chứng minh: sin2\(\alpha\)= 1/1+tan2\(\alpha\)
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12 tháng 8 2022
Câu 2:
a: Xét ΔBAC có \(BC^2=AB^2+AC^2\)
nên ΔBAC vuông tại A
b: Xét ΔBAC vuông tại A có sin B=AC/BC=4/5
nên góc B=53 độ
=>góc C=37 độ
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 2 2019
\(\dfrac{1+cos2a-sin2a}{1+cos2a+sin2a}=\dfrac{2cos^2a-2sina.cosa}{2cos^2a+2sinacosa}\)
\(=\dfrac{2cosa\left(cosa-sina\right)}{2cosa\left(cosa+sina\right)}=\dfrac{cosa-sina}{cosa+sina}=\dfrac{\sqrt{2}sin\left(\dfrac{\pi}{4}-a\right)}{\sqrt{2}cos\left(\dfrac{\pi}{4}-a\right)}=tan\left(\dfrac{\pi}{4}-a\right)\)
\(\dfrac{1+cos2a-cosa}{sin2a-sina}=\dfrac{2cos^2a-cosa}{2sina.cosa-sina}=\dfrac{cosa\left(2cosa-1\right)}{sina\left(2cosa-1\right)}=\dfrac{cosa}{sina}=cota\)
AH
Akai Haruma
Giáo viên
19 tháng 4 2021
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.
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
20 tháng 4 2019
\(sin^6a+cos^6a=\left(sin^2x\right)^3+\left(cos^2x\right)^3\)
\(=\left(sin^2x+cos^2x\right)\left(sin^4x+cos^4x-sin^2x.cos^2x\right)\)
\(=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2x.cos^2x\)
\(=\left(sin^2x+cos^2x\right)^2-\frac{3}{4}.\left(2sinx.cosx\right)^2\)
\(=1-\frac{3}{4}sin^22x=1-\frac{3}{4}\left(\frac{1}{2}-\frac{1}{2}cos4x\right)=\frac{5}{8}+\frac{3}{8}cos4x\)
2/
\(\frac{1+sin2a-cos2a}{1+cos2a}=\frac{1+2sina.cosa-\left(1-2sin^2a\right)}{1+2cos^2a-1}=\frac{2sina.cosa+2sin^2a}{2cos^2a}\)
\(=\frac{2sina.cosa}{2cos^2a}+\frac{2sin^2a}{2cos^2a}=tana+tan^2a\)
\(\dfrac{1}{1+tan^2\alpha}=\dfrac{1}{1+\left(\dfrac{sin\alpha}{cos\alpha}\right)^2}\)
\(=\dfrac{1}{1+\dfrac{sin^2\alpha}{cos^2\alpha}}\)
\(=1:\dfrac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}=\dfrac{cos^2\alpha}{cos^2\alpha+sin^2\alpha}=cos^2\alpha\)
Đề bài sai, sửa thành c/m \(\sin^2\alpha=\dfrac{1}{1+\cot^2\alpha}\)
\(\dfrac{1}{1+\cot^2\alpha}=\dfrac{1}{1+\dfrac{\cos^2\alpha}{\sin^2\alpha}}=\dfrac{1}{\dfrac{\sin^2\alpha+\cos^2\alpha}{\sin^2\alpha}}=\dfrac{1}{\dfrac{1}{\sin^2\alpha}}=\sin^2\alpha\)