Question

Bài 4: Chứng minh rằng: 

a)  \(sin^4a+cos^4a=1-2sin^2acos^2a\)

b) \(sin^6a+cos^6a=1-3sin^2acos^2a\)

Bài 5: Tính giá trị của

1) A = \(\dfrac{2}{tanx-1}+\dfrac{cotx+1}{cotx-1}\)

2) B = \(2cos^4x-sin^4x+sin^2x.cos^2x+3sin^2x\)

Answer 1

`a, sin^4 alpha + 2 sin^2 alpha cos^2 alpha + cos^4 alpha = 1`

`<=> (sin^2 alpha + cos^2 alpha) ^2 = 1`

`<=> 1^2 = 1`.

`b, sin^6 + cos^6 = 1 - 3 sin^2 cos^2`

`<=> (sin^2 + cos^2)(sin^4 - sin^2 cos^2 + cos^4) = 1 - 3 sin^2 cos ^2`

`<=> sin^4 - sin^2 cos^2 + cos^4 = 1-3 sin^2 cos ^2`

`<=>sin^4 + 2 sin^2 cos^2 + cos^4 = 1`

`<=> (sin^2+cos^2)^2 = 1`

`<=> 1^2 = 1`

Answer 2

Bài 4:

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

\(=1-2\cdot sin^2a\cdot cos^2a\)

b: \(sin^6a+cos^6a=\left(sin^2a+cos^2a\right)^3-3\cdot sin^2a\cdot cos^2a\left(sin^2a+cos^2a\right)\)

\(=1-3\cdot sin^2a\cdot cos^2a\)

Bài 5:

1: \(A=\dfrac{2}{tanx-1}+\dfrac{cotx+1}{cotx-1}\)

\(=\dfrac{2}{tanx-1}+\dfrac{\dfrac{cosx}{sinx}+1}{\dfrac{cosx}{sinx}-1}\)

\(=\dfrac{2}{tanx-1}+\dfrac{cosx+sinx}{cosx-sinx}\)

\(=\dfrac{2}{tanx-1}+\dfrac{\dfrac{cosx}{cosx}+\dfrac{sinx}{cosx}}{\dfrac{cosx}{cosx}-\dfrac{sinx}{cosx}}\)

\(=\dfrac{2}{tanx-1}+\dfrac{1+tanx}{1-tanx}=\dfrac{2}{tanx-1}-\dfrac{tanx+1}{tanx-1}\)

\(=\dfrac{2-tanx-1}{tanx-1}=\dfrac{1-tanx}{tanx-1}=-1\)

2: \(B=2\cdot cos^4x-sin^4x+sin^2x\cdot cos^2x+3\cdot sin^2x\)

\(=cos^4x+\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)+sin^2x\cdot cos^2x+3\cdot sin^2x\)

\(=cos^4x+cos^2x-sin^2x+3\cdot sin^2x+sin^2x\cdot cos^2x\)

\(=cos^4x+cos^2x+2\cdot sin^2x+sin^2x\cdot cos^2x\)

\(=cos^4x+sin^2x+sin^2x\cdot cos^2x+1\)

\(=cos^2x\left(sin^2x+cos^2x\right)+sin^2x+1=1+1=2\)