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NV
10 tháng 4 2021

\(=3\left(sin^4x+cos^4x\right)\left(sin^2x-cos^2x\right)+4cos^6x-8sin^6x+6sin^4x\)

\(=3\left(sin^4x+cos^4x\right)\left(sin^2x-cos^2x\right)+4cos^6x-2sin^6x+6sin^4x\left(1-sin^2x\right)\)

\(=sin^6x+3sin^4x.cos^2x+3cos^2x.sin^4x+cos^6x\)

\(=\left(sin^2x+cos^2x\right)^3=1\)

9 tháng 5 2016

a) P = cos(\(\frac{\Pi}{2}\) + x) + cos(2π - x) + cos(3π + x)   = -sinx + cosx - cosx = -sinx

NV
4 tháng 6 2020

\(A=\frac{2sin^2x-5sinx.cosx+cos^2x}{2sin^2x+sinx.cosx+cos^2x}=\frac{\frac{2sin^2x}{cos^2x}-\frac{5sinx.cosx}{cos^2x}+\frac{cos^2x}{cos^2x}}{\frac{2sin^2x}{cos^2x}+\frac{sinx.cosx}{cos^2x}+\frac{cos^2x}{cos^2x}}\)

\(=\frac{2tan^2x-5tanx+1}{2tan^2x+tanx+1}=\frac{2.3^2-5.3+1}{2.3^2+3+1}=...\)

5 tháng 7 2021

1,\(A=3\left(sin^4x+cos^4x\right)-2\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)\)

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

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

Vậy...

2,\(B=cos^6x+2sin^4x\left(1-sin^2x\right)+3\left(1-cos^2x\right)cos^4x+sin^4x\)

\(=-2cos^6x+3sin^4x-2sin^6x+3cos^4x\)

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

\(=-2\left(sin^4x-sin^2x.cos^2x+cos^4x\right)+3\left(cos^4x+sin^4x\right)\)\(=cos^4x+sin^4x+2sin^2x.cos^2x=1\)

Vậy...

3,\(C=\dfrac{1}{2}\left[cos\left(-\dfrac{7\pi}{12}\right)+cos\left(2x-\dfrac{\pi}{12}\right)\right]+\dfrac{1}{2}\left[cos\left(-\dfrac{7\pi}{12}\right)+cos\left(2x+\dfrac{11\pi}{12}\right)\right]\)

\(=cos\left(-\dfrac{7\pi}{12}\right)+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)+cos\left(2x+\dfrac{11\pi}{12}\right)\right]\)\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)+cos\left(2x-\dfrac{\pi}{12}+\pi\right)\right]\)

\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)-cos\left(2x-\dfrac{\pi}{12}\right)\right]\)\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}\)

Vậy...

4, \(D=cos^2x+\left(-\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx\right)^2+\left(-\dfrac{1}{2}.cosx+\dfrac{\sqrt{3}}{2}.sinx\right)^2\)

\(=cos^2x+\dfrac{1}{4}cos^2x+\dfrac{\sqrt{3}}{4}cosx.sinx+\dfrac{3}{4}sin^2x+\dfrac{1}{4}cos^2x-\dfrac{\sqrt{3}}{4}cosx.sinx+\dfrac{3}{4}sin^2x\)

\(=\dfrac{3}{2}\left(cos^2x+sin^2x\right)=\dfrac{3}{2}\)

Vậy...

5, Xem lại đề

6,\(F=-cosx+cosx-tan\left(\dfrac{\pi}{2}+x\right).cot\left(\pi+\dfrac{\pi}{2}-x\right)\)

\(=tan\left(\pi-\dfrac{\pi}{2}-x\right).cot\left(\dfrac{\pi}{2}-x\right)\)\(=tan\left(\dfrac{\pi}{2}-x\right).cot\left(\dfrac{\pi}{2}-x\right)\)\(=cotx.tanx=1\)

Vậy...

AH
Akai Haruma
Giáo viên
26 tháng 10 2018

a)

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

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

b)

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

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

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

c)

\(\tan ^2x-\sin ^2x=(\frac{\sin x}{\cos x})^2-\sin ^2x\)

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

\(=\sin ^2x\left(\frac{\sin x}{\cos x}\right)^2=\sin ^2x\tan ^2x\)

AH
Akai Haruma
Giáo viên
26 tháng 10 2018

d)

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

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

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

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

\(=1-3\sin ^2x\cos ^2x\) (theo kq phần b)

e)

\(\sin x\cos x(1+\tan x)(1+\cot x)=\sin x\cos x(1+\frac{\sin x}{\cos x})(1+\frac{\cos x}{\sin x})\)

\(=\sin x\cos x.\frac{\cos x+\sin x}{\cos x}.\frac{\sin x+\cos x}{\sin x}\)

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

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

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P/s: Nói chung cứ bám vào công thức \(\sin ^2x+\cos ^2x=1\)

NV
18 tháng 4 2020

\(C=2cosx+cosx-cosx-sinx=2cosx-sinx\)

NV
7 tháng 6 2020

\(A=sin^6x+sin^4x.cos^2x+2\left(sin^2x.cos^4x+sin^4x.cos^2x\right)+cos^4x\)

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

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

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

NV
18 tháng 2 2020

\(P=sin^4x+2sin^2x.cos^2x+cos^4x=\left(sin^2x+cos^2x\right)^2=1\)

Vậy P ko phụ thuộc vào x