A = sin π + x - cos π 2 - x + tan 3 π 2 - x + c o t 2 π - x = - s i n x - sin x + tan π + π 2 - x + c o t - x = - 2 sin x + c o t x - c o t x = - 2 sin x

Chọn B.

A = sin 2 x + sin 3 x + sin 4 x cos 2 x + cos 3 x + cos 4 x = sin 2 x + sin 4 x + sin 3 x cos 2 x + cos 4 x + cos 3 x = 2 sin 3 x . cos x + sin 3 x 2 cos 3 x . cos x + cos 3 x = sin 3 x 2 cos x + 1 cos 3 x 2 cos x + 1 = sin 3 x cos 3 x = tan 3 x

Chọn B.

cos - π 4 . cos 3 π 4 + sin - π 4 . sin 3 π 4 = cos - π 4 - 3 π 4 = cos - π = cosπ = - 1

\(P=sin^2x+3cos^2x=1-cos^2x+3cos^2x=1+2cos^2x=1+2.\left(\dfrac{1}{4}\right)^2=\dfrac{9}{8}\)

Lời giải:

Sử dụng công thức lượng giác:

\(\cos a-\cos b=(-2)\sin \frac{a+b}{2}\sin \frac{a-b}{2}\) ta có:

\(\cos \frac{2\pi}{3}-\cos 2x=-2\sin \left(\frac{\pi}{3}+x\right)\sin \left(\frac{\pi}{3}-x \right)\)

Suy ra:

\(\sin \left(\frac{\pi}{3}+x\right)\sin \left(\frac{\pi}{3}-x \right)=\frac{\cos \frac{2\pi}{3}-\cos 2x}{-2}=\frac{1+2\cos 2x}{4}\)

\(\Rightarrow \text{VT}=4\sin x\sin \left(\frac{\pi}{3}+x\right)\sin \left(\frac{\pi}{3}-x \right)=\sin x(1+2\cos 2x)\)

\(=\sin x(1+\cos 2x+\cos ^2x-\sin ^2x)\)

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

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

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

Do đó ta có đpcm.

a: \(A=2\left(\sin^6x+cos^6x\right)-3\cdot\left(\sin^4x+cos^4x\right)\)

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

\(=2\left\lbrack1-3\cdot sin^2x\cdot cos^2x\right\rbrack-3\cdot\left\lbrack1-2\cdot\sin^2x\cdot cos^2x\right\rbrack\)

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

=2-3

=-1

c: \(C=\frac{\sin^2x}{1+\cot x}+\frac{cos^2x}{1+\tan x}+\sin x\cdot cosx\)

\(=\frac{\sin^2x}{1+\frac{cosx}{\sin x}}+\frac{cos^2x}{1+\frac{\sin x}{cosx}}+\sin x\cdot cosx=\sin^2x:\frac{\sin x+cosx}{\sin x}+cos^2x:\frac{\sin x+cosx}{cosx}+\sin x\cdot cosx\)

\(=\frac{\sin^3x+cos^3x}{\sin x+cosx}+\sin x\cdot cosx\)

\(=\frac{\left(\sin x+cosx\right)\left(\sin^2x-\sin x\cdot cosx+cos^2x\right)}{\sin x+cosx}+\sin x\cdot cosx\)

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

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

d: \(D=\frac{\cot^2x-cos^2x}{cot^2x}+\frac{\sin x\cdot cosx}{\cot x}\)

\(=\left(\frac{cos^2x}{\sin^2x}-cos^2x\right):\frac{cos^2x}{sin^2x}+\frac{\sin x\cdot cosx}{\frac{cosx}{\sin x}}\)

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

\(=\frac{1-\sin^2x}{\sin^2x}\cdot\sin^2x+\sin^2x=1-\sin^2x+\sin^2x=1\)

sin π 12 . sin 7 π 12 = 1 2 cos π 12 - 7 π 12 - cos π 12 + 7 π 12 = 1 2 . cos - π 2 - cos 2 π 3 = 1 2 . 0 - - 1 2 = 1 4 ⇒ sin π 4 . sin π 12 . sin 7 π 12 = 2 2 . 1 4 = 2 8