Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\1-x^2\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ne0\\-1\le x\le1\end{matrix}\right.\)
b/ ĐKXĐ: \(\left\{{}\begin{matrix}x^2-4>0\\x+1\ge0\end{matrix}\right.\) \(\Rightarrow x>2\)
c/ ĐKXĐ: \(\left\{{}\begin{matrix}1+x\ge0\\x-3\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge-1\\x\ne3\end{matrix}\right.\)
d/ ĐKXĐ: \(\left\{{}\begin{matrix}x^2-4x+3>0\\x\ge-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>3\\x< 1\end{matrix}\right.\\x\ge-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>3\\-1\le x< 1\end{matrix}\right.\)
\(A=cos10+cos170+cos40+cos140+cos70+cos110\)
\(A=cos10+cos\left(180-10\right)+cos40+cos\left(180-40\right)+cos70+cos\left(180-70\right)\)
\(A=cos10-cos10+cos40-cos40+cos70-cos70\)
\(A=0\)
\(B=sin5+sin355+sin10+sin350+...+sin175+sin185+sin360\)
\(B=sin5+sin\left(360-5\right)+sin10+sin\left(360-10\right)+...+sin175+sin\left(360-175\right)+sin360\)
\(B=sin5-sin5+sin10-sin10+...+sin175-sin175+sin360\)
\(B=sin360=0\)
\(C=cos^22+cos^288+cos^24+cos^284+...+cos^244+cos^246\)
\(C=cos^22+cos^2\left(90-2\right)+cos^24+cos^2\left(90-4\right)+...+cos^244+cos^2\left(90-44\right)\)
\(C=cos^22+sin^22+cos^24+sin^24+...+cos^244+sin^244\)
\(C=1+1+...+1\) (có \(\frac{44-2}{2}+1=22\) số 1)
\(\Rightarrow C=22\)
Ta có:
\(A=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}.\frac{4.4}{4.5}=\frac{1.1.2.2.3.3.4.4}{1.2.2.3.3.4.4.5}=\frac{1}{5}\)
\(\frac{a^3}{\sqrt{b^2+3}}+\frac{a^3}{\sqrt{b^2+3}}+\frac{b^2+3}{8}\ge\frac{3}{2}a^2\)\(\Leftrightarrow\)\(\frac{a^3}{\sqrt{b^2+3}}\ge\frac{3}{4}a^2-\frac{1}{16}b^2-\frac{3}{16}\)
\(P=\Sigma\frac{a^3}{\sqrt{b^2+3}}\ge\frac{3}{4}\left(a^2+b^2+c^2\right)-\frac{1}{16}\left(a^2+b^2+c^2\right)-\frac{9}{16}=\frac{3}{2}\)
Dấu "=" xảy ra khi a=b=c=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 \(\Leftrightarrow\frac{x^2+1}{8}=\frac{1}{\sqrt{x^2+1}}\Leftrightarrow x=\pm\sqrt{3}\)