Cho tanα = 2. Tính P=\(\dfrac{\tan\left(8\pi-\alpha\right)+2\cot\left(\pi+\alpha\right)}{3\tan\left(\dfrac{3\pi}{2}+\alpha\right)}\)
Đơn giản các biểu thức sau:
G = \(cos\left(\alpha-5\pi\right)+sin\left(-\dfrac{3\pi}{2}+\alpha\right)-tan\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\dfrac{3\pi}{2}-\alpha\right)\)
H = \(cot\left(\alpha-2\pi\right).cos\left(\alpha-\dfrac{3\pi}{2}\right)+cos\left(\alpha-6\pi\right)-2sin\left(\alpha-\pi\right)\)
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 : \(H=cot\left(\alpha\right).cos\left(\alpha+\dfrac{\pi}{2}\right)+cos\left(\alpha\right)-2sin\left(\alpha-\pi\right)\) \(H=cot\left(\alpha\right).cos\left(\dfrac{\pi}{2}-\left(-\alpha\right)\right)+cos\left(\alpha\right)+2sin\left(\pi-\alpha\right)\) \(H=-cot\left(\alpha\right).sin\left(\alpha\right)+cos\left(\alpha\right)+2sin\left(\alpha\right)\) \(H=-cos\alpha+cos\alpha+2sin\alpha=2sin\alpha\)
Chứng minh đẳng thức: \(\dfrac{tan\left(\alpha-\dfrac{\pi}{2}\right).cos\left(\dfrac{3\pi}{2}+\alpha\right)-sin^3\left(\dfrac{7\pi}{2}-\alpha\right)}{cos\left(\alpha-\dfrac{\pi}{2}\right).tan\left(\dfrac{3\pi}{2}+\alpha\right)}=sin^2\alpha\)
\(VT=\dfrac{-tan\left(\dfrac{\pi}{2}-a\right)cos\left(2\pi-\dfrac{\pi}{2}+a\right)-sin^3\left(4\pi-\dfrac{\pi}{2}-a\right)}{cos\left(\dfrac{\pi}{2}-a\right)tan\left(2\pi-\dfrac{\pi}{2}+a\right)}\)
\(=\dfrac{-cota.sina+sin^3\left(\dfrac{\pi}{2}+a\right)}{sina.\left(-cota\right)}=\dfrac{-cosa+cos^3a}{-cosa}=1-cos^2a=sin^2a\)
Đơn giản biểu thức sau:
\(F=sin\left(\pi+\alpha\right)-cos\left(\dfrac{\pi}{2}-\alpha\right)+cot\left(2\pi-\alpha\right)+tan\left(\dfrac{3\pi}{2}-\alpha\right)\)
Lời giải:
Theo công thức lượng giác:
\(F=\sin (\pi +a)-\cos (\frac{\pi}{2}-a)+\cot (2\pi -a)+\tan (\frac{3\pi}{2}-a)\)
\(=-\sin a-\sin a+\cot (\pi -a)+\tan (\frac{\pi}{2}-a)\)
\(=-2\sin a-\cot a+\cot a=-2\sin a\)
Đơn giản biểu thức sau:
\(G=Cos\left(\alpha-5\pi\right)+sin\left(-\dfrac{3\pi}{2}+\alpha\right)-tan\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\dfrac{3\pi}{2}-\alpha\right)\)
.
G = \(cos\left(a+\pi-6\text{}\text{}\pi\right)+sin\left(-2\pi+\dfrac{\pi}{2}+a\right)-tan\left(\dfrac{\pi}{2}+a\right)\cdot cot\left(\pi+\dfrac{\pi}{2}-a\right)\)
= \(cos\left(a+\pi\right)+sin\left(\dfrac{\pi}{2}+a\right)-tan\left(\dfrac{\pi}{2}+a\right)\cdot cot\left(\dfrac{\pi}{2}-a\right)\)
= \(-cosa+cosa-\left(-cota\cdot tana\right)=1\)
Chứng minh rằng với mọi \(\alpha\), ta luôn có :
a) \(\sin\left(\alpha+\dfrac{\pi}{2}\right)=\cos\alpha\)
b) \(\cos\left(\alpha+\dfrac{\pi}{2}\right)=-\sin\alpha\)
c) \(\tan\left(\alpha+\dfrac{\pi}{2}\right)=-\cot\alpha\)
d) \(\cot\left(\alpha+\dfrac{\pi}{2}\right)=-\tan\alpha\)
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) \(cot\left(\alpha+\dfrac{\pi}{2}\right)=\dfrac{cos\left(\alpha+\dfrac{\pi}{2}\right)}{sin\left(\alpha+\dfrac{\pi}{2}\right)}=\dfrac{-sin\alpha}{cos\alpha}=-tan\alpha\).
Cho \(\pi< \alpha< \dfrac{3\pi}{2}\). Xác định dấu của các giá trị lượng giác sau :
a) \(\cos\left(\alpha-\dfrac{\pi}{2}\right)\)
b) \(\sin\left(\dfrac{\pi}{2}+\alpha\right)\)
c) \(\tan\left(\dfrac{3\pi}{2}-\alpha\right)\)
d) \(\cot\left(\alpha+\pi\right)\)
Do \(\pi< \alpha< \dfrac{3\pi}{2}\) nên \(sin\alpha,cos\alpha< 0;tan\alpha,cot\alpha< 0\).
\(cos\left(\alpha-\dfrac{\pi}{2}\right)=cos\left(\dfrac{\pi}{2}-\alpha\right)=sin\alpha< 0\).
\(sin\left(\dfrac{\pi}{2}+\alpha\right)=cos\alpha< 0\).
\(tan\left(\dfrac{3\pi}{2}-\alpha\right)=tan\left(\dfrac{3\pi}{2}-\alpha-2\pi\right)\)\(=tan\left(-\dfrac{\pi}{2}-\alpha\right)\)\(=-tan\left(\dfrac{\pi}{2}+\alpha\right)=cot\left(\alpha\right)>0\).
\(cot\left(\alpha+\pi\right)=cot\left(\alpha\right)>0\).
a) Rút gọn biểu thức
\(A=\dfrac{\sin4x+2\sin2x}{\sin4x-2\sin2x}.\cot\left(\dfrac{3\pi}{2}-x\right)\) (khi biểu thức có nghĩa)
b) Cho \(\cot\alpha=\dfrac{4}{3},3\pi< \alpha< \dfrac{7\pi}{2}\). Tính \(\cos\left(\dfrac{2\pi}{3}-\alpha\right)\)
Cho 2tanα-cotα=1. Tính P=\(\dfrac{\text{tan ( 8 π − α ) + 2 cot ( π + α )}}{3\tan\left(\dfrac{3\pi}{2}+\alpha\right)}\)
2tan a-cot a=1
=>2tana-1/tan a=1
=>\(\dfrac{2tan^2a-1}{tana}=1\)
=>2tan^2a-tana-1=0
=>(tan a-1)(2tana+1)=0
=>tan a=-1/2 hoặc tan a=1
\(P=\dfrac{tan\left(-a\right)+2\cdot cota}{3\cdot tan\left(\dfrac{pi}{2}+a\right)}=\dfrac{-tana+2\cdot cota}{-3\cdot cota}\)
TH1: tan a=-1/2
\(P=\dfrac{\dfrac{1}{2}+2\cdot\left(-2\right)}{-3\cdot\left(-2\right)}=-\dfrac{7}{2}:6=-\dfrac{7}{12}\)
TH2: tan a=1
=>cot a=1
\(P=\dfrac{-1+2}{-3}=\dfrac{1}{-3}=-\dfrac{1}{3}\)
Ta có :
\(2tan\alpha-cot\alpha=1\)
\(\Leftrightarrow2tan\alpha-\dfrac{1}{tan\alpha}=1\)
\(\Leftrightarrow2tan\alpha-\dfrac{1}{tan\alpha}-1=0\)
\(\Leftrightarrow\dfrac{2tan^2\alpha-tan\alpha-1}{tan\alpha}=0\left(tan\alpha\ne0\right)\)
\(\Leftrightarrow2tan^2\alpha-tan\alpha-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tan\alpha=1\\tan\alpha=-\dfrac{1}{2}\end{matrix}\right.\)
\(P=\dfrac{tan\left(8\pi-\alpha\right)+2cot\left(\pi+\alpha\right)}{3tan\left(\dfrac{3\pi}{2}+\alpha\right)}\)
\(\Leftrightarrow P=\dfrac{tan\left(4.2\pi-\alpha\right)+2cot\alpha}{3tan\left(2\pi-\dfrac{\pi}{2}+\alpha\right)}\)
\(\Leftrightarrow P=\dfrac{tan\left(-\alpha\right)+2cot\alpha}{3tan\left[-\left(\dfrac{\pi}{2}-\alpha\right)\right]}\)
\(\Leftrightarrow P=\dfrac{-tan\alpha+2cot\alpha}{-3tan\left(\dfrac{\pi}{2}-\alpha\right)}\)
\(\Leftrightarrow P=\dfrac{-tan\alpha+2cot\alpha}{-3cot\alpha}\)
- Với \(tan\alpha=1\Rightarrow cot\alpha=1\)
\(\Leftrightarrow P=\dfrac{-1+2.1}{-3.1}=-\dfrac{1}{3}\)
- Với \(tan\alpha=-\dfrac{1}{2}\Rightarrow cot\alpha=-2\)
\(\Leftrightarrow P=\dfrac{\dfrac{1}{2}+2.\left(-2\right)}{-3.\left(-2\right)}=\dfrac{-\dfrac{7}{2}}{6}=-\dfrac{7}{12}\)
Chứng minh các đẳng thức :
a) \(\tan3\alpha-\tan2\alpha-\tan\alpha=\tan\alpha\tan2\alpha\tan3\alpha\)
b) \(\dfrac{4\tan\alpha\left(1-\tan^2\alpha\right)}{\left(1+\tan^2\alpha\right)^2}=\sin4\alpha\)
c) \(\dfrac{1+\tan^4\alpha}{\tan^2\alpha+\cot^2\alpha}=\tan^2\alpha\)
d) \(\dfrac{\cos\alpha\sin\left(\alpha-3\right)-\sin\alpha\cos\left(\alpha-3\right)}{\cos\left(3-\dfrac{\pi}{6}\right)-\dfrac{1}{2}\sin3}=-\dfrac{2\tan3}{\sqrt{3}}\)
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).
c) \(\dfrac{1+tan^4\alpha}{tan^2\alpha+cot\alpha}=\left(1+tan^4\alpha\right):\left(tan^2\alpha+cot^2\alpha\right)\)
\(=\left(1+\dfrac{sin^4\alpha}{cos^4\alpha}\right):\left(\dfrac{sin^2\alpha}{cos^2\alpha}+\dfrac{cos^2\alpha}{sin^2\alpha}\right)\)
\(=\dfrac{sin^4\alpha+cos^4\alpha}{cos^4\alpha}:\dfrac{sin^4\alpha+cos^4\alpha}{cos^2\alpha sin^2\alpha}\)
\(=\dfrac{sin^2\alpha}{cos^2\alpha}=tan^2\alpha\) (Đpcm).