`sin^2x+cos^2x=1`

`<=>sin^2x+(1/2)^2=1`

`<=> sinx=\pm \sqrt3/2`

• `sinx=\sqrt3/2 => P=3. (\sqrt3/2)^2 +1=13/4`

• `sinx=-\sqrt3/2 => P = 3.(-\sqrt3/2) +1=13/4`

`=>` A.

\(P=3sin^2x+1=3\left(1-cos^2x\right)+1=3\left(1-\dfrac{1}{4}\right)+1=\dfrac{13}{4}\)

\(sin\left(\dfrac{\pi}{2}-x\right)+1=cosx+1=\dfrac{4}{5}+1=\dfrac{9}{5}\)

\(C=2\left(sin^4x+cos^4x+sin^2xcos^2x\right)^2-\left(sin^8x+cos^8x\right)\)

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

\(=2\left(1-sin^2xcos^2x\right)^2-\left(\left(\left(sin^2x+cos^2x\right)^2-2sin^2xcos^2x\right)^2-2sin^4xcos^4x\right)\)

\(=2\left(1-2sin^2xcos^2x+sin^4xcos^4x\right)-\left(1-4sin^2xcos^2x+4sin^4xcos^4x-2sin^4xcos^4x\right)\)

\(=1\)

co cach giai trac nghiem cau nay nhanh k ak

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

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

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

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

Bạn tham khảo

Giải giúp mình với ạ Giá trị nhỏ nhất của biểu thức P=x+4/x với x >0 là A 8 B 3 C 4 D 2

=>X=4 thay vào nha

Là sao chứ 

Cho biết \(cosx=-\dfrac{1}{2}\)

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

\(\Rightarrow sin^2x=1-\dfrac{1}{4}=\dfrac{3}{4}\)

\(S=4sin^2x+8tan^2x\)

\(\Rightarrow S=4\left(sin^2x+2\dfrac{sin^2x}{cos^2x}\right)\)

\(\Rightarrow S=4\left(\dfrac{3}{4}+2\dfrac{\dfrac{3}{4}}{\dfrac{1}{4}}\right)\)

\(\Rightarrow S=4\left(\dfrac{3}{4}+6\right)\)

\(\Rightarrow S=4.\dfrac{27}{4}=27\)