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\(\Rightarrow \tan A+\tan C=2\tan B\)
\(\Leftrightarrow \frac{\sin\left ( A+C \right )}{\cos A\cos C}=2\cdot\frac{\sin\left ( A+C \right )}{\cos B}\\\)
\(\Rightarrow \cos B=2\cos A\cos C\)
\(\Leftrightarrow 2\cos B=\cos(A-C)\)
\(\left (\cos A+\cos C \right )^2=\cos^2 A+\cos^2 C+2\cos A\cos C\\=\frac{\cos2A+\cos2C}{2}+1+\cos B\\=-\cos(B)\cos(A-C)+1+\cos B \\=-2\cos^2B+\cos B+1 \le \frac{9}{8}\\\Rightarrow \cos A+\cos C\le \frac{3\sqrt2}{4}\)
Chứng minh hoàn tất.
Bài tập này áp dụng công thức phụ - chéo:
cot(a)=tan(\(\dfrac{\Pi}{2}\)-a) (cái này chắc bạn không quên chứ hihi)
Điều kiện: cos(2x+\(\dfrac{\Pi}{4}\))\(\ne\)0<=>x\(\ne\)\(\dfrac{\Pi}{8}\)+\(\dfrac{k\Pi}{2}\)
cos(\(\Pi\)-\(\dfrac{x}{2}\))\(\ne\)0<=>x\(\ne\)\(\Pi\)-2\(\Pi\)
PT<=>tan(2x+\(\dfrac{\Pi}{4}\))=\(\dfrac{1}{tan\left(\Pi-\dfrac{x}{2}\right)}\)
<=>tan(2x+\(\dfrac{\Pi}{4}\))=cot(\(\Pi\)-\(\dfrac{x}{2}\))
<=>tan(2x+\(\dfrac{\Pi}{4}\))=tan(\(\dfrac{\Pi}{2}\)-\(\Pi\)+\(\dfrac{x}{2}\))
<=>2x+\(\dfrac{\Pi}{4}\)=\(\dfrac{\Pi}{2}\)-\(\Pi\)+\(\dfrac{x}{2}\)
<=>x=-\(\dfrac{\Pi}{2}\)+k\(\dfrac{2\Pi}{3}\)(k\(\in\)Z)
Chúc bạn học tốt. Thân!
\(\dfrac{\Pi}{4}\)\(\Pi\)\(\Pi\)
a) \(x=-45^0+k90^0,k\in\mathbb{Z}\)
b) \(x=-\dfrac{\pi}{6}+k\pi,k\in\mathbb{Z}\)
c) \(x=\dfrac{3\pi}{4}+k2\pi,k\in\mathbb{Z}\)
d) \(x=300^0+k540^0,k\in\mathbb{Z}\)
ĐKXĐ: \(cosx\ne0\Rightarrow x\ne\dfrac{\pi}{2}+k\pi\)
\(\dfrac{tan^2x+tanx}{tan^2x+1}=\dfrac{\sqrt{2}}{2}sin\left(\dfrac{\pi}{4}+x\right)\)
\(\Leftrightarrow cos^2x\left(tan^2x+tanx\right)=\dfrac{\sqrt{2}}{2}\left(sin\dfrac{\pi}{4}.cosx+cos\dfrac{\pi}{4}.sinx\right)\)
\(\Leftrightarrow sin^2x+sinxcosx=\dfrac{1}{2}\left(sinx+cosx\right)\)
\(\Leftrightarrow sinx\left(sinx+cosx\right)-\dfrac{1}{2}\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx-\dfrac{1}{2}\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\sinx+cosx=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\\sqrt{2}.sin\left(x+\dfrac{\pi}{4}\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\\x=\dfrac{-\pi}{4}+k\pi\end{matrix}\right.\)
có thể giải thích rõ ở dấu tương đương 1 và 2 cho em hiểu làm sao để rút gọn nó thành như vậy được không ạ
mình làm cách này là cách khj nào mà ko cách nào khác ms làm vậy thôi, áp dụng định lí sin và cosin trong tam giác
Theo giả thiết ta có 3 góc: \(\alpha;\beta=\alpha+\dfrac{\pi}{3};\gamma=\alpha+\dfrac{2\pi}{3}\).
Ta có:
\(tan\alpha.tan\left(\alpha+\dfrac{\pi}{3}\right)+tan\left(\alpha+\dfrac{\pi}{3}\right).tan\left(\alpha+\dfrac{2\pi}{3}\right)+\)\(tan\left(\alpha+\dfrac{2\pi}{3}\right).tan\alpha\)
\(=tan\alpha\left[tan\left(\alpha+\dfrac{\pi}{3}\right)+tan\left(\alpha+\dfrac{2\pi}{3}\right)\right]\)\(+tan\left(a+\dfrac{\pi}{3}\right)tan\left(\alpha+\dfrac{2\pi}{3}\right)\)
\(=tan\alpha\dfrac{sin\left(2\alpha+\pi\right)}{cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)\(+\dfrac{sin\left(\alpha+\dfrac{\pi}{3}\right)sin\left(\alpha+\dfrac{2\pi}{3}\right)}{cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)
\(=tan\alpha\dfrac{-sin2\alpha}{cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)\(+\dfrac{cos\dfrac{\pi}{3}-cos\left(2\alpha+\pi\right)}{2cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)
\(=\dfrac{-2sin^2\alpha}{cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)\(+\dfrac{\dfrac{1}{2}+cos2\alpha}{2cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)
\(=\dfrac{\dfrac{1}{2}-4sin^2\alpha+cos2\alpha}{2cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)
\(=\dfrac{\dfrac{1}{2}-4\left(1-cos^2\alpha\right)+2cos^2\alpha-1}{cos\dfrac{\pi}{3}+cos\left(2\alpha+\pi\right)}\)
\(=\dfrac{6cos^2\alpha-\dfrac{9}{2}}{\dfrac{1}{2}-cos2\alpha}\)
\(=\dfrac{3\left(2cos^2\alpha-\dfrac{3}{2}\right)}{\dfrac{1}{2}-\left(2cos^2\alpha-1\right)}=\dfrac{3\left(2cos^2\alpha-\dfrac{3}{2}\right)}{\dfrac{3}{2}-2cos^2\alpha}=-3\).
\(4cos\alpha.cos\beta cos\gamma=4cos\alpha cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)\)
\(=4cos\alpha.\dfrac{1}{2}\left(cos\dfrac{\pi}{3}+cos\left(2\alpha+\pi\right)\right)\)
\(=4cos\alpha.\dfrac{1}{2}\left(\dfrac{1}{2}-cos2\alpha\right)\)
\(=cos\alpha-2cos\alpha.cos2\alpha\)
\(=cos\alpha-\left(cos\alpha+cos3\alpha\right)\)
\(=-cos3\alpha\)
\(=cos\left(\pi+3\alpha\right)\)
\(=cos3\left(\dfrac{\pi}{3}+\alpha\right)\)
\(=cos3\beta\) (đpcm).
Ta có:
\(\dfrac{tanA}{tan^3B}=\dfrac{tanA}{tanB}.\dfrac{1}{tan^2B}=\dfrac{\dfrac{sinA}{cosA}}{\dfrac{sinB}{cosB}}.\dfrac{cos^2B}{sin^2B}\)
\(=\dfrac{sinA}{sinB}.\dfrac{cosB}{cosA}.\dfrac{cos^2B}{sin^2B}\)
\(=\dfrac{a}{b}.\dfrac{\dfrac{a^2+c^2-b^2}{2ac}}{\dfrac{b^2+c^2-a^2}{2bc}}.\dfrac{\left(\dfrac{a^2+c^2-b^2}{2ac}\right)^2}{1-\left(\dfrac{a^2+c^2-b^2}{2ac}\right)^2}\)
\(=\dfrac{a^2+c^2-b^2}{b^2+c^2-a^2}.\dfrac{\left(a^2+c^2-b^2\right)^2}{\left(2ac\right)^2-\left(a^2+c^2-b^2\right)^2}\)
\(=\dfrac{\left(a^2+c^2-b^2\right)^3}{b^2+c^2-a^2}.\dfrac{1}{\left[\left(a+c\right)^2-b^2\right]\left[b^2-\left(a-c\right)^2\right]}\)
\(=\dfrac{\left(a^2+c^2-b^2\right)^3}{b^2+c^2-a^2}.\dfrac{1}{\left(a+b+c\right)\left(a+c-b\right)\left(b+c-a\right)\left(a+b-c\right)}\)
Biến đổi tương tự, ta có BĐT tương đương với BĐT đã cho:
\(\dfrac{\left(a^2+c^2-b^2\right)^3}{b^2+c^2-a^2}+\dfrac{\left(a^2+b^2-c^2\right)^3}{a^2+c^2-b^2}+\dfrac{\left(b^2+c^2-a^2\right)^3}{a^2+b^2-c^2}\ge\left(a+b+c\right)\left(b+c-a\right)\left(a+c-b\right)\left(a+b-c\right)\)
Ta có BĐT phụ sau:
\(\dfrac{x^3}{y}+\dfrac{y^3}{z}+\dfrac{z^3}{x}\ge xy+yz+xz\left(\text{*}\right)\) với \(x,y,z>0\)
Chứng minh:
Áp dụng BĐT cộng mẫu:
\(\dfrac{x^3}{y}+\dfrac{y^3}{z}+\dfrac{z^3}{x}=\dfrac{x^4}{xy}+\dfrac{y^4}{yz}+\dfrac{z^4}{xz}\)
\(\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{xy+yz+xz}\ge\dfrac{\left(xy+yz+xz\right)^2}{xy+yz+xz}=xy+yz+xz\)(đpcm)
Đẳng thức xảy ra khi và chỉ khi \(x=y=z\)
Áp dụng BĐT \(\left(\text{*}\right)\), với đk \(\Delta ABC\) có ba góc nhọn, ta có:
\(\dfrac{\left(a^2+c^2-b^2\right)^3}{b^2+c^2-a^2}+\dfrac{\left(a^2+b^2-c^2\right)^3}{a^2+c^2-b^2}+\dfrac{\left(b^2+c^2-a^2\right)^3}{a^2+b^2-c^2}\ge\left(a^2+c^2-b^2\right)\left(a^2+b^2-c^2\right)+\left(a^2+b^2-c^2\right)\left(b^2+c^2-a^2\right)+\left(b^2+c^2-a^2\right)\left(a^2+c^2-b^2\right)\)
Ta chứng minh được:
\(\left(a^2+c^2-b^2\right)\left(a^2+b^2-c^2\right)+\left(a^2+b^2-c^2\right)\left(b^2+c^2-a^2\right)+\left(b^2+c^2-a^2\right)\left(a^2+c^2-b^2\right)=\left(a+b+c\right)\left(b+c-a\right)\left(a+c-b\right)\left(a+b-c\right)\)
\(=-a^4-b^4-c^4+2a^2b^2+2b^2c^2+2a^2c^2\)
Vậy ta có BĐT cần chứng minh, đẳng thức xảy ra khi và chỉ khi \(\widehat{A}=\widehat{B}=\widehat{C}=60^0\)