2tan(3x)−3tan(2x)−tan2(2x)tan(3x)=0
Cho tan α + cot α = 3. Tính sin α.cos α và tan2 α + cot2 α
Cho tan α + cot α = 3. Tính sin α.cos α và tan2 α + cot2 α
Ta có: \(tana+cota=3\Rightarrow\dfrac{sina}{cosa}+\dfrac{cosa}{sina}=3\)
\(\Rightarrow\dfrac{sin^2a+cos^2a}{sina\cdot cosa}=3\Rightarrow sina\cdot cosa=\dfrac{1}{3}\)
Ta có: \(\left(tana+cota\right)^2=9\)\(\Rightarrow tan^2a+cot^2a=9-2tana\cdot cota=9-2=7\)
chứng minh công thức nhân đôi
\(\sin2\alpha=2.\sin\alpha.\cos\alpha\)
\(\cos2\alpha=\cos^2\alpha-\sin^2\alpha\)
\(\tan2\alpha=\dfrac{2\tan\alpha}{1-\tan^2\alpha}\)
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).
4km 15m=..........km
12 tấn 4kg=........tấn
5ha 500m2=.......ha
giúp mk,mk tick cho
4 km 15 m=4,015 km
12 tấn 4 kg=12,004 tấn
5 ha 500 m2=5,05 ha
4km 15m= 4,015 km
12 tấn 4kg= 12,004 tấn
5ha 500m2 = 5,05 ha
~Std well~
#Sharon
4km 15m = 4,015 km; 12 tấn 4kg = 12,004 tấn; 5ha 500m2 = 5,05 ha
Tui có sai cứ việc kiện cáo, chỉ biết giúp được là tui vui rồi hén?
Cho tam giác ABC vuông tại A, góc C = \(\alpha< 45^o\) . Chứng minh rằng:
\(tan2\alpha=\frac{2.tan\alpha}{1-tan^2\alpha}\)
VIET THEO THU TU TU BE DEN LON
2,03 TA ;202 KG; 0,21 TAN ; 20,4 YEN
Sắp xếp :
202 kg; 2,03 tạ; 20,4 yến; 0,21 tấn
tk nha!
202 kg ; 2,03 ta ; 20,4 yen ; 0,21 tan
\(\dfrac{tan2\alpha}{tan4\alpha-tan2\alpha}=cos4\alpha\)
chứng minh cong thức lượng giavs
Lời giải:
\(\frac{\tan 2a}{\tan 4a-\tan 2a}=\frac{\tan 2a}{\frac{2\tan 2a}{1-\tan ^22a}-\tan 2a}=\frac{1}{\frac{2}{1-\tan ^22a}-1}=\frac{1-\tan ^22a}{1+\tan ^22a}\)
\(=\frac{1-\frac{\sin ^22a}{\cos ^22a}}{1+\frac{\sin ^22a}{\cos ^22a}}=\frac{\cos ^22a-\sin ^22a}{\sin ^22a+\cos ^22a}=\cos ^22a-\sin ^22a=\cos 4a\)
Ta có đpcm.
