Lời giải.

Ta có:

$$\begin{gathered} \sin \left(a–17°\right).\cos \left(a+13°\right) \\ –\sin \left(a+13°\right).\cos \left(a–17°\right) \\ =\sin \left[\left(a-17°\right)-\left(a+13°\right)\right] \\ =\sin \left(-30°\right)=-\frac{1}{2}. \end{gathered}$$

Lời giải.

$$\begin{gathered} \cos \frac{37\pi }{12}=\cos \left(2\pi +\pi +\frac{\pi }{12}\right) \\ =\cos \left(\pi +\frac{\pi }{12}\right)=-\cos \left(\frac{\pi }{12}\right) \\ =-\cos \left(\frac{\pi }{3}-\frac{\pi }{4}\right) \\ =-\left(\cos \frac{\pi }{3}.\cos \frac{\pi }{4}+\sin \frac{\pi }{3}.sin\frac{\pi }{4}\right) \\ =-\frac{\sqrt{6}+\sqrt{2}}{4} \end{gathered}$$

Lời giải.

$$\begin{gathered} \sin \frac{47\pi }{6}=\sin \left(8\pi -\frac{\pi }{6}\right) \\ =\sin \left(-\frac{\pi }{6}+4.2\pi \right) \\ =\sin \left(-\frac{\pi }{6}\right)=-\frac{1}{2} \end{gathered}$$

Lời giải.

$$\begin{gathered} A=\frac{1}{2\sin {10}^0}-2\sin {70}^0 \\ =\frac{1-4\sin {10}^0.\sin {70}^0}{2\sin {10}^0} \\ =\frac{2\sin {80}^0}{2\sin {10}^0}=\frac{2\sin {10}^0}{2\sin {10}^0}=1 \end{gathered}$$

Lời giải.

Chọn C.

$$\begin{gathered} \cos 10°.\cos 30°.\cos 50°.\cos 70° \\ =\cos 10°.\cos 30°.\frac{1}{2}\left(\cos {120}^{\text{o}}+\cos {20}^{\text{o}}\right) \\ =\frac{\sqrt{3}}{4}\left(-\frac{\cos 10°}{2}+\frac{\cos 30°+\cos 10°}{2}\right) \\ =\frac{\sqrt{3}}{4}.\frac{1}{4}=\frac{\sqrt{3}}{16} \end{gathered}$$

Lời giải.

$$\begin{gathered} \cos \frac{\pi }{7}.\cos \frac{4\pi }{7}.\cos \frac{5\pi }{7} \\ =\frac{\sin \frac{2\pi }{7}.\cos \frac{4\pi }{7}.\cos \frac{5\pi }{7}}{2\sin \frac{\pi }{7}} \\ =-\frac{\sin \frac{2\pi }{7}.\cos \frac{2\pi }{7}.\cos \frac{4\pi }{7}}{2\sin \frac{\pi }{7}} \\ =-\frac{\sin \frac{4\pi }{7}.\cos \frac{4\pi }{7}}{4\sin \frac{\pi }{7}} \\ =-\frac{\sin \frac{8\pi }{7}}{8\sin \frac{\pi }{7}}=\frac{1}{8} \end{gathered}$$

Lời giải.

$$\begin{gathered} A=\frac{\tan 30°+\tan 40°+\tan 50°+\tan 60°}{\cos 20°} \\ =\frac{\frac{\sin 70°}{\cos 30°.\cos 40°}+\frac{\sin 110°}{\cos 50°.\cos 60°}}{\cos 20°} \\ =\frac{1}{\cos 30°.\cos 40°}+\frac{1}{\cos 50°.\cos 60°} \\ =\frac{2}{\sqrt{3}\cos 40°}+\frac{2}{\cos 50°} \\ =2\left(\frac{\cos 50°+\sqrt{3}\cos 40°}{\sqrt{3}\cos 40°.\cos 50°}\right) \\ =2\left(\frac{\sin 40°+\sqrt{3}\cos 40°}{\sqrt{3}\cos 40°.\cos 50°}\right) \\ =4\frac{\sin 100°}{\frac{\sqrt{3}}{2}\left(\cos 10°+\cos 90°\right)} \\ =\frac{8\cos 10°}{\sqrt{3}\cos 10°}=\frac{8}{\sqrt{3}} \end{gathered}$$

Lời giải.

$$\begin{gathered} A=\tan 9°+\cot 9°+\tan 15° \\ +\cot 15°–\tan 27°–\cot 27° \\ =\tan 9°+\cot 9°–\tan 27° \\ –\cot 27°+\tan 15°+\cot 15° \\ =\tan 9°+tan81°–\tan 27° \\ –tan63°+\tan 15°+\cot 15° \end{gathered}$$

Ta có

$$\begin{gathered} \tan 9°–\tan 27°+tan81°–tan63° \\ =\frac{-\sin 18°}{\cos 9°.\cos 27°} \\ +\frac{\sin 18°}{\cos 81°.\cos 63°} \\ =\sin 18°\left(\frac{\cos 9°.\cos 27°-\cos 81°.\cos 63°}{\cos 81°.\cos 63°.\cos 9°.\cos 27°}\right) \\ =\frac{\sin 18°\left(\cos 9°.\cos 27°-\sin 9°.\sin 27°\right)}{\cos 81°.\cos 63°.\cos 9°.\cos 27°} \\ =\frac{4\sin 18°.\cos 36°}{\left(\mathrm{co}s72°+\cos 90°\right)\left(\cos 36°+\cos 90°\right)} \\ =\frac{4\sin 18°}{\cos 72°}=4 \\ \tan 15°+\cot 15° \\ =\frac{{\sin }^{2}15°+{\cos }^{2}15°}{\sin 15°.\cos 15°} \\ =\frac{2}{\sin 30°}=4 \end{gathered}$$

Vậy $A=8$.

Lời giải.

$$\begin{gathered} \tan \left(A+B+C\right) \\ =\frac{\tan \left(A+B\right)+\tan C}{1-\tan \left(A+B\right).\tan C} \\ =\frac{\frac{\tan A+\tan B}{1-\tan A.\tan B}+\tan C}{\frac{\tan A+\tan B}{1-\tan A.\tan B}.\tan C}=1 \end{gathered}$$

Suy ra $A+B+C=\frac{\pi }{4}$.

Lời giải.

$$\begin{gathered} \tan \left(a+b\right) \\ =\frac{\tan a+\tan b}{1-\tan a.\tan b}=1 \end{gathered}$$

suy ra $a+b=\frac{\pi }{4}$

Lời giải.

Ta có:

$$\begin{gathered} \tan \left(x+y\right)=\frac{\tan x+\tan y}{1-\tan x.\tan y} \\ =\frac{\frac{4}{3}+7}{1-\frac{4}{3}.7}=-1 \end{gathered}$$

suy ra $x+y=\frac{3\pi }{4}$

Lời giải.

$$\begin{gathered} \Rightarrow \frac{1}{{\sin }^{2}a}=226 \\ \Rightarrow \left\{\begin{array}{l}{\sin }^{2}a=\frac{1}{226}\\ {\cos }^{2}a=\frac{225}{226}\end{array}\right. \\ \Rightarrow \sin 2a=\pm \frac{15}{113} \end{gathered}$$

Lời giải.

Ta có:

$$\begin{gathered} \left\{\begin{array}{l}0<a<\frac{\pi }{2}\\ \sin a=\frac{1}{3}\end{array}\right.\Rightarrow \cos a=\frac{2\sqrt{2}}{3}; \\ \left\{\begin{array}{l}0<b<\frac{\pi }{2}\\ \sin b=\frac{1}{2}\end{array}\right.\Rightarrow \cos b=\frac{\sqrt{3}}{2} \end{gathered}$$

$$\begin{gathered} \sin 2\left(a+b\right) \\ =2\sin \left(a+b\right).\cos \left(a+b\right) \\ =2\left(\sin a.\cos b+\sin b.\cos a\right) \\ \left(\cos a.\cos b+\sin a.\sin b\right) \\ =\frac{4\sqrt{2}+7\sqrt{3}}{18} \end{gathered}$$

Lời giải.

Ta có:

$$\begin{gathered} A={\cos }^{2}x+{\cos }^2\left(\frac{\pi }{3}+x\right) \\ +{\cos }^2{\left(\frac{\pi }{3}-x\right)}^2 \\ ={\cos }^{2}x+{\left(\frac{\sqrt{3}}{2}\cos x-\frac{1}{2}\sin x\right)}^2 \\ +\left(\frac{\sqrt{3}}{2}\cos x+\frac{1}{2}\sin x\right) \\ =\frac{3}{2} \end{gathered}$$

Lời giải.

Công thức đúng là $\tan 2x=\frac{2\tan x}{1-{\tan }^{2}x}$

Lời giải.

Ta có:$\cos 2a={\cos }^{2}a–{\sin }^{2}a$

$=2{\cos }^{2}a-1=1-2{\sin }^{2}a.$

Lời giải.

Ta có :

$$\begin{gathered} \sin a+1 \\ =2\sin \frac{a}{2}\cos \frac{a}{2}+{\sin }^2\frac{a}{2}+{\cos }^2\frac{a}{2} \\ ={\left(\sin \frac{a}{2}+\cos \frac{a}{2}\right)}^2 \\ =2{\sin }^2\left(\frac{a}{2}+\frac{\pi }{4}\right) \\ =2\sin \left(\frac{a}{2}+\frac{\pi }{4}\right)\cos \left(\frac{\pi }{4}-\frac{a}{2}\right) \\ =2\sin \left(\frac{a}{2}+\frac{\pi }{4}\right)\cos \left(\frac{a}{2}-\frac{\pi }{4}\right) \end{gathered}$$

Lời giải.

Ta có:

$\alpha +\beta +\gamma =\frac{\pi }{2}$suy ra:

$$\begin{gathered} \cot \beta =\tan \left(\alpha +\gamma \right) \\ =\frac{\tan \alpha +\tan \gamma }{1-\tan \alpha \tan \gamma } \\ =\frac{\cot \alpha +cot\gamma }{\cot \alpha cot\gamma -1} \\ =\frac{2\cot \beta }{\cot \alpha \cot \gamma -1} \\ \Rightarrow \cot \alpha \cot \gamma =3. \end{gathered}$$

Lời giải.

Ta có :

$$\begin{gathered} {\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C \\ =\frac{{\displaystyle 1+\cos 2A}}{{\displaystyle 2}}+\frac{{\displaystyle 1+\cos 2B}}{{\displaystyle 2}}+{\cos }^{2}C \\ =1+\cos \left(A+B\right)\cos \left(A-B\right)+{\cos }^{2}C \\ =1-\cos C\cos \left(A-B\right)-\cos C\cos \left(A+B\right) \\ =1-\cos C\left[\cos \left(A-B\right)+\cos \left(A+B\right)\right] \\ =1+2\cos A\cos B\cos C. \end{gathered}$$

Lời giải.

Ta có:

$\sin \left(a–b\right)=\sin a.\cos b-\cos a.\sin b.$

Lời giải.

Ta có $\tan \left(a+b\right)$

$=\frac{\tan a+\tan b}{1-\tan a\tan b}.$ 

Lời giải.

Ta có $\sin a\cos b$

$=\frac{1}{2}\left[\sin \left(a–b\right)+\sin \left(a+b\right)\right].$

Lời giải

Ta có :

$5\sin \alpha =3\sin \left(\alpha +2\beta \right)$

$\iff 5\sin \left[\left(\alpha +\beta \right)-\beta \right]$

$=3\sin \left[\left(\alpha +\beta \right)+\beta \right]$

$\iff 5\sin \left(\alpha +\beta \right)\cos \beta -5\cos \left(\alpha +\beta \right)\sin \beta$

$=3\sin \left(\alpha +\beta \right)\cos \beta +3\cos \left(\alpha +\beta \right)\sin \beta$

$\iff 2\sin \left(\alpha +\beta \right)\cos \beta$

$=8\cos \left(\alpha +\beta \right)\sin \beta$

$\iff \frac{{\displaystyle \sin \left(\alpha +\beta \right)}}{{\displaystyle \cos \left(\alpha +\beta \right)}}=4\frac{{\displaystyle \sin \beta }}{{\displaystyle \cos \beta }}$

$\iff \tan \left(\alpha +\beta \right)=4\tan \beta$

Lời giải.

Ta có:

$$\begin{gathered} \left\{\begin{array}{l}\cos a=\frac{3}{4}\\ \sin a>0\end{array}\right.\Rightarrow \sin a \\ =\sqrt{1-{\cos }^{2}a}=\frac{\sqrt{7}}{4} \\ \left\{\begin{array}{l}\sin b=\frac{3}{5}\\ \cos b<0\end{array}\right.\Rightarrow \cos b \\ =-\sqrt{1-{\sin }^{2}b}=-\frac{4}{5} \\ \cos \left(a+b\right) \\ =\cos a\cos b-\sin a\sin b \\ =\frac{3}{4}.\left(-\frac{4}{5}\right)-\frac{\sqrt{7}}{4}.\frac{3}{5} \\ =-\frac{3}{5}\left(1+\frac{\sqrt{7}}{4}\right). \end{gathered}$$

Lời giải.

Ta có :

$$\begin{gathered} \left\{\begin{array}{l}\cos \left(a-\frac{b}{2}\right)=\frac{1}{2}\\ \sin \left(a-\frac{b}{2}\right)>0\end{array}\right. \\ \Rightarrow \sin \left(a-\frac{b}{2}\right) \\ =\sqrt{1-{\cos }^2\left(a-\frac{b}{2}\right)}=\frac{\sqrt{3}}{2} \\ \left\{\begin{array}{l}\sin \left(\frac{a}{2}-b\right)=\frac{3}{5}\\ \cos \left(\frac{a}{2}-b\right)\end{array}\right. \\ \Rightarrow \cos \left(\frac{a}{2}-b\right) \\ =\sqrt{1-{\sin }^2\left(\frac{a}{2}-b\right)}=\frac{4}{5} \\ cos\frac{a+b}{2} \\ =\cos \left(a-\frac{b}{2}\right)\cos \left(\frac{a}{2}-b\right) \\ +\sin \left(a-\frac{b}{2}\right)\sin \left(\frac{a}{2}-b\right) \\ =\frac{{\displaystyle 1}}{{\displaystyle 2}}.\frac{{\displaystyle 4}}{{\displaystyle 5}}+\frac{{\displaystyle 3}}{{\displaystyle 5}}.\frac{{\displaystyle \sqrt{3}}}{{\displaystyle 2}}=\frac{{\displaystyle 3\sqrt{3}+4}}{{\displaystyle 10}} \\ \cos \left(a+b\right)=2{\cos }^2\frac{{\displaystyle a+b}}{{\displaystyle 2}}-1 \\ =\frac{{\displaystyle 24\sqrt{3}-7}}{{\displaystyle 50}}. \end{gathered}$$

Lời giải.

$$\begin{gathered} \cos \left(120°–x\right)+\cos \left(120°+x\right)–\cos x \\ =-\frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x \\ -\frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x-\cos x \\ =-2\cos x \end{gathered}$$

Lời giải.

Ta có :

$$\begin{gathered} A={\sin }^2\left(a+b\right)–{\sin }^{2}a–{\sin }^{2}b \\ ={\sin }^2\left(a+b\right)-\frac{1-\cos 2a}{2} \\ -\frac{1-\cos 2b}{2} \\ ={\sin }^2\left(a+b\right)-1 \\ +\frac{{\displaystyle 1}}{{\displaystyle 2}}\left(\cos 2a+\cos 2b\right) \\ =-{\cos }^2\left(a+b\right) \\ +\cos \left(a+b\right)\cos \left(a-b\right) \\ =\cos \left(a+b\right)\left[\cos \left(a-b\right)-\cos \left(a+b\right)\right] \\ =\cos \left(a+b\right)\left[\cos \left(a-b\right)-\cos \left(a+b\right)\right] \\ =2\sin a\sin b\cos \left(a+b\right). \end{gathered}$$

Lời giải.

Ta có :

$$\begin{gathered} \left\{\begin{array}{l}\sin a=\frac{3}{5}\\ \cos a<0\end{array}\right.\Rightarrow \cos a \\ =-\sqrt{1-{\sin }^{2}a}=-\frac{4}{5} \\ \left\{\begin{array}{l}\cos b=\frac{3}{4}\\ \sin b>0\end{array}\right.\Rightarrow \sin b \\ =\sqrt{1-{\cos }^{2}b}=\frac{\sqrt{7}}{4} \\ \sin \left(a-b\right) \\ =\sin a\cos b-\cos a\sin b \\ =\frac{3}{5}.\frac{3}{4}-\left(-\frac{4}{5}\right).\frac{\sqrt{7}}{4} \\ =\frac{1}{5}\left(\sqrt{7}+\frac{9}{4}\right) \end{gathered}$$