Đáp án A.

Chọn B.

Chọn A.

Ta có:

+ sin⁴x + cos⁴x = (sin²x + cos²x)² - 2sin²x.cos²x = 1 - 2sin²x.cos²x.

+ sin⁴x + cos⁴x = 1 - 3sin²x.cos²x.

Do đó

A = 3(1 - 2sin²x^.cos²x) - 2(1 - 3sin²x.cos²x) = 1.

\(A=3\left[\left(sin^2x+cos^2x\right)^2-2\cdot sin^2x\cdot cos^2x\right]-2\left[\left(sin^2x+cos^2x\right)^3-3\cdot sin^2x\cdot cos^2x\left(sin^2x+cos^2x\right)\right]\)
 

\(=3\left[1-2\cdot sin^2x\cdot cos^2x\right]-2\left[1-3\cdot sin^2x\cdot cos^2x\right]\)

\(=3-6\cdot sin^2x\cdot cos^2x-2+6\cdot sin^2x\cdot cos^2x\)

=1

\(D=\frac{sin4x+sin5x+sin6x}{cos4x+cos5x+cos6x}\)

\(=\frac{\left(sin4x+sin6x\right)+sin5x}{\left(cos4x+cos6x\right)+cos5x}\)

\(=\frac{2sin\frac{4x+6x}{2}.cos\frac{4x-6x}{2}+sin5x}{2cos\frac{4x+6x}{2}.cos\frac{4x-6x}{2}+cos5x}\)

\(=\frac{2sin5x.cos\left(-x\right)+sin5x}{2cos5x.cos\left(-x\right)+cos5x}=\frac{sin5x\left(2.cos\left(-x\right)+1\right)}{cos5x\left(2.cos\left(-x\right)+1\right)}=\frac{sin5x}{cos5x}=tan5x\)

\(P=\dfrac{1+2sin3xcos3x-\left(1-2sin^23x\right)}{1+2sin3xcos3x+2cos^2x-1}=\dfrac{2sin3xcos3x+2sin^23x}{2sin3xcos3x+2cos^23x}=\dfrac{sin3x}{cos3x}=tan3x\)

\(x=\dfrac{7\pi}{4}\Rightarrow P=tan\dfrac{21\pi}{4}=tan\dfrac{\pi}{4}=1\)

Đáp án: C

Ta có:

sin 6 x  +  c o s 6 x  = ( sin 2 x ) 3  + ( cos 2 x ) 3

= ( sin 2 x  +  c o s 2 x )( sin 4 x  -  sin 2 x cos 2 x  +  c o s 4 x )

=  sin 4 x  -  sin 2 x cos 2 x  +  c o s 4 x

= ( sin 2 x  +  cos 2 x ) 2  - 3  sin 2 x cos 2 x

= 1 - 3 sin 2 x cos 2 x

= 1 - (3/4)  sin 2 2 x

Vì

Vậy giá trị nhỏ nhất của  sin 6 x  +  c o s 6 x  là 1/4

Dấu “=” xảy ra ⇔  sin 2 2 x  = 1 ⇔ sin⁡2x = 1 hoặc sin⁡2x = -1