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\(B=\sin^6\alpha+\cos^6\alpha+3\sin^2\alpha.\cos^2\alpha\)
\(B=\left(\sin^2\alpha\right)^3+\left(\cos^2\alpha\right)^3+3\sin^2\alpha.\cos^2\alpha\)
\(B=\left(\sin^2\alpha+\cos^2\alpha\right)\left(\sin^4\alpha+\cos^4\alpha-\sin^2\alpha.\cos^2\alpha\right)+3\sin^2\alpha.\cos^2\alpha\)
\(B=\sin^4\alpha+\cos^4\alpha-\sin^2\alpha.\cos^2\alpha+3\sin^2\alpha.\cos^2\alpha\)(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
mình ko bt cách viết phân số nên đường gạch ngang mờ mờ mà các bạn nhìn là phân số nhé
Tính:
a) A= cos2 20 độ + cos2 40 độ + cos2 50 độ + cos2 70 độ
b) B= sin4 a + cos4 a + 2sin2 a . cos2 a
A= \(\left(\sin^2a\right)^3+\left(cos^2a\right)^3+3sin^2acos^2a\)
=\(\left(sin^2a+cos^2a\right)\left(sin^4a-cos^2asin^2a+cos^4a\right)+3sin^2acos^2a\)
\(sin^4a+2sin^2acos^2a+cos^4a=\left(sin^2+cos^2\right)^2=1^2=1\)
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)
\(C=\left(sin^2a+cos^2a\right)^3-3.sin^2a.cos^2a\left(sin^2a+cos^2a\right)+3sin^2a.cos^2a\)
\(=1-3sin^2a.cos^2a\left(1-1\right)=1\)
\(a,A=\left(\cos^220^0+\cos^270^0\right)+\left(\cos^240^0+\cos^250^0\right)\\ A=\left(\cos^220^0+\sin^220^0\right)+\left(\cos^240^0+\sin^240^0\right)=1+1=2\\ b,B=\left(\cos^2\alpha\right)^3+\left(\sin^2\alpha\right)^3+3\sin^2\alpha\cdot\cos^2\alpha\cdot\left(\sin^2\alpha+\cos^2\alpha\right)\\ B=\left(\sin^2\alpha+\cos^2\alpha\right)^3=1^3=1\)