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Ta có:
\(A=\dfrac{\cos10^0-\sqrt{3}\sin10^0}{\sin10^0\cos10^0}\)
\(=\dfrac{4\left(\dfrac{1}{2}cos10^0-\dfrac{\sqrt{3}}{2}sin10^0\right)}{2sin10^0cos10^0}=\dfrac{4\left(s\text{in3}0^0cos10^0-cos30^0s\text{in}10^0\right)}{sin20^0}=\dfrac{4sin\left(30^0-10^0\right)}{s\text{in2}0^0}=4\)
Câu 3:
\(A=cos\frac{\pi}{7}.cos\frac{5\pi}{7}.cos\frac{4\pi}{7}=cos\frac{\pi}{7}.cos\left(\pi-\frac{2\pi}{7}\right).cos\frac{4\pi}{7}\)
\(A=-cos\frac{\pi}{7}.cos\frac{2\pi}{7}.cos\frac{4\pi}{7}\)
\(\Rightarrow sin\frac{\pi}{7}.A=-\frac{1}{2}.2sin\frac{\pi}{7}.cos\frac{\pi}{7}.cos\frac{2\pi}{7}.cos\frac{4\pi}{7}\)
\(\Rightarrow sin\frac{\pi}{7}.A=-\frac{1}{2}.sin\frac{2\pi}{7}.cos\frac{2\pi}{7}.cos\frac{4\pi}{7}\)
\(\Rightarrow sin\frac{\pi}{7}.A=-\frac{1}{4}sin\frac{4\pi}{7}.cos\frac{4\pi}{7}\)
\(\Rightarrow sin\frac{\pi}{7}.A=-\frac{1}{8}sin\frac{8\pi}{7}=-\frac{1}{8}sin\left(\pi+\frac{\pi}{7}\right)=\frac{1}{8}sin\frac{\pi}{7}\)
\(\Rightarrow A=\frac{1}{8}\)
Câu 4:
Đầu tiên ta chứng minh công thức:
\(tana+tanb=\frac{sina}{cosa}+\frac{sinb}{cosb}=\frac{sina.cosb+cosa.sinb}{cosa.cosb}=\frac{sin\left(a+b\right)}{cosa.cosb}\)
Áp dụng để biến đổi tử số:
\(tan30+tan60+tan40+tan50=\frac{sin90}{cos30.cos60}+\frac{sin90}{cos40.cos50}=\frac{1}{cos30.cos60}+\frac{1}{cos40.cos50}\)
\(=\frac{2}{cos90+cos30}+\frac{2}{cos90+cos10}=\frac{2}{cos30}+\frac{2}{cos10}=2\left(\frac{cos30+cos10}{cos30.cos10}\right)\)
\(=2\left(\frac{2cos20.cos10}{cos30.cos10}\right)=\frac{4.cos20}{cos30}=\frac{8\sqrt{3}}{3}.cos20\)
\(\Rightarrow A=\frac{\frac{8\sqrt{3}}{3}cos20}{cos20}=\frac{8\sqrt{3}}{3}\)
Câu 5:
\(cos54.cos4-cos36.cos86=cos54.cos4-cos\left(90-54\right).cos\left(90-4\right)\)
\(=cos54.cos4-sin54.sin4=cos\left(54+4\right)=cos58\)
Câu 1:
\(A=\frac{1}{2sin10}-2sin70=\frac{1-4sin10.sin70}{2sin10}=\frac{1+2\left(cos80-cos60\right)}{2sin10}\)
\(=\frac{1+2cos80-1}{2sin10}=\frac{2cos80}{2sin10}=\frac{sin10}{sin10}=1\)
Câu 2:
\(cos10.cos30.cos50.cos70=cos10.cos30.\frac{1}{2}\left(cos120+cos20\right)\)
\(=\frac{1}{2}cos30\left(cos10.cos120+cos10.cos20\right)\)
\(=\frac{1}{2}cos30\left(cos10.cos120+\frac{1}{2}\left(cos30+cos10\right)\right)\)
\(=\frac{1}{2}cos30\left(cos10.cos120+\frac{1}{2}cos30+\frac{1}{2}cos10\right)\)
\(=\frac{1}{2}.\frac{\sqrt{3}}{2}\left(-\frac{1}{2}cos10+\frac{1}{2}\frac{\sqrt{3}}{2}+\frac{1}{2}cos10\right)\)
\(=\frac{3}{16}\)
A=sin240+cos210+2sin40cos10-cos240-sin210-2sin10cos40+cos(90+50)
A=(sin240-cos240)+(cos210-sin210)+2(sin40cos10-cos40sin10)-sin50
A=(sin40-cos40)(sin40+cos40)-(sin10-cos10)(sin10+cos10)+1-sin50
A=\(\sqrt{2}\) sin(40-\(\frac{\pi}{4}\))\(\sqrt{2}\) cos(40-\(\frac{\pi}{4}\))-\(\sqrt{2}\)sin(10-\(\frac{\pi}{4}\))\(\sqrt{2}\) cos(10-\(\frac{\pi}{4}\))+1-sin50
A=-2sin5cos5+2sin35cos35+1-sin50
A= - sin10+sin70+1-sin50
A= 2cos40sin30-sin(90-40)+1
A=cos40-cos40+1 =1
\(P=\frac{x^2+1}{8}+\frac{1}{\sqrt{x^2+1}}+\frac{1}{\sqrt{x^2+1}}\ge3\sqrt[3]{\frac{x^2+1}{8\left(x^2+1\right)}}=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\frac{x^2+1}{8}=\frac{1}{\sqrt{x^2+1}}\Leftrightarrow x=\pm\sqrt{3}\)
