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KT
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
0
5 tháng 7 2019
\(A=\left(\sin\alpha+\cos\alpha+\sin\alpha-\cos\alpha\right)^2-2\left(\sin\alpha+\cos\alpha\right)\left(\sin\alpha-\cos\alpha\right)\)
\(=4\sin^2\alpha-2\sin^2\alpha+2\cos^2\alpha=2\left(\sin^2\alpha+\cos^2\alpha\right)=2\)
\(B=\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right)=\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha\)
\(=\left(\sin^2\alpha+\cos^2\alpha\right)^2-1=0\)
\(C=3\left(\sin^4\alpha+\cos^4\alpha\right)-2\sin^2\alpha.\cos^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right)=3\left(\sin^4\alpha+\cos^4\alpha\right)-2\sin^2\alpha.\cos^2\alpha\)
\(=3\left(\sin^2\alpha+\cos^2\alpha-\frac{1}{9}\right)^2-\frac{1}{9}=\frac{61}{27}\)
NH
2
1 tháng 7 2017
... \(=\left(sin^2a\right)^2+2\cdot sin^2a\cdot cos^2+\left(cos^2a\right)^2=\left(sin^2a+cos^2a\right)^2=1^2=1\)
2 tháng 7 2019
\(=\left(\sin^2\alpha+\cos^2\alpha\right)^2=1\)
D
0
TT
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
0
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) \(tan^2\alpha\left(2cos^2\alpha+sin^2\alpha-1\right)\)
= \(tan^2\alpha\left[cos^2\alpha+\left(cos^2\alpha+sin^2\alpha\right)-1\right]\)
= \(tan^2\alpha\left(cos^2\alpha+1-1\right)\)
= \(tan^2\alpha.cos^2\alpha=sin^2\alpha\)
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)
\(C=\left(sin^2a+cos^2a\right)^3-3.sin^2a.cos^2a\left(sin^2a+cos^2a\right)+3sin^2a.cos^2a\)
\(=1-3sin^2a.cos^2a\left(1-1\right)=1\)
4 tháng 8 2018
a) ta có \(VT=cos^4\alpha-sin^4\alpha=\left(cos^4\alpha+sin^4\alpha\right)-2sin^4\alpha\)
\(=\left(\left(sin^2\alpha+cos^2\alpha\right)^2-2sin^2\alpha.cos^2\alpha\right)-2sin^4\alpha\)
\(=1-2sin^2\alpha.cos^2\alpha-2sin^4\alpha=1-2sin^2\alpha\left(cos^2\alpha+sin^2\alpha\right)\)
\(=1-2sin^2\alpha=VP\left(đpcm\right)\)
b) ta có : \(VP=\dfrac{1}{cos^2\alpha}=\dfrac{sin^2\alpha+cos^2\alpha}{cos^2\alpha}=1+tan^2\alpha=VT\left(đpcm\right)\)
\(cos^4a-sin^4a+2sin^2a\)
\(=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)+2sin^2a\)
\(=cos^2a\left(cos^2a+sin^2a\right)+2sin^2a\)
Bài làm này chắc ổn hơn bài làm trước ✔
\(cos^4a-sin^4a+2sin^2a\)
\(=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)+2sin^2a\)
\(=\left(cos^2a-sin^2a\right)1+2sin^2a\)
\(=cos^2a-sin^2a+2sin^2a\)
\(=cos^2a+sin^2a\)
\(=1\)