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\(tan2A=tan\left[\left(A+B\right)+\left(A-B\right)\right]=\frac{tan\left(A+B\right)+tan\left(A-B\right)}{1-tan\left(A+B\right).tan\left(A-B\right)}=\frac{5+3}{1-5.3}=-\frac{4}{7}\)
Câu a)
Ta sử dụng 2 công thức:
\(\bullet \tan (180-\alpha)=-\tan \alpha\)
\(\bullet \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha.\tan \beta}\)
Áp dụng vào bài toán:
\(\text{VT}=\tan A+\tan B+\tan C=\tan A+\tan B+\tan (180-A-B)\)
\(=\tan A+\tan B-\tan (A+B)=\tan A+\tan B-\frac{\tan A+\tan B}{1-\tan A.\tan B}\)
\(=(\tan A+\tan B)\left(1+\frac{1}{1-\tan A.\tan B}\right)=(\tan A+\tan B).\frac{-\tan A.\tan B}{1-\tan A.\tan B}\)
\(=-\tan A.\tan B.\frac{\tan A+\tan B}{1-\tan A.\tan B}=-\tan A.\tan B.\tan (A+B)\)
\(=\tan A.\tan B.\tan (180-A-B)\)
\(=\tan A.\tan B.\tan C=\text{VP}\)
Do đó ta có đpcm
Tam giác $ABC$ có ba góc nhọn nên \(\tan A, \tan B, \tan C>0\)
Áp dụng BĐT Cauchy ta có:
\(P=\tan A+\tan B+\tan C\geq 3\sqrt[3]{\tan A.\tan B.\tan C}\)
\(\Leftrightarrow P=\tan A+\tan B+\tan C\geq 3\sqrt[3]{\tan A+\tan B+\tan C}\)
\(\Rightarrow P\geq 3\sqrt[3]{P}\)
\(\Rightarrow P^3\geq 27P\Leftrightarrow P(P^2-27)\geq 0\)
\(\Rightarrow P^2-27\geq 0\Rightarrow P\geq 3\sqrt{3}\)
Vậy \(P_{\min}=3\sqrt{3}\). Dấu bằng xảy ra khi \(\angle A=\angle B=\angle C=60^0\)
Câu b)
Ta sử dụng 2 công thức chính:
\(\bullet \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha.\tan \beta}\)
\(\bullet \tan (90-\alpha)=\frac{1}{\tan \alpha}\)
Áp dụng vào bài toán:
\(\text{VT}=\tan \frac{A}{2}.\tan \frac{B}{2}+\tan \frac{B}{2}.\tan \frac{C}{2}+\tan \frac{C}{2}.\tan \frac{A}{2}\)
\(=\tan \frac{A}{2}.\tan \frac{B}{2}+\tan \frac{C}{2}(\tan \frac{A}{2}+\tan \frac{B}{2})\)
\(=\tan \frac{A}{2}.\tan \frac{B}{2}+\tan (90-\frac{A+B}{2})(\tan \frac{A}{2}+\tan \frac{B}{2})\)
\(=\tan \frac{A}{2}.\tan \frac{B}{2}+\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{\tan (\frac{A+B}{2})}\)
\(=\tan \frac{A}{2}.\tan \frac{B}{2}+\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2}.\tan \frac{B}{2}}}\)
\(=\tan \frac{A}{2}.\tan \frac{B}{2}+1-\tan \frac{A}{2}.\tan \frac{B}{2}=1=\text{VP}\)
Ta có đpcm.
Cũng giống phần a, ta biết do ABC là tam giác nhọn nên
\(\tan A, \tan B, \tan C>0\)
Đặt \(\tan A=x, \tan B=y, \tan C=z\). Ta có: \(xy+yz+xz=1\)
Và \(T=x+y+z\)
\(\Rightarrow T^2=x^2+y^2+z^2+2(xy+yz+xz)\)
Theo hệ quả quen thuộc của BĐT Cauchy:
\(x^2+y^2+z^2\geq xy+yz+xz\)
\(\Rightarrow T^2\geq 3(xy+yz+xz)=3\)
\(\Rightarrow T\geq \sqrt{3}\Leftrightarrow T_{\min}=\sqrt{3}\)
Dấu bằng xảy ra khi \(x=y=z=\frac{1}{\sqrt{3}}\Leftrightarrow \angle A=\angle B=\angle C=60^0\)
\(\frac{\pi}{2}< a< \pi\Rightarrow\pi< 2a< 2\pi\)
Mà \(tan2a< 0\) \(\Rightarrow\frac{3\pi}{2}< 2a< 2\pi\Rightarrow cos2a>0\)
\(\Rightarrow cos2a=\frac{1}{\sqrt{1+tan^22a}}=\frac{3}{5}\)
\(tan\left(2a+\frac{\pi}{4}\right)=\frac{tan2a+tan\frac{\pi}{4}}{1-tan2a.tan\frac{\pi}{4}}=\frac{-\frac{4}{3}+1}{1+\frac{4}{3}}=...\)
\(cosa=\sqrt{1-sin^2a}=\frac{\sqrt{5}}{3}\)
\(tan\left(\frac{5\pi}{2}-b\right)=tan\left(2\pi+\frac{\pi}{2}-b\right)=tan\left(\frac{\pi}{2}-b\right)=cotb\)
\(\Rightarrow cotb=\frac{3}{4}\Rightarrow sinb=\frac{1}{\sqrt{1+cot^2b}}=\frac{4}{5}\)
\(\Rightarrow cosb=\sqrt{1-sin^2b}=\frac{3}{5}\)
\(A=sina.cosb+cosa.sinb=...\)
\(B=cosa.cosb+sina.sinb=...\)
\(cosa=\sqrt{1-sin^2a}=\frac{15}{17}\)
\(cosb=\frac{1}{\sqrt{1+tan^2b}}=\frac{12}{13}\)
\(sinb=\sqrt{1-cos^2b}=\frac{5}{13}\)
\(tanb=\frac{sinb}{cosb}=\frac{5}{12}\)
\(sin\left(a-b\right)=sina.cosb-cosa.sinb=\frac{8}{17}.\frac{12}{13}-\frac{15}{17}.\frac{5}{13}=...\)
\(cos\left(a+b\right)=cosa.cosb-sina.sinb=...\)
\(tan\left(a+b\right)=\frac{tana+tanb}{1-tana.tanb}=...\)
Bạn tự thay số và bấm máy
a) \(sin6\alpha cot3\alpha cos6\alpha=2.sin3\alpha.cos3\alpha\dfrac{cos3\alpha}{sin3\alpha}-cos6\alpha\)
\(=2cos^23\alpha-\left(2cos^23\alpha-1\right)=1\) (Không phụ thuộc vào x).
b) \(\left[tan\left(90^o-\alpha\right)-cot\left(90^o+\alpha\right)\right]^2\)\(-\left[cot\left(180^o+\alpha\right)+cot\left(270^o+\alpha\right)\right]^2\)
\(=\left[cot\alpha+cot\left(90^o-\alpha\right)\right]^2\)\(-\left[cot\alpha+cot\left(90^o+\alpha\right)\right]^2\)
\(=\left[cot\alpha+tan\alpha\right]^2-\left[cot\alpha-tan\alpha\right]^2\)
\(=4tan\alpha cot\alpha=4\). (Không phụ thuộc vào \(\alpha\)).
a) \(tan3\alpha-tan2\alpha-tan\alpha=\left(tan3\alpha-tan\alpha\right)-tan2\alpha\)
\(=\left(\dfrac{sin3\alpha}{cos3\alpha}-\dfrac{sin\alpha}{cos\alpha}\right)-\dfrac{sin2\alpha}{cos2\alpha}\)\(=\dfrac{sin3\alpha cos\alpha-cos3\alpha sin\alpha}{cos3\alpha cos\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\)
\(=\dfrac{sin2\alpha}{cos3\alpha cos\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\)
\(=sin2\alpha.\left(\dfrac{1}{cos3\alpha cos\alpha}-\dfrac{1}{cos2\alpha}\right)\)
\(=sin2\alpha.\dfrac{cos2\alpha-cos3\alpha cos\alpha}{cos3\alpha cos\alpha cos2\alpha}\)
\(=sin2\alpha.\dfrac{cos2\alpha-\dfrac{1}{2}\left(cos4\alpha+cos2\alpha\right)}{cos3\alpha cos2\alpha cos\alpha}\)
\(=sin2\alpha.\dfrac{cos2\alpha-cos4\alpha}{2cos3\alpha cos2\alpha cos\alpha}\)
\(=\dfrac{sin2\alpha.2sin3\alpha.sin\alpha}{2cos3\alpha cos2\alpha cos\alpha}\)
\(=tan3\alpha tan2\alpha tan\alpha\) (Đpcm).
b) \(\dfrac{4tan\alpha\left(1-tan^2\alpha\right)}{\left(1+tan^2\right)^2}=4tan\alpha\left(1-tan^2\alpha\right):\left(\dfrac{1}{cos^2\alpha}\right)^2\)
\(=4tan\alpha\left(1-tan^2\alpha\right)cos^4\alpha\)
\(=4\dfrac{sin\alpha}{cos\alpha}\left(1-\dfrac{sin^2\alpha}{cos^2\alpha}\right)cos^4\alpha\)
\(=4sin\alpha\left(cos^2\alpha-sin^2\alpha\right)cos\alpha\)
\(=4sin\alpha cos\alpha.cos2\alpha\)
\(=2.sin2\alpha.cos2\alpha=sin4\alpha\) (Đpcm).
\(tan2a=tan\left[\left(a+b\right)+\left(a-b\right)\right]=\dfrac{tan\left(a+b\right)+tan\left(a-b\right)}{1-tan\left(a+b\right)tan\left(a-b\right)}\)
\(\Rightarrow\dfrac{tan\left(a+b\right)+tan\left(a-b\right)}{1-tan\left(a+b\right)tan\left(a-b\right)}=\dfrac{5+4}{1-5.4}=-\dfrac{9}{19}\)
Vậy \(tan2a=-\dfrac{9}{19}\)
mình làm cách khác cũng ra kết quả nhưng hơi dài