Chứng minh: \(\tan\alpha.\tan\left(\frac{\pi}{3}-\alpha\right).\tan\left(\frac{\pi}{3}+\alpha\right)=\tan3\alpha\)
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BT
11 tháng 5 2017
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).
BT
11 tháng 5 2017
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).
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 4 2021
\(VT=\dfrac{-tan\left(\dfrac{\pi}{2}-a\right)cos\left(2\pi-\dfrac{\pi}{2}+a\right)-sin^3\left(4\pi-\dfrac{\pi}{2}-a\right)}{cos\left(\dfrac{\pi}{2}-a\right)tan\left(2\pi-\dfrac{\pi}{2}+a\right)}\)
\(=\dfrac{-cota.sina+sin^3\left(\dfrac{\pi}{2}+a\right)}{sina.\left(-cota\right)}=\dfrac{-cosa+cos^3a}{-cosa}=1-cos^2a=sin^2a\)
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 6 2020
\(\left(tana+cota\right)^2=16\)
\(\Leftrightarrow tan^2a+cot^2a+2=16\)
\(\Rightarrow tan^2a+cot^2a=14\)
\(tan^2\left(a+3\pi\right)+tan^2\left(a+\frac{3\pi}{2}\right)=tan^2a+cot^2a=14\)
HQ
Hà Quang Minh
Giáo viên
25 tháng 8 2023
\(a,\dfrac{1}{tan\alpha+1}+\dfrac{1}{cot\alpha+1}\\ =\dfrac{cot\alpha+1+tan\alpha+1}{\left(tan\alpha+1\right)\left(cot\alpha+1\right)}\\ =\dfrac{tan\alpha+cot\alpha+2}{tan\alpha\cdot cot\alpha+tan\alpha+cot\alpha+1}\\ =\dfrac{tan\alpha+cot\alpha+2}{tan\alpha+cot\alpha+2}\\ =1\)
\(b,cos\left(\dfrac{\pi}{2}-\alpha\right)-sin\left(\pi+\alpha\right)\\ =sin\alpha+sin\alpha\\ =2sin\alpha\)
\(c,sin\left(\alpha-\dfrac{\pi}{2}\right)+cos\left(-\alpha+6\pi\right)-tan\left(\alpha+\pi\right)cot\left(3\pi-\alpha\right)\\ =-sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\alpha\right)-tan\left(\alpha\right)cot\left(\pi-\alpha\right)\\ =-cos\left(\alpha\right)+cos\left(\alpha\right)+tan\left(\alpha\right)\cdot cot\left(\alpha\right)\\ =1\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 5 2020
\(\frac{a}{2}\in\left(\frac{\pi}{2};\frac{3\pi}{4}\right)\Rightarrow tan\frac{a}{2}< 0\) ; \(sin\frac{a}{2}>0;cos\frac{a}{2}< 0\)
Đặt \(tan\frac{a}{2}=x< 0\)
\(\frac{2x}{1-x^2}=3\Leftrightarrow3x^2+2x-3=0\Rightarrow tan\frac{a}{2}=x=\frac{-1-\sqrt{10}}{3}\)
\(tan2a=\frac{2tana}{1-tan^2a}=\frac{6}{1-9}=-\frac{3}{4}\)
\(tan4a=\frac{2tan2a}{1-tan^22a}=-\frac{24}{7}\)
\(cos\frac{a}{2}=-\frac{1}{\sqrt{1+tan^2\frac{a}{2}}}=\) số thật kinh khủng
\(sin\frac{a}{2}=\sqrt{1-cos^2\frac{a}{2}}=...\)
\(sin\left(\frac{a}{2}+\frac{\pi}{2}\right)=\sqrt{2}\left(sin\frac{a}{2}+cos\frac{a}{2}\right)=...\)
TN
17 tháng 6 2016
a)\(tan3A=tan\left(A+2A\right)\)
\(=\frac{tanA+tan2A}{1-tanAtan2A}\)
\(=\frac{\frac{tanA+2tanA}{1-tan^2A}}{\frac{1-2tan^2A}{1-tan^2A}}\)
\(=\frac{\left(tanA-tan^3A+2tanA\right)}{1-tan^2A-2tan^2A}\)
\(=\frac{3tanA-tan^3A}{1-3tan^2A}\)
b)\(VT=cos^6A+sin^6A\)
\(=\left(cos^2A\right)^3+\left(sin^2A\right)^3\)
\(=\left(cos^2A+sin^2A\right)^3-3cos^2Asin^2A\left(cos^2A+sin^2A\right)^2\)
\(=1^3-3cos^2Asin^2A\left(1\right)^2\).Từ đó,\(sin^2A+cos^2A=1\)
\(=1-3cos^2Asin^2A=VP\)
13 tháng 9 2023
\(P=\dfrac{tan\left(-a\right)+2\cdot cota}{3\cdot tan\left(\dfrac{pi}{2}+a\right)}=\dfrac{-tana+2\cdot\dfrac{1}{2}}{3\cdot\left(-cota\right)}\)
\(=\dfrac{-2+1}{3\cdot\dfrac{-1}{2}}=-1:\dfrac{-3}{2}=\dfrac{2}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 6 2019
\(0< a< \frac{\pi}{2}\Rightarrow sina;cosa;tana>0\)
\(tana+\frac{1}{tana}=3\Leftrightarrow tan^2a-3tana+1=0\) \(\Rightarrow\left[{}\begin{matrix}tana=\frac{3-\sqrt{5}}{2}\\tana=\frac{3+\sqrt{5}}{2}\end{matrix}\right.\)
- Với \(tana=\frac{3-\sqrt{5}}{2}\)
\(\Rightarrow cota=\frac{1}{tana}=\frac{3+\sqrt{5}}{2}\)
\(1+tan^2a=\frac{1}{cos^2a}\Rightarrow cosa=\frac{1}{\sqrt{1+tan^2a}}=\frac{2}{\sqrt{18-6\sqrt{5}}}\)
\(sina=\sqrt{1-cos^2a}=\frac{2}{\sqrt{18+6\sqrt{5}}}\)
\(cos\left(\frac{3\pi}{2}-a\right)=cos\left(2\pi-\frac{\pi}{2}-a\right)=-sina=...\)
\(sin\left(2\pi+a\right)=sina=...\)
\(tan\left(\pi-a\right)=-tana=...\)
\(cot\left(\pi+a\right)=cota=...\)
TH2: \(tana=\frac{3+\sqrt{5}}{2}\)
Tương tự như trên
\(tan\left(\frac{\pi}{3}-a\right)tan\left(\frac{\pi}{3}+a\right)=\frac{sin\left(\frac{\pi}{3}-a\right)sin\left(\frac{\pi}{3}+a\right)}{cos\left(\frac{\pi}{3}-a\right)cos\left(\frac{\pi}{3}+a\right)}\)
\(=\frac{cos2a-cos\frac{2\pi}{3}}{cos2a+cos\frac{2\pi}{3}}=\frac{cos2a+\frac{1}{2}}{cos2a-\frac{1}{2}}=\frac{2cos2a+1}{2cos2a-1}\)
\(\Rightarrow tana.tan\left(\frac{\pi}{3}-a\right)tan\left(\frac{\pi}{3}+a\right)=\frac{sina\left(2cos2a+1\right)}{cosa\left(2cos2a-1\right)}=\frac{2sina.cos2a+sina}{2cos2a.cosa-cosa}\)
\(=\frac{sin3a-sina+sina}{cos3a+cosa-cosa}=\frac{sin3a}{cos3a}=tan3a\)