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NH
2
1 tháng 7 2017
... \(=\left(sin^2a\right)^2+2\cdot sin^2a\cdot cos^2+\left(cos^2a\right)^2=\left(sin^2a+cos^2a\right)^2=1^2=1\)
29 tháng 10 2018
a) 1- \(sin^2\alpha\)= \(cos^2\alpha\)
b) (\(1-cos\alpha\))(\(1+cos\alpha\)) = 1 - cos2\(\alpha\) = sin2\(\alpha\)
c) 1 + cos2\(\alpha\) + sin2\(\alpha\) = \(1+1=2\)
d) sin\(\alpha\) - sin\(\alpha.cos^2\alpha\)
= \(sin\alpha\left(1-cos^2\alpha\right)=sin\alpha.sin^2\alpha=sin^3\alpha\)
e) \(sin^4\alpha+cos^4\alpha+2sin^2\alpha.cos^2\alpha\)
= \(\left(sin^2\alpha\right)^2+2sin^2\alpha.cos^2\alpha+\left(cos^2\alpha\right)^2\)
= \(\left(sin^2\alpha+cos^2\alpha\right)^2=1^2=1\)
f) \(tan^2\alpha-sin^2\alpha.tan^2\alpha\)
= \(tan^2\alpha\left(1-sin^2\alpha\right)=tan^2\alpha.cos^2\alpha=sin^2\alpha\)
g) \(cos^2\alpha+tan^2\alpha.cos^2\alpha\)
= \(cos^2\alpha\left(1+tan^2\alpha\right)=cos^2\alpha.\dfrac{1}{cos^2\alpha}=1\)
h) \(tan^2\alpha\left(2cos^2\alpha+sin^2\alpha-1\right)\)
= \(tan^2\alpha\left[cos^2\alpha+\left(cos^2\alpha+sin^2\alpha\right)-1\right]\)
= \(tan^2\alpha\left(cos^2\alpha+1-1\right)\)
= \(tan^2\alpha.cos^2\alpha=sin^2\alpha\)
17 tháng 8 2019
a.\(1-\sin^2\alpha=\cos^2\alpha\)
b.\(\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha=\left(\sin^2\alpha+\cos^2\alpha\right)^2=1\)
c.\(\left(1-\cos\alpha\right)\left(1+\cos\alpha\right)=1-\cos^2\alpha=\sin^2\alpha\)
d.\(1+\sin^2\alpha+\cos^2\alpha=1+1=2\)
e.\(\tan^2\alpha-\sin^2\alpha.\tan^2\alpha=\tan^2\alpha\left(1-\sin^2\alpha\right)=\tan^2\alpha.\cos^2\alpha=\sin^2\alpha\)
g.\(\cos^2\alpha+\cos^2\alpha.\tan^2\alpha=\cos^2\alpha\left(1+\tan^2\alpha\right)=\cos^2\alpha.\frac{1}{\cos^2\alpha}=1\)
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 8 2020
\(cos^4\alpha-sin^4\alpha=\left(cos^2\alpha-sin^2\alpha\right)\left(cos^2\alpha+cos^2a\right)\)
\(=cos^2\alpha-sin^2\alpha=\left(1-sin^2\alpha\right)-sin^2\alpha\)
\(=1-2sin^2\alpha=1-2a\)
AH
Akai Haruma
Giáo viên
22 tháng 7 2018
Lời giải:
Ta có:
\(A=4\sin ^4a\cos ^2a+(\sin ^2a-\cos ^2a)^2+4\cos ^4a\sin ^2a\)
\(=4\sin ^2a\cos ^2a(\sin ^2a+\cos ^2a)+(\sin ^2a-\cos ^2a)^2\)
\(=4\sin ^2a\cos ^2a+(\sin ^2a-\cos ^2a)^2\)
\(=4\sin ^2a\cos ^2a+\sin ^4a+\cos ^4a-2\sin ^2a\cos ^2a\)
\(=2\sin ^2a\cos ^2a+\sin ^4a+\cos ^4a=(\sin ^2a+\cos ^2a)^2\)
\(=1^2=1\)
Vậy biểu thức có giá trị không phụ thuộc vào $a$
B
24 tháng 7 2020
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)
\(C=\left(sin^2a+cos^2a\right)^3-3.sin^2a.cos^2a\left(sin^2a+cos^2a\right)+3sin^2a.cos^2a\)
\(=1-3sin^2a.cos^2a\left(1-1\right)=1\)