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NV
7 tháng 6 2020

\(\left(tana+cota\right)^2=16\)

\(\Leftrightarrow tan^2a+cot^2a+2=16\)

\(\Rightarrow tan^2a+cot^2a=14\)

\(tan^2\left(a+3\pi\right)+tan^2\left(a+\frac{3\pi}{2}\right)=tan^2a+cot^2a=14\)

25 tháng 7 2018

bài 1) ta có : \(G=cos\left(\alpha-5\pi\right)+sin\left(\dfrac{-3\pi}{2}+\alpha\right)-tan\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\dfrac{3\pi}{2}-\alpha\right)\)

\(G=cos\left(\alpha-\pi\right)+sin\left(\dfrac{\pi}{2}+\alpha\right)-tan\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\dfrac{\pi}{2}-\alpha\right)\)

\(G=cos\left(\pi-\alpha\right)+sin\left(\dfrac{\pi}{2}-\left(-\alpha\right)\right)-tan\left(\pi+\alpha-\dfrac{\pi}{2}\right).cot\left(\dfrac{\pi}{2}-\alpha\right)\) \(G=cos\left(\alpha\right)+cos\left(\alpha\right)+tan\left(\dfrac{\pi}{2}-\alpha\right).cot\left(\dfrac{\pi}{2}-\alpha\right)=2cos\alpha+1\) bài 2) ta có : \(H=cot\left(\alpha\right).cos\left(\alpha+\dfrac{\pi}{2}\right)+cos\left(\alpha\right)-2sin\left(\alpha-\pi\right)\) \(H=cot\left(\alpha\right).cos\left(\dfrac{\pi}{2}-\left(-\alpha\right)\right)+cos\left(\alpha\right)+2sin\left(\pi-\alpha\right)\) \(H=-cot\left(\alpha\right).sin\left(\alpha\right)+cos\left(\alpha\right)+2sin\left(\alpha\right)\) \(H=-cos\alpha+cos\alpha+2sin\alpha=2sin\alpha\)

NV
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

AH
Akai Haruma
Giáo viên
3 tháng 7 2018

Lời giải:

Theo công thức lượng giác:

\(F=\sin (\pi +a)-\cos (\frac{\pi}{2}-a)+\cot (2\pi -a)+\tan (\frac{3\pi}{2}-a)\)

\(=-\sin a-\sin a+\cot (\pi -a)+\tan (\frac{\pi}{2}-a)\)

\(=-2\sin a-\cot a+\cot a=-2\sin a\)

NV
18 tháng 4 2021

\(VT=\dfrac{-tan\left(\dfrac{\pi}{2}-a\right)cos\left(2\pi-\dfrac{\pi}{2}+a\right)-sin^3\left(4\pi-\dfrac{\pi}{2}-a\right)}{cos\left(\dfrac{\pi}{2}-a\right)tan\left(2\pi-\dfrac{\pi}{2}+a\right)}\)

\(=\dfrac{-cota.sina+sin^3\left(\dfrac{\pi}{2}+a\right)}{sina.\left(-cota\right)}=\dfrac{-cosa+cos^3a}{-cosa}=1-cos^2a=sin^2a\)

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).

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).

2 tháng 6 2018

G = \(cos\left(a+\pi-6\text{​​}\text{​​}\pi\right)+sin\left(-2\pi+\dfrac{\pi}{2}+a\right)-tan\left(\dfrac{\pi}{2}+a\right)\cdot cot\left(\pi+\dfrac{\pi}{2}-a\right)\)

= \(cos\left(a+\pi\right)+sin\left(\dfrac{\pi}{2}+a\right)-tan\left(\dfrac{\pi}{2}+a\right)\cdot cot\left(\dfrac{\pi}{2}-a\right)\)

= \(-cosa+cosa-\left(-cota\cdot tana\right)=1\)

NV
5 tháng 5 2020

\(tan\left(\frac{\pi}{3}-a\right)tan\left(\frac{\pi}{3}+a\right)=\frac{sin\left(\frac{\pi}{3}-a\right)sin\left(\frac{\pi}{3}+a\right)}{cos\left(\frac{\pi}{3}-a\right)cos\left(\frac{\pi}{3}+a\right)}\)

\(=\frac{cos2a-cos\frac{2\pi}{3}}{cos2a+cos\frac{2\pi}{3}}=\frac{cos2a+\frac{1}{2}}{cos2a-\frac{1}{2}}=\frac{2cos2a+1}{2cos2a-1}\)

\(\Rightarrow tana.tan\left(\frac{\pi}{3}-a\right)tan\left(\frac{\pi}{3}+a\right)=\frac{sina\left(2cos2a+1\right)}{cosa\left(2cos2a-1\right)}=\frac{2sina.cos2a+sina}{2cos2a.cosa-cosa}\)

\(=\frac{sin3a-sina+sina}{cos3a+cosa-cosa}=\frac{sin3a}{cos3a}=tan3a\)

11 tháng 5 2017

a)\(sin\left(\alpha+\dfrac{\pi}{2}\right)=cos\left[\dfrac{\pi}{2}-\left(\alpha+\dfrac{\pi}{2}\right)\right]=cos\left(-\alpha\right)=cos\alpha\).
b) \(cos\left(x+\dfrac{\pi}{2}\right)=sin\left[\dfrac{\pi}{2}-\left(x+\dfrac{\pi}{2}\right)\right]=sin\left(-x\right)=-sinx\).
c) \(tan\left(\alpha+\dfrac{\pi}{2}\right)=\dfrac{sin\left(\alpha+\dfrac{\pi}{2}\right)}{cos\left(\alpha+\dfrac{\pi}{2}\right)}=\dfrac{cos\alpha}{-sin\alpha}=-cot\alpha\).
d) \(cot\left(\alpha+\dfrac{\pi}{2}\right)=\dfrac{cos\left(\alpha+\dfrac{\pi}{2}\right)}{sin\left(\alpha+\dfrac{\pi}{2}\right)}=\dfrac{-sin\alpha}{cos\alpha}=-tan\alpha\).

NV
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

\(sin\frac{a}{2}=\sqrt{1-cos^2\frac{a}{2}}=...\)

\(sin\left(\frac{a}{2}+\frac{\pi}{2}\right)=\sqrt{2}\left(sin\frac{a}{2}+cos\frac{a}{2}\right)=...\)