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\(=\left(\sin^2\alpha+\cos^2\alpha\right)^2=1\)
rút gọn biểu thức
a. 1 - sin2 2
b. (1+cos2) (1 - cos2)
c. sin4 2 + cos4 2 + 2sin2 2 cos2 2
giúp mình với
... \(=\left(sin^2a\right)^2+2\cdot sin^2a\cdot cos^2+\left(cos^2a\right)^2=\left(sin^2a+cos^2a\right)^2=1^2=1\)
a) \(\frac{1+2sina.cosa}{cos^2a-sin^2a}=\frac{1+sin2a}{cos2a}\)
b) \(B=\left(1+tan^2a\right)\left(1-sin^2a\right)-\left(1+cot^2a\right)\left(1-cos^2a\right)\)
\(=\left(1+\frac{sin^2a}{cos^2a}\right)\left(sin^2a+cos^2a-sin^2a\right)-\left(1+\frac{cos^2a}{sin^2a}\right)\left(cos^2a+sin^2a-cos^2a\right)\)
\(=\left(\frac{cos^2a+sin^2a}{cos^2a}\right).cos^2a-\left(\frac{sin^2a+cos^2a}{sin^2a}\right).sin^2a\)
\(=\frac{1}{cos^2a}.cos^2a-\frac{1}{sin^2a}.sin^2a=1-1=0\)
c)
\(C=\left(sin^2a+cos^2a\right)^3-3.sin^2a.cos^2a\left(sin^2a+cos^2a\right)+3sin^2a.cos^2a\)
\(=1-3sin^2a.cos^2a\left(1-1\right)=1\)
\(B=\sin^6\alpha+\cos^6\alpha+3\sin^2\alpha.\cos^2\alpha\)
\(B=\left(\sin^2\alpha\right)^3+\left(\cos^2\alpha\right)^3+3\sin^2\alpha.\cos^2\alpha\)
\(B=\left(\sin^2\alpha+\cos^2\alpha\right)\left(\sin^4\alpha+\cos^4\alpha-\sin^2\alpha.\cos^2\alpha\right)+3\sin^2\alpha.\cos^2\alpha\)
\(B=\sin^4\alpha+\cos^4\alpha-\sin^2\alpha.\cos^2\alpha+3\sin^2\alpha.\cos^2\alpha\)(vì \(\sin^2\alpha+\cos^2\alpha=1\))
\(B=\left(\sin^2\alpha\right)^2+\left(\cos^2\alpha\right)^2+2.\sin^2\alpha.\cos^2\alpha\)
\(B=\left(\sin^2\alpha+\cos^2\alpha\right)^2=1\)(vì \(\sin^2\alpha+\cos^2\alpha=1\))
Vậy B = 1
A= \(\left(\sin^2a\right)^3+\left(cos^2a\right)^3+3sin^2acos^2a\)
=\(\left(sin^2a+cos^2a\right)\left(sin^4a-cos^2asin^2a+cos^4a\right)+3sin^2acos^2a\)
\(sin^4a+2sin^2acos^2a+cos^4a=\left(sin^2+cos^2\right)^2=1^2=1\)
$\sin^4 a-cos^4 a+2\sin^2 a.\cos^2 a\\=(\sin^4 a-\cos^4 a)+2\sin^2 a.\cos^2 a\\=(\sin^2 a+\cos^2 a)(\sin^2-\cos ^2 )+2\sin^2 a.\cos^2 a\\=\sin^2 a-\cos^2 a+2\sin^2 a.\cos^2 a$