Chứng minh rằng giá trị của biểu thức sau không phụ thuộc vào giá trị của góc nhọn \(\alpha\)
a) A = \(\frac{\cot^2\alpha-\cos^2\alpha}{\cot^2\alpha}-\frac{\sin\alpha.\cos\alpha}{\cot\alpha}\)
b) B = \(\left(\cos\alpha-\sin\alpha\right)^2+\left(\cos\alpha+\sin\alpha\right)^2+\cos^4\alpha-\sin^4\alpha-2\cos^2\alpha\)
c) C = \(\sin^6x+\cos^6x+3\sin^2x.\cos^2x\)
a/ \(A=\frac{cot^2a-cos^2a}{cot^2a}-\frac{sina.cosa}{cota}\)
\(=\frac{\frac{cos^2a}{sin^2a}-cos^2a}{\frac{cos^2a}{sin^2a}}-\frac{sina.cosa}{\frac{cosa}{sina}}\)
\(=\left(1-sin^2a\right)-sin^2a=1\)
b/ \(B=\left(cosa-sina\right)^2+\left(cosa+sina\right)^2+cos^4a-sin^4a-2cos^2a\)
\(=cos^2a-2cosa.sina+sin^2a+cos^2a+2cosa.sina+sin^2a+\left(cos^2a+sin^2a\right)\left(cos^2a-sin^2a\right)-2cos^2a\)
\(=2+\left(cos^2a-sin^2a\right)-2cos^2a\)
\(=2-sin^2a-cos^2a=2-1=1\)
c/ \(C=sin^6x+cos^6x+3sin^2x.cos^2x\)
\(=\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)+3sin^2x.cos^2x\)
\(=sin^4x-sin^2x.cos^2x+cos^4x+3sin^2x.cos^2x\)
\(=sin^4x+cos^4x+2sin^2x.cos^2x\)
\(=\left(sin^2x+cos^2x\right)^2=1\)
Chứng minh rằng giá trị của các biểu thức sau không phụ thuộc vào giá trị
của góc nhọn a
\(\left(\sqrt{\dfrac{1+\sin\alpha}{1-\sin\alpha}}+\sqrt{\dfrac{1-\sin\alpha}{1+\sin\alpha}}\right)\dfrac{1}{\sqrt{1+\tan^2\alpha}}\)
\(\left(\sqrt{\dfrac{1+sin\alpha}{1-sin\alpha}}+\sqrt{\dfrac{1-sin\alpha}{1+sin\alpha}}\right).\dfrac{1}{\sqrt{1+tan^2\alpha}}\)
\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{\left(1-sin\alpha\right)\left(1+sin\alpha\right)}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{\left(1+sin\alpha\right)\left(1-sin\alpha\right)}}\right).\dfrac{1}{\sqrt{1+\left(\dfrac{sin\alpha}{cos\alpha}\right)^2}}\)
\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{1-sin^2\alpha}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{1-sin^2\alpha}}\right).\dfrac{1}{\sqrt{\dfrac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}}}\)
\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{cos^2\alpha}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{cos^2\alpha}}\right).\dfrac{1}{\sqrt{\dfrac{1}{cos^2\alpha}}}\)
\(=\left(\dfrac{1+sin\alpha}{cos\alpha}+\dfrac{1-sin\alpha}{cos\alpha}\right).\dfrac{1}{\dfrac{1}{cos\alpha}}=\dfrac{2}{cos\alpha}.cos\alpha=2\)
Chứng minh các biểu thức sau không phụ thuộc vào \(\alpha\)
\(A=\left(\sin\alpha+\cos\alpha\right)^2+\left(\sin\alpha-\cos\alpha\right)^2\)
\(B=\sin^4\alpha\left(1+2\cos^2\alpha\right)+\cos^4\alpha\left(1+2\sin^2\alpha\right)\)
\(C=\sin^4\alpha\left(3-2\sin^2\alpha\right)+\cos^4\alpha\left(3-2\cos^2\alpha\right)\)
Giúp tớ điii
Chứng minh các biểu thức sau không phụ thuộc vào các góc nhọn \(\alpha\)
a) \(C=\cos^4\alpha+\sin^2\alpha.\cos^2\alpha+\sin^2\alpha\)
b) \(D=\sin^2\alpha.\sin^2\beta+\sin^2\alpha.\cos^2\beta+\cos^2\alpha\)
c) E=\(\sin^6\alpha+\sin^6\beta+3.\sin^2\alpha.\cos^2\alpha\)
d) \(M=\frac{\left(\cos\alpha-\sin\alpha\right)^2-\left(\cos\alpha+\sin\alpha\right)^2}{\cos\alpha.\sin\alpha}\)
Chứng minh rằng biểu thức sau đây không phụ thuộc vào \(\alpha\) :
a) \(A=\left(\sin\alpha+\cos\alpha\right)^2+\left(\sin\alpha-\cos\alpha\right)^2\)
b) \(B=\sin^4\alpha-\cos^4\alpha-2\sin^2\alpha+1\)
a)
\(A=\left(sin\alpha+cos\alpha\right)^2+\left(sin\alpha-cos\alpha\right)^2\)
\(=1+2sin\alpha cos\alpha+1-2sin\alpha cos\alpha=2\) (không phụ thuộc vào \(\alpha\)).
b)
\(B=sin^4\alpha-cos^4\alpha-2sin^2\alpha+1\)
\(=\left(sin^2\alpha+cos^2\alpha\right)\left(sin^2\alpha-cos^2\alpha\right)-2sin^2\alpha+1\)
\(=sin^2\alpha-cos^2\alpha-2sin^2\alpha+1\)
\(=-sin^2\alpha-cos^2\alpha+1\)
\(=-\left(sin^2\alpha+cos^2\alpha\right)+1=-1+1=0\).
2. Chứng minh rằng mỗi biểu thức sau ko phụ thuộc vào biến
A= \(\left(\sin\alpha+\cos\alpha\right)^2-2\sin\alpha.\cos\alpha-1\)
B= \(3\left(\sin^4\alpha+\cos^4\alpha\right)-2\left(\sin^6\alpha+\cos^6\alpha\right)\)
CM các biểu thức sau ko phụ thuộc vào giá trị góc \(\alpha\)(0< \(\alpha\)< 90')
\(A=\sin^6\alpha+\cos^6\alpha+3\sin^2\alpha.\cos^2\alpha
\)
\(B=\left(\cos\alpha-\sin\alpha\right)^2+\left(\cos\alpha+\sin\alpha\right)^2\)
\(C=\frac{\left(\cos\alpha-\sin\alpha\right)^2-\left(\cos\alpha+\sin\alpha\right)^2}{\sin\alpha.\cos\alpha}\)
các nm làm ơn giải giúp mk nhé
mk camon nheii lắm ạ !!!! ^_^ ^_^
\(A=\sin^6\alpha+cos^6\alpha+3\sin^2\alpha\cos^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right).\)vì\(\sin^2\alpha+\cos^2\alpha=1\)
\(=\left(\sin^2\alpha+\cos^2\alpha\right)^3=1^3=1\)
\(B=2\left(\cos^2\alpha+\sin^2\alpha\right)=2.1=2\)
\(C=\frac{-4\cos\alpha\sin\alpha}{\sin\alpha\cos\alpha}=-4\)
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= \(cos^2a\left(tg^2a+1\right)=cos^2a.\dfrac{1}{cos^2a}=1\)
Chứng minh rằng các biểu thức sau là những hằng số không phụ thuộc \(\alpha\) :
a) \(A=2\left(\sin^6\alpha+\cos^6\alpha\right)-3\left(\sin^4\alpha+\cos^4\alpha\right)\)
b) \(B=4\left(\sin^4\alpha+\cos^4\alpha\right)-\cos4\alpha\)
c) \(C=8\left(\cos^8\alpha-\sin^8\alpha\right)-\cos6\alpha-7\cos2\alpha\)
a) \(A=2\left(sin^6\alpha+cos^6\alpha\right)-3\left(sin^4\alpha+cos^4\alpha\right)\)
\(=2\left(sin^2\alpha+cos^2\alpha\right)\left(sin^4\alpha-sin^2\alpha cos^2\alpha+cos^4\alpha\right)\)\(-3\left(sin^4\alpha+cos^4\alpha\right)\)
\(=2\left(sin^4\alpha+cos^4\alpha-sin^2\alpha cos^2\alpha\right)-3\left(sin^4\alpha+cos^4\alpha\right)\)
\(=-\left(sin^4\alpha+cos^4\alpha+2sin^2\alpha cos^2\alpha\right)\)
\(=-\left(sin^2\alpha+cos^2\alpha\right)^2=-1\) (Không phụ thuộc vào \(\alpha\)).
b) \(B=4\left(sin^4\alpha+cos^4\alpha\right)-cos4\alpha\)
\(=4\left(sin^4\alpha+cos^4\alpha+2sin^2\alpha cos^2\alpha\right)-8sin^2\alpha cos^2\alpha\)\(-\left(1-2sin^22\alpha\right)\)
\(=4.\left(sin^2\alpha+cos^2\alpha\right)^2-2sin^22\alpha-1+2sin^22\alpha\)
\(=4-1=3\).
c) \(8\left(cos^8\alpha-sin^8\alpha\right)-cos6\alpha-7cos2\alpha\)
\(=8\left(cos^4\alpha-sin^4\alpha\right)\left(sin^4\alpha+cos^4\alpha\right)-cos6\alpha-7cos2\alpha\)
\(=8\left(cos^2\alpha-sin^2\alpha\right)\left(sin^4\alpha+cos^4\alpha\right)-cos6\alpha-7cos2\alpha\)
\(=8cos2\alpha\left(sin^4\alpha+cos^4\alpha\right)-cos6\alpha-7cos2\alpha\)
\(=8cos2\alpha\left(sin^4\alpha+cos^4\alpha\right)-8cos2\alpha+cos2\alpha-cos6\alpha\)
\(=8cos2\alpha\left(sin^4\alpha+cos^4-1\right)+sin4\alpha sin2\alpha\)
\(=8cos2\alpha\left[\left(sin^4\alpha-sin^2\alpha\right)+\left(cos^4\alpha-cos^2\alpha\right)\right]+\)\(sin4\alpha sin2\alpha\)
\(=8cos2\alpha.\left[sin^2\alpha\left(sin^2\alpha-1\right)+cos^2\alpha\left(cos^2\alpha-1\right)\right]\)\(+sin4\alpha sin2\alpha\)
\(=8.cos2\alpha.\left(-2sin^2\alpha cos^2\alpha\right)+2sin2\alpha cos2\alpha sin2\alpha\)
\(=-2cos2\alpha.\left(sin2\alpha\right)^2+2cos2\alpha.\left(sin2\alpha\right)^2\)
\(=sin^22\alpha\left(-cos2\alpha+cos2\alpha\right)=sin^22\alpha.0=0\) (không phụ thuộc vào \(\alpha\)).
