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22 tháng 6 2019

\(\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha=\left(\sin^2\alpha+\cos^2\alpha\right)^2=1\)

22 tháng 6 2019

\(\tan^2\alpha\left(2.\cos^2\alpha+\sin^2\alpha-1\right)=\tan^2\alpha\left(\cos^2\alpha+\left(\sin^2\alpha+\cos^2\alpha\right)-1\right)\)\(=\tan^2\alpha.\cos^2\alpha=\left(\frac{1}{\cos^2\alpha}-1\right)\cos^2\alpha=1-\cos^2\alpha=\sin^2\alpha\)

9 tháng 8 2016

http://olm.vn/thanhvien/kangta

9 tháng 8 2016

a/ sina

b/ sin2 a

c/ 1

d/ sina

12 tháng 10 2018

a) ta có : \(sin\alpha.cos\alpha\left(tan\alpha+cot\alpha\right)=sin\alpha.cos\alpha\left(\dfrac{sin\alpha}{cos\alpha}+\dfrac{cos\alpha}{sin\alpha}\right)\)

\(=sin^2\alpha+cos^2\alpha=1\)

b) ta có : \(\left(sin^2\alpha+cos^2\alpha\right)^2+\left(sin\alpha-cos\alpha\right)^2\)

\(=1^2+1-2sin\alpha.cos=2\left(1-2sin\alpha.cos\alpha\right)\)

c) ta có : \(tan^2\alpha-sin^2\alpha.tan^2\alpha=tan^2\alpha\left(1-sin^2\alpha\right)\)

\(=\dfrac{sin^2\alpha}{cos^2\alpha}.cos^2\alpha=sin^2\alpha\)

a, = \(\sin^2\alpha+2\sin\alpha.\cos\alpha+\cos^2\alpha\)\(\sin^2\alpha-2\sin\alpha\cos\alpha+\cos^2\alpha\)

\(2\sin^2\alpha+2\cos^2\alpha\)= 4

b,=\(\sin\alpha\cos\alpha\)(\(\frac{\sin\alpha}{\cos\alpha}+\frac{\cos\alpha}{\sin\alpha}\))

\(\sin\alpha\cos\alpha.\frac{\sin^2\alpha+\cos^2\alpha}{\sin\alpha\cos\alpha}\)

=1

#mã mã#

21 tháng 6 2017

đáp án :

a) \(cos^2\alpha\)

b) 1

c) \(sin^2\alpha\)

d) \(sin^2\alpha\)

e) 2

g) 1

h) \(sin^3\alpha\)

i) \(sin^2\alpha\)

25 tháng 6 2019

a) \(\left(sin\alpha+cos\alpha\right)^2+\left(sin\alpha-cos\alpha\right)^2\)

\(=sin^2\alpha+2sin\alpha\cdot cos\alpha+cos^2\alpha+sin^2\alpha-2sin\alpha\cdot cos\alpha+cos^2\alpha\)

\(=2\left(sin^2\alpha+cos^2\alpha\right)\)

\(=2\)

b) Vẽ hình minh họa cho dễ nhìn nè :

A B C α

\(sin\alpha\cdot cos\alpha\cdot\left(tan\alpha+cot\alpha\right)\)

\(=\frac{AC}{BC}\cdot\frac{AB}{BC}\cdot\left(\frac{AC}{AB}+\frac{AB}{AC}\right)\)

\(=\frac{AC\cdot AB\cdot AC}{BC\cdot BC\cdot AB}+\frac{AC\cdot AB\cdot AB}{BC\cdot BC\cdot AC}\)

\(=\left(\frac{AC}{BC}\right)^2+\left(\frac{AB}{BC}\right)^2\)

\(=sin^2\text{α}+cos^2\text{α}\)

\(=1\)

15 tháng 7 2019
\(\left(1-\cos\alpha\right)\left(1+\cos\alpha\right)=1-\cos^2\alpha=\left(\sin^2\alpha+\cos^2\alpha\right)-\cos^2\alpha\\ =\sin^2\alpha\)

\(1+\sin^2\alpha+\cos^2\alpha=1+1=2\)

\(\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha\cdot\cos^2\alpha\\ =\left(\sin^2\alpha\right)^2+2\sin^2\alpha\cdot\cos^2\alpha+\left(\cos^2\alpha\right)^2\\ =\left(\sin^2\alpha+\cos^2\alpha\right)^2\\ =1^2=1\)

15 tháng 7 2019

\(\tan^2\alpha-\sin^2\alpha\cdot\tan^2\alpha\\ =\tan^2\alpha\left(1-\sin^2\alpha\right)\\ =\left(\frac{\sin\alpha}{\cos\alpha}\right)^2\cdot\cos^2\alpha\\ =\frac{\sin^2\alpha}{\cos^2\alpha}\cdot\cos^2\alpha\\ =\sin^2\alpha\)

\(\cos^2\alpha+\tan^2\alpha\cdot\cos^2\alpha\\ =\cos^2\alpha+\left(\frac{\sin\alpha}{\cos\alpha}\right)^2\cdot\cos^2\alpha\\ =\cos^2\alpha+\frac{\sin^2\alpha}{\cos^2\alpha}\cdot\cos^2\alpha\\ =\cos^2\alpha+\sin^2\alpha\\ =1\)

\(\tan^2\alpha\cdot\left(2\cos^2\alpha+\sin^2\alpha-1\right)\\ =\tan^2\alpha\cdot\left(2\cos^2\alpha+\sin^2\alpha-\sin^2\alpha-\cos^2\alpha\right)\\ =\tan^2\alpha\cdot\cos^2\alpha\\ =\frac{\sin^2\alpha}{\cos^2\alpha}\cdot\cos^2\alpha=\sin^2\alpha\)

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) \(tan^2\alpha\left(2cos^2\alpha+sin^2\alpha-1\right)\)

= \(tan^2\alpha\left[cos^2\alpha+\left(cos^2\alpha+sin^2\alpha\right)-1\right]\)

= \(tan^2\alpha\left(cos^2\alpha+1-1\right)\)

= \(tan^2\alpha.cos^2\alpha=sin^2\alpha\)