K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

25 tháng 6 2023

Ta có:

`sin^4 \alpha + cos^4 \alpha -sin^6 \alpha- cos^6\alpha`

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

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

`=sin^2\alpha cos^2\alpha(ĐPCM)`

a: \(VT=\dfrac{\left(sina+cosa\right)^3-3\cdot sina\cdot cosa\left(sina+cosa\right)}{sina+cosa}\)

=(sina+cosa)^2-3*sina*cosa

=sin^2a+cos^2a-sina*cosa

=1-sina*cosa=VP

c: VT=(sin^2a+cos^2a)^2-2*sin^2a*cos^2a-(sin^2a+cos^2a)^3+3*sin^2a*cos^2a*(sin^2a+cos^2a)

=1-2sin^2a*cos^2a-1+3*sin^2a*cos^2a

=sin^2a*cos^2a=VP

a: (sina+cosa)^2

=sin^2a+cos^2a+2*sina*cosa

=1+sin2a

b: \(cos^4a-sin^4a=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)\)

\(=cos^2a-sin^2a=cos2a\)

c: 2(sin^6a+cos^6a)+1

=2[(sin^2a+cos^2a)^3-3*sin^2a*cos^2a]+1

=2-6sin^2acos^2a+1

=3-6*sin^2a*cos^2a

=3(sin^4a+cos^4a)

a:

Sửa đề: =-tana*tanb

 \(VT=\left(\dfrac{sina}{cosa}-\dfrac{sinb}{cosb}\right):\left(\dfrac{cosa}{sina}-\dfrac{cosb}{sinb}\right)\)

\(=\dfrac{sina\cdot cosb-sinb\cdot cosa}{cosa\cdot cosb}:\dfrac{cosa\cdot sinb-cosb\cdot sina}{sina\cdot sinb}\)

\(=\dfrac{sin\left(a-b\right)}{cosa\cdot cosb}\cdot\dfrac{sina\cdot sinb}{sin\left(b-a\right)}\)

\(=-tana\cdot tanb\)

=VP

cos^4a-sin^4a

=(cos^2a-sin^2a)(cos^2a+sin^2a)

=cos^2a-sin^2a

=cos2a

25 tháng 6 2023

\(a,cos^4a-sin^4a=2cos^2a-1\\ VT=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)\\ =cos^2a-sin^2a\\ =cos2a=2cos^2a-1\)

\(b,VT=\dfrac{cos^2a+\dfrac{sin^2a}{cos^2a}-1}{sin^2a}\\ =\dfrac{\dfrac{cos^4a+sin^2a-cos^2a}{cos^2a}}{sin^2a}\\ =\dfrac{\dfrac{cos^4a+\left(1-cos^2a\right)-cos^2a}{cos^2a}}{sin^2a}\\ =\dfrac{\dfrac{cos^4a+1-2cos^2a}{cos^2a}}{sin^2a}\\ =\dfrac{\dfrac{\left(1-cos^2a\right)^2}{cos^2a}}{sin^2a}\\ =\dfrac{sin^4a}{cos^2a}:sin^2a\\ =\dfrac{sin^4a}{cos^2a}\times\dfrac{1}{sin^2a}\\ =\dfrac{sin^2a}{cos^2a}=tan^2a\)

a: VT=sin^2a(sin^2a+cos^2a)+cos^2a

=sin^2a+cos^2a

=1=VP

b: \(VT=\dfrac{sina+sina\cdot cosa+sina-sina\cdot cosa}{1-cos^2a}=\dfrac{2sina}{sin^2a}=\dfrac{2}{sina}=VP\)

c: \(VT=\dfrac{sin^2a+1+2cosa+cos^2a}{sina\left(1+cosa\right)}\)

\(=\dfrac{2\left(cosa+1\right)}{sina\left(1+cosa\right)}=\dfrac{2}{sina}=VP\)

28 tháng 2 2018

22 tháng 2 2017

26 tháng 1 2019