chứng minh các biểu thức sau :
a) \(\dfrac{cos\alpha}{1-sin\alpha}=\dfrac{1+sin\alpha}{cos\alpha}\)
b) \(\dfrac{\left(sin\alpha+cos\alpha\right)^2-\left(sin\alpha-cos\alpha\right)^2}{sin\alpha+cos\alpha}\)
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)
\((1+\cot a)\sin ^3a+(1+\tan a)\cos ^3a\)
\(=(\sin ^3a+\cos ^3a)+\cot a.\sin ^3a+\tan a.\cos^3a\)
\(=(\sin a+\cos a)(\sin ^2a-\sin a\cos a+\cos ^2a)+\frac{\cos a}{\sin a}.\sin ^3a+\frac{\sin a}{\cos a}.\cos ^3a\)
\(=(\sin a+\cos a)(1-\sin a\cos a)+\cos a\sin ^2a+\sin a\cos ^2a\)
\(=\sin a+\cos a-\sin a\cos a(\sin a+\cos a)+\cos a\sin a(\sin a+\cos a)\)
\(=\sin a+\cos a\)
Cho góc nhọn α
a) Rút gọn biểu thức S=\(\cos^2\alpha+tg^2.\cos^2\alpha\)
b) Chứng minh:
\(\dfrac{\left(\sin\alpha+\cos\alpha\right)^2-\left(\sin\alpha-\cos\alpha\right)^2}{\sin\alpha.\cos\alpha}=4\)
Help me plsssssssssss
\(\dfrac{\left(sina+cosa\right)^2-\left(sina-cosa\right)^2}{sina.cosa}=4\\ VT=\dfrac{sin^2a+2sinacosa+cos^2a-sin^2a+2sinacosa-cos^2a}{sinacosa}\\ =\dfrac{4sinacosa}{sinacosa}=4=VP\)
a: \(S=cos^2a\left(1+tan^2a\right)=cos^2a\cdot\dfrac{1}{cos^2a}=1\)
b: \(VP=\dfrac{1+sin2a-1+sin2a}{\dfrac{1}{2}\cdot sin2a}=\dfrac{2\cdot sin2a}{\dfrac{1}{2}\cdot sin2a}=4=VT\)
a) S= \(cos^2a\left(tg^2a+1\right)=cos^2a.\dfrac{1}{cos^2a}=1\)
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 $
\(\dfrac{\left(sin\alpha+cos\alpha\right)^2-\left(sin\alpha-cos\alpha\right)^2}{sin\alpha-cos\alpha}=4\)
Hãy chứng minh
Đề sai em
Đề đúng: \(\dfrac{\left(sina+cosa\right)^2-\left(sina-cosa\right)^2}{sina.cosa}=4\)
\(\dfrac{\left(sina+cosa\right)^2-\left(sina-cosa\right)^2}{sina.cosa}=\dfrac{sin^2a+cos^2a+2sina.cosa-\left(sin^2a+cos^2a-2sina.cosa\right)}{sina.cosa}\)
\(=\dfrac{4sina.cosa}{sina.cosa}=4\)
chứng minh các biểu thức sau không phụ thuộc vào α
A=\(\dfrac{\sin^4\alpha+\cos^4\alpha-1}{\sin^6\alpha+\cos^6\alpha+3\cos^4\alpha-1}\)
B=\(\cot^230\left(\sin^8\alpha-\cos^8\alpha\right)+4\cos60\left(\cos^6\alpha-\sin^6\alpha\right)-\sin^6\left(90-\alpha\right)\left(\tan^2-1\right)^3\)
Bạn xem lại biểu thức A. Biểu thức $A$ sau khi rút gọn thì \(A=\frac{-2\sin ^2a}{3\cos 2a}\) vẫn phụ thuộc vào $a$
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Sử dụng công thức: \(\sin (90-a)=\cos a; \cot (90-a)=\tan a\), ta có:
\(B=\tan ^260(\sin ^8a-\cos ^8a)+4\cos 60(\cos ^6a-\sin ^6a)-\cos ^6a(\tan ^2a-1)^3\)
\(=3(\sin ^8a-\cos ^8a)+2(\cos ^6a-\sin ^6a)-\cos ^6a\left(\frac{\sin ^2a}{\cos ^2a}-1\right)^3\)
\(=3(\sin ^8a-\cos ^8a)+2(\cos ^6a-\sin ^6a)-(\sin ^2a-\cos ^2a)^3\)
\(=3(\sin ^2a-\cos ^2a)(\sin ^2a+\cos ^2a)(\sin ^4a+\cos ^4a)+2(\cos ^2a-\sin ^2a)(\cos ^4a+\sin ^2a\cos ^2a+\sin ^4a)-(\sin ^2a-\cos ^2a)^3\)
\(=3(\sin ^2-\cos ^2a)(\sin ^4a+\cos ^4a)-2(\sin ^2a-\cos ^2a)(\cos ^4a+\sin ^2a\cos ^2a+\sin ^4a)-(\sin ^2a-\cos ^2a)^3\)
\(=(\sin ^2a-\cos ^2a)[3(\sin ^4a+\cos ^4a)-2(\cos ^4a+\sin ^2a\cos ^2a+\sin ^4a)-(\sin ^2a-\cos ^2a)^2]\)
\(=(\sin ^2a-\cos ^2a).0=0\). Do đó giá trị của biểu thức không phụ thuộc vào $a$
Rút gọn các biểu thức :
a) \(\dfrac{\sin2\alpha+\sin\alpha}{1+\cos2\alpha+\cos\alpha}\)
b) \(\dfrac{4\sin^2\alpha}{1-\cos^2\dfrac{\alpha}{2}}\)
c) \(\dfrac{1+\cos\alpha-\sin\alpha}{1-\cos\alpha-\sin\alpha}\)
d) \(\dfrac{1+\sin\alpha-2\sin^2\left(45^0-\dfrac{\alpha}{2}\right)}{4\cos\dfrac{\alpha}{2}}\)
a) \(\dfrac{\sin2\text{a}+\cos a}{1+\cos2\text{a}+\cos a}=2\tan a\)
a) \(\dfrac{sin2\alpha+sin\alpha}{1+cos2\alpha+cos\alpha}=\dfrac{2sin\alpha cos\alpha+sin\alpha}{2cos^2\alpha+cos\alpha}\)\(=\dfrac{sin\alpha\left(2cos\alpha+1\right)}{cos\alpha\left(2cos\alpha+1\right)}=\dfrac{sin\alpha}{cos\alpha}=tan\alpha\).
b) \(\dfrac{4sin^2\alpha}{1-cos^2\dfrac{\alpha}{2}}=\dfrac{4sin^2\alpha}{sin^2\dfrac{\alpha}{2}}=\dfrac{4.sin^2\dfrac{\alpha}{2}.cos^2\dfrac{\alpha}{2}}{sin^2\dfrac{\alpha}{2}}=4sin^2\dfrac{\alpha}{2}\).
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=\(\dfrac{cosa}{1+sina}+tga=\dfrac{cosa}{1+sina}+\dfrac{sina}{cosa}=\dfrac{cos^2a+sin^2a+sina}{\left(1+sina\right)cosa}=\dfrac{1+sina}{\left(1+sina\right)cosa}=\dfrac{1}{cosa}=vp\left(dpcm\right)\)
Rút gọn .
\(A=\dfrac{1+2\sin\alpha\cos\alpha}{\sin\alpha+\cos\alpha}\)
\(B=\left(\sin\alpha+\cos\alpha\right)^2-\left(\cos\alpha-\sin\alpha\right)^2\)
\(C=\dfrac{\left(\sin\alpha-\cos\alpha\right)^2-\left(\sin\alpha+\cos\alpha\right)}{\sin\alpha\cos\alpha}\)
Mấy bạn giúp đỡ được phần nào thì giúp , giúp hết thì tốt quá .
\(B=\left(sina+cosa\right)^2-\left(cosa-sina\right)^2=\left(sin^2a+2sinacosa+cos^2a\right)-\left(cos^2a-2cosasina+sin^2a\right)=sin^2a+2sinacosa+cos^2a-cos^2a+2cosasina-sin^2a=4sinacosa\)\(A=\dfrac{1+2sinacosa}{sina+cosa}=\dfrac{sin^2a+cos^2a+2cosasina}{sina+cosa}=\dfrac{\left(sina+cosa\right)^2}{sina+cosa}=sina+cosa\)
C mik bó tay
Rút gọn biểu thức:
a, A = \(\dfrac{4\sin^2\alpha}{1-\cos\dfrac{\alpha}{2}}\)
b, B = \(\dfrac{1+\cos\alpha-\sin\alpha}{1-\cos\alpha-\sin\alpha}\)
c, C = \(\dfrac{1+\sin\alpha-2\sin^2\left(45^o-\dfrac{\pi}{2}\right)}{4\cos\dfrac{\alpha}{2}}\)