Chứng minh rằng: \(\dfrac{\sin\alpha+\cos\alpha-1}{1-\cos\alpha}\)=\(\dfrac{2\cdot\cos\alpha}{\sin\alpha-\cos\alpha+1}\)
Chứng minh:
a)\(cot^2\alpha-cos^2\alpha\cdot cot^2\alpha=cos^2\alpha\)
b)\(tan^2\alpha-sin^2\alpha\cdot tan^2\alpha=sin^2\alpha\)
c) \(\dfrac{1-cos^2}{sin\alpha}\) = \(\dfrac{sin\alpha}{1+cos\alpha}\)
d)\(tan^2\alpha-sin^2\alpha=tan^2\cdot sin^2\alpha\)
e) \(\sin^6\alpha+cos^6\alpha+3sin^2\cdot cos^2\alpha=1\)
chứng minh các đẳng thức sau
a) \(\dfrac{1-cos\alpha}{sin\alpha}=\dfrac{sin\alpha}{1+cos\alpha}\)
b)\(\dfrac{cos\alpha}{1+sin\alpha}+tg\alpha=\dfrac{1}{cos\alpha}\)
a) Cần chứng minh \(\dfrac{1-cos\alpha}{sin\alpha}=\dfrac{sin\alpha}{1+cos\alpha}\)
\(\Rightarrow sin^2\alpha=\left(1-cos\alpha\right)\left(1+cos\alpha\right)\Rightarrow sin^2\alpha=1-cos^2\alpha\)
\(\Rightarrow sin^2\alpha+cos^2\alpha=1\)
Giả sử tam giác ABC vuông tại A
Ta có: \(\left\{{}\begin{matrix}sin^2B=\dfrac{AC^2}{BC^2}\\cos^2B=\dfrac{AB^2}{BC^2}\end{matrix}\right.\Rightarrow sin^2B+cos^2B=\dfrac{AC^2+AB^2}{BC^2}=\dfrac{BC^2}{BC^2}=1\)
a)\(\dfrac{1-cosa}{sina}=\dfrac{sina}{1+cosa}\)
<=>\(\left(1-cosa\right)\left(1+cosa\right)=sin^2a\)
<=>\(1-cos^2a=sin^2a\) (lđ)
b)Ta có VT=\(\dfrac{cosa}{1+sina}+tga=\dfrac{cosa}{1+sina}+\dfrac{sina}{cosa}=\dfrac{cos^2a+sin^2a+sina}{\left(1+sina\right)cosa}=\dfrac{1+sina}{\left(1+sina\right)cosa}=\dfrac{1}{cosa}=vp\left(dpcm\right)\)
Chứng minh rằng : \(\dfrac{1 + cos \alpha}{1-cos\alpha} - \dfrac{1-cos\alpha}{1+cos\alpha} = \dfrac{4cot\alpha}{sin\alpha}\)
Lời giải:
\(\frac{1+\cos a}{1-\cos a}-\frac{1-\cos a}{1+\cos a}=\frac{(1+\cos a)^2-(1-\cos a)^2}{(1-\cos a)(1+\cos a)}=\frac{1+2\cos a+\cos ^2a-(1-2\cos a+\cos ^2a)}{1-\cos ^2a}\)
\(=\frac{4\cos a}{\sin ^2a}=\frac{\frac{4\cos a}{\sin a}}{\sin a}=\frac{4\cot a}{\sin a}\) (đpcm)
chứng minh các biểu thức sau :
a) \(\dfrac{cos\alpha}{1-sin\alpha}=\dfrac{1+sin\alpha}{cos\alpha}\)
b) \(\dfrac{\left(sin\alpha+cos\alpha\right)^2-\left(sin\alpha-cos\alpha\right)^2}{sin\alpha+cos\alpha}\)
a, Sử dụng tích chéo:
Ta có:
+/ \(\cos\alpha.\cos\alpha=\cos^2\alpha\) (1)
+/ \(\left(1+\sin\alpha\right)\left(1-\sin\alpha\right)=1-\sin^2\alpha\)
Mà \(\sin^2\alpha+\cos^2\alpha=1\)
\(\Rightarrow1-\sin^2\alpha=\cos^2\alpha\)
hay \(\left(1+\sin\alpha\right)\left(1-\sin\alpha\right)=\cos^2\alpha\) (2)
Từ (1), (2)
\(\Rightarrow\)\(\cos\alpha.\cos\alpha=\)\(\left(1+\sin\alpha\right)\left(1-\sin\alpha\right)\)
\(\Rightarrow\)\(\dfrac{\cos\alpha}{1-\sin\alpha}=\dfrac{1+\sin\alpha}{\cos\alpha}\) (đpcm)
b/ xem lại đề
Chứng minh các đẳng thức sau:
a, \(\sin^4\alpha-\cos^4\alpha+1=2\sin^2\alpha\)
b,\(\dfrac{\sin^2\alpha+2\cos^2\alpha-1}{\cot^2\alpha}=\sin^2\alpha\)
c, \(\dfrac{1-\sin^2\alpha.\cos^2\alpha}{\cos^2\alpha}-\cos^2\alpha=\tan^2\alpha\)
d, \(\dfrac{\sin^2\alpha-\tan^2\alpha}{\cos^2\alpha-\cot^2\alpha}=\tan^6\alpha\)
e, \(\left(1+\cot\alpha\right)\sin^3\alpha+\left(1+\tan\alpha\right)\cos^3\alpha=\sin\alpha.\cos\alpha\)
f,\(\dfrac{\left(\sin\alpha+\cos\alpha\right)^2-1}{\cot\alpha-\sin\alpha.\cos\alpha}=2\tan^2\alpha\)
a)
\(\sin ^4a-\cos ^4a+1=(\sin ^2a-\cos ^2a)(\sin ^2a+\cos^2a)+1\)
\(=(\sin ^2a-\cos ^2a).1+1=\sin ^2a-\cos ^2a+\sin ^2a+\cos ^2a\)
\(=2\sin ^2a\)
b) \(\sin ^2a+2\cos ^2a-1=(\sin ^2a+\cos^2a)+\cos ^2a-1\)
\(=1+\cos ^2a-1=\cos ^2a\)
\(\Rightarrow \frac{\sin ^2a+2\cos ^2a-1}{\cot ^2a}=\frac{\cos ^2a}{\cot ^2a}=\frac{\cos ^2a}{\frac{\cos ^2a}{\sin ^2a}}=\sin ^2a\)
c)
\(\frac{1-\sin ^2a\cos ^2a}{\cos ^2a}-\cos ^2a=\frac{1}{\cos ^2a}-\sin ^2a-\cos ^2a\)
\(=\frac{1}{\cos ^2a}-(\sin ^2a+\cos ^2a)=\frac{1}{\cos ^2a}-1\)
\(=\frac{1-\cos ^2a}{\cos ^2a}=\frac{\sin ^2a}{\cos ^2a}=\tan ^2a\)
d)
\(\frac{\sin ^2a-\tan ^2a}{\cos ^2a-\cot ^2a}=\frac{\sin ^2a-\frac{\sin ^2a}{\cos ^2a}}{\cos ^2a-\frac{\cos ^2a}{\sin ^2a}}\) \(=\frac{\sin ^2a(1-\frac{1}{\cos ^2a})}{\cos ^2a(1-\frac{1}{\sin ^2a})}\)
\(=\frac{\sin ^2a.\frac{\cos ^2a-1}{\cos ^2a}}{\cos ^2a.\frac{\sin ^2a-1}{\sin ^2a}}\) \(=\frac{\sin ^2a.\frac{-\sin ^2a}{\cos ^2a}}{\cos ^2a.\frac{-\cos ^2a}{\sin ^2a}}=\frac{\sin ^6a}{\cos ^6a}=\tan ^6a\)
f)
\(\frac{(\sin a+\cos a)^2-1}{\cot a-\sin a\cos a}=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\frac{\cos a}{\sin a}-\sin a\cos a}\)
\(=\sin a.\frac{1+2\sin a\cos a-1}{\cos a-\cos a\sin ^2a}\)
\(=\sin a. \frac{2\sin a\cos a}{\cos a(1-\sin ^2a)}=\sin a. \frac{2\sin a\cos a}{\cos a. \cos^2 a}=\frac{2\sin ^2a}{\cos ^2a}=2\tan ^2a\)
e)
\((1+\cot a)\sin ^3a+(1+\tan a)\cos ^3a\)
\(=(\sin ^3a+\cos ^3a)+\cot a.\sin ^3a+\tan a.\cos^3a\)
\(=(\sin a+\cos a)(\sin ^2a-\sin a\cos a+\cos ^2a)+\frac{\cos a}{\sin a}.\sin ^3a+\frac{\sin a}{\cos a}.\cos ^3a\)
\(=(\sin a+\cos a)(1-\sin a\cos a)+\cos a\sin ^2a+\sin a\cos ^2a\)
\(=\sin a+\cos a-\sin a\cos a(\sin a+\cos a)+\cos a\sin a(\sin a+\cos a)\)
\(=\sin a+\cos a\)
1. Cho tam giác $ABC$. Chứng minh rằng $\sin ^{2} A+\sin ^{2} B-\sin ^{2} C=2\sin A.\sin B.\cos C$.
2. Chứng minh rằng:
a. $\sin \alpha .\sin \left(\dfrac{\pi }{3} -\alpha \right).\sin \left(\dfrac{\pi }{3} +\alpha \right)=\dfrac{1}{4} \sin 3\alpha $
b. $\sin 5\alpha -2\sin \alpha \left({\rm cos} {\rm 4}\alpha +\cos 2\alpha \right)=\sin \alpha $
Chứng minh các công thức sau :
\(Tan\alpha=\dfrac{sin\alpha}{cos\alpha}\)
\(Cot\alpha=\dfrac{cos\alpha}{sin\alpha}\)
\(sin^2\alpha+cos^2\alpha=1\)
\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\)
\(1+cos^2\alpha=\dfrac{1}{sin^2\alpha}\)
\(cos^4\alpha-sin^4\alpha=2cos^2\alpha-1\)
Ta có:
\(sin=\dfrac{doi}{huyen}\); \(cos=\dfrac{ke}{chuyen}\);\(tan=\dfrac{doi}{ke}\); \(cot=\dfrac{ke}{doi}\)
Dùng cái này làm được hết mấy câu đó.
nếu bn thấy dùng cách của hùng có hới dài thì bn chỉ cần sử dụng cách đó cho 3 ý trên thôi . còn 3 ý dưới bn có thể sử dụng công thức \(sin^2x+cos^2x=1\) vừa chứng minh xong để giải quyết .
Bài 1: Rút gọn:
A= \(\dfrac{sin2\alpha+sin\alpha}{1+cos2\alpha+cos2\alpha}\)
B= \(\dfrac{4sin^2\alpha}{1-cos^2\dfrac{\alpha}{2}}\)
C= \(\dfrac{1+cos\alpha-sin\alpha}{1-cos\alpha-sin\alpha}\)
D = \(\dfrac{\left(\sin\alpha+\cos\alpha\right)^2-\left(\sin\alpha-\cos\alpha\right)^2}{\sin\alpha\cdot\cos\alpha}\)
Giúp mk vs please
`D=(sin^2 alpha + 2sin alpha . cos alpha + cos^2 alpha - sin^2 alpha + 2 sin alpha . cos alpha + cos^2 alpha ) / (sin alpha . cos alpha )`
`D=(4 sin alpha . cos alpha )/(sin alpha . cos alpha )`
`D=4`
`D=(sin^2 alpha + 2sin alpha . cos alpha + cos^2 alpha - (sin^2 alpha - 2 sin alpha . cos alpha + cos^2 alpha )) / (sin alpha . cos alpha )`
`D=(sin^2 alpha + 2sin alpha . cos alpha + cos^2 alpha - sin^2 alpha + 2 sin alpha . cos alpha - cos^2 alpha ) / (sin alpha . cos alpha )`
`D=((sin^2 alpha - sin^2 alpha ) + (2sin alpha . cos alpha + cos^2 alpha + 2 sin alpha . cos alpha) +( cos^2 alpha - cos^2 alpha )) / (sin alpha . cos alpha )`
`D=(4 sin alpha . cos alpha )/(sin alpha . cos alpha )`
`D=4`
Ta có: \(D=\dfrac{\left(\sin\alpha+\cos\alpha\right)^2-\left(\sin\alpha-\cos\alpha\right)^2}{\sin\alpha\cdot\cos\alpha}\)
\(=\dfrac{4\cdot\sin\alpha\cdot\cos\alpha}{\sin\alpha\cdot\cos\alpha}\)
=4