circ}+\cos 100^{\circ}\right)$

$ \begin{aligned} &=\left(\cos 0^{\circ}-\cos 0^{\circ}\right)+\left(\cos 20^{\circ}-\cos 20^{\circ}\right)+\ldots+\left(\cos 80^{\circ}-\cos 80^{\circ}\right) \\ &=0. \end{aligned} $

c)  $ \begin{aligned} C &=\left(\tan 5^{\circ} \tan 85^{\circ}\right)\left(\tan 15^{\circ} \tan 75^{\circ}\right) \ldots\left(\tan 45^{\circ} \tan 45^{\circ}\right) \\ &=\left(\tan 5^{\circ} \cot 5^{\circ}\right)\left(\tan 15^{\circ} \cot 5^{\circ}\right) \ldots\left(\tan 45^{\circ} \cot 5^{\circ}\right) \\ &=1. \end{aligned} $

Hướng dẫn giải:

a) $\sin ^{4} x+\cos ^{4} x=\sin ^{4} x+\cos ^{4} x+2 \sin ^{2} x \cos ^{2} x-2 \sin ^{2} x \cos ^{2} x$
$\begin{aligned}&=\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x \\&=1-2 \sin ^{2} x \cos ^{2} x\end{aligned}$

b) $\dfrac{1+\cot x}{1-\cot x}=\dfrac{1+\dfrac{1}{\tan x}}{1-\dfrac{1}{\tan x}}=\dfrac{\dfrac{\tan x+1}{\tan x}}{\dfrac{\tan x-1}{\tan x}}=\dfrac{\tan x+1}{\tan x-1}$

c) $\dfrac{\cos x+\sin x}{\cos ^{3} x}=\dfrac{1}{\cos ^{2} x}+\dfrac{\sin x}{\cos ^{3} x}=\tan ^{2} x+1+\tan x\left(\tan ^{2} x+1\right)$
$=\tan ^{3} x+\tan ^{2} x+\tan x+1$

Hướng dẫn giải:

Vì $A+B+C=180^{\circ}$ nên $V T=\dfrac{\sin ^{3} \dfrac{B}{2}}{\cos \left(\dfrac{180^{\circ}-B}{2}\right)}+\dfrac{\cos ^{3} \dfrac{B}{2}}{\sin \left(\dfrac{180^{\circ}-B}{2}\right)}-\dfrac{\cos \left(180^{\circ}-B\right)}{\sin B} \cdot \tan B$.

$V T=\dfrac{\sin ^{3} \dfrac{B}{2}}{\cos \left(\dfrac{180^{\circ}-B}{2}\right)}+\dfrac{\cos ^{3} \dfrac{B}{2}}{\sin \left(\dfrac{180^{\circ}-B}{2}\right)}-\dfrac{\cos \left(180^{\circ}-B\right)}{\sin B} \cdot \tan B$ $=\dfrac{\sin ^{3} \dfrac{B}{2}}{\sin \dfrac{B}{2}}+\dfrac{\cos ^{3} \dfrac{B}{2}}{\cos \dfrac{B}{2}}-\dfrac{-\cos B}{\sin B} \cdot \tan B=\sin ^{2} \dfrac{B}{2}+\cos ^{2} \dfrac{B}{2}+1=2=V P$

Suy ra điều phải chứng minh.

Hướng dẫn giải:

a) $A=\cos x-\cos x+\sin ^{2} x \cdot \dfrac{1}{\cos ^{2} x}-\tan ^{2} x=0$

b) $B=\dfrac{1}{\sin x} \cdot \sqrt{\dfrac{1-\cos x+1+\cos x}{(1-\cos x)(1+\cos x)}}-\sqrt{2}$

$\begin{aligned}&=\dfrac{1}{\sin x} \cdot \sqrt{\dfrac{2}{1-\cos ^{2} x}}-\sqrt{2}=\dfrac{1}{\sin x} \cdot \sqrt{\dfrac{2}{\sin ^{2} x}}-\sqrt{2} \\&=\sqrt{2}\left(\dfrac{1}{\sin ^{2} x}-1\right)=\sqrt{2} \cot ^{2} x\end{aligned}$

Hướng dẫn giải:

$\begin{aligned}P=& \sqrt{\left(1-\cos ^{2} x\right)^{2}+6 \cos ^{2} x+3 \cos ^{4} x}+\sqrt{\left(1-\sin ^{2} x\right)^{2}+6 \sin ^{2} x+3 \sin ^{4} x} \\&=\sqrt{4 \cos ^{4} x+4 \cos ^{2} x+1}+\sqrt{4 \sin ^{4} x+4 \sin ^{2} x+1} \\&=\sqrt{\left(2 \cos ^{2} x+1\right)^{2}}+\sqrt{\left(2 \sin ^{2} x+1\right)^{2}} \\&=2 \cos ^{2} x+1+2 \sin ^{2} x+1 \\&=3\end{aligned}$

Vậy $P$ không phụ thuộc vào $x$.

Hướng dẫn giải:

a) Vì $90^{\circ}<\alpha<180^{\circ}$ nên $\cos \alpha<0$ mặt khác $\sin ^{2} \alpha+\cos ^{2} \alpha=1$ suy ra $\cos \alpha=-\sqrt{1-\sin ^{2} \alpha}=-\sqrt{1-\dfrac{1}{9}}=-\dfrac{2 \sqrt{2}}{3}$.

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

b) Vì $\sin ^{2} \alpha+\cos ^{2} \alpha=1$ nên $\sin \alpha=\sqrt{1-\cos ^{2} \alpha}=\sqrt{1-\dfrac{4}{9}}=\dfrac{\sqrt{5}}{3}$ và $\cot \alpha=\dfrac{\cos \alpha}{\sin \alpha}=\dfrac{-\dfrac{2}{3}}{\dfrac{\sqrt{5}}{3}}=-\dfrac{2}{\sqrt{5}}$.

c) Vì $\tan \gamma=-2 \sqrt{2}<0 \Rightarrow \cos \alpha<0$ mặt khác $\tan ^{2} \alpha+1=\dfrac{1}{\cos ^{2} \alpha}$ nên $\cos \alpha=-\sqrt{\dfrac{1}{\tan ^{2}+1}}=-\sqrt{\dfrac{1}{8+1}}=-\dfrac{1}{3}$.
Ta có $\tan \alpha=\dfrac{\sin \alpha}{\cos \alpha} \Rightarrow \sin \alpha=\tan \alpha \cdot \cos \alpha=-2 \sqrt{2} \cdot\left(-\dfrac{1}{3}\right)=\dfrac{2 \sqrt{2}}{3}$ $\Rightarrow \cot \alpha=\dfrac{\cos \alpha}{\sin \alpha}=\dfrac{-\dfrac{1}{3}}{\dfrac{2 \sqrt{2}}{3}}=-\dfrac{1}{2 \sqrt{2}}$.

Hướng dẫn giải:

a) Ta có $A=\dfrac{\tan \alpha+3 \dfrac{1}{\tan \alpha}}{\tan \alpha+\dfrac{1}{\tan \alpha}}=\dfrac{\tan ^{2} \alpha+3}{\tan ^{2} \alpha+1}=\dfrac{\dfrac{1}{\cos ^{2} \alpha}+2}{\dfrac{1}{\cos ^{2} \alpha}}=1+2 \cos ^{2} \alpha$ Suy ra $A=1+2 \cdot \dfrac{9}{16}=\dfrac{17}{8}$.

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

Suy ra $B=\dfrac{\sqrt{2}(2+1)-(2+1)}{2 \sqrt{2}+3+2 \sqrt{2}(2+1)}=\dfrac{3(\sqrt{2}-1)}{3+8 \sqrt{2}}$.

Hướng dẫn giải:

a) Ta có $(\sin x+\cos x)^{2}=\sin ^{2} x+2 \sin x \cos x+\cos ^{2} x=1+2 \sin x \cos x$ (*)
Mặt khác $\sin x+\cos x=m$ nên $m^{2}=1+2 \sin \alpha \cos \alpha$ hay $\sin \alpha \cos \alpha=\dfrac{m^{2}-1}{2}$
Đặt $A=\left|\sin ^{4} x-\cos ^{4} x\right|$. Ta có
$A=\left|\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{2} x-\cos ^{2} x\right)\right|=|(\sin x+\cos x)(\sin x-\cos x)|$
$\Rightarrow A^{2}=(\sin x+\cos x)^{2}(\sin x-\cos x)^{2}=(1+2 \sin x \cos x)(1-2 \sin x \cos x)$

$\Rightarrow A^{2}=\left(1+\dfrac{m^{2}-1}{2}\right)\left(1-\dfrac{m^{2}-1}{2}\right)=\dfrac{3+2 m^{2}-m^{4}}{4}$
Vậy $A=\dfrac{\sqrt{3+2 m^{2}-m^{4}}}{2}$

b) Ta có $2 \sin x \cos x \leq \sin ^{2} x+\cos ^{2} x=1$ kết hợp với $(*)$ suy ra

$(\sin x+\cos x)^{2} \leq 2 \Rightarrow|\sin x+\cos x| \leq \sqrt{2}$

Vậy $|m| \leq \sqrt{2}$.