1. Rút gọn biểu thức $P=cos^4x-sin^4x$$A.P=cos2x$ $B.P=\dfrac{3}{4} \dfrac{1}{4}cos4x$ $C.P=\dfrac{1}{4} \dfrac{3}{4}cos4x$ $D.P=\dfrac{3}{4}-\dfrac{1}{4}cos4x$2.Đơn giản biểu thức \(D...

1

LT

Lê Thị Thục Hiền

6 tháng 7 2021

1.Ý A

$P=cos^4x-sin^4x=\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)=cos2x$

2. Ý B

$D=sin\left(\dfrac{5\pi}{2}-\alpha\right)+cos\left(13\pi+\alpha\right)-3sin\left(\alpha-5\pi\right)$

$=sin\left(2\pi+\dfrac{\pi}{2}-\alpha\right)+cos\left(\pi+\alpha+12\pi\right)-3sin\left(\alpha+\pi-6\pi\right)$

$=sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\pi+\alpha\right)-3sin\left(\alpha+\pi\right)$

$=cos\alpha-cos\alpha+3sin\alpha=3sin\alpha$

Chứng minh đẳng thức:1 ,$tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)+cot\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)=\dfrac{2}{cosx}$2 ,$sin^8x-cos^8x=-\left(\dfrac{7}{8}cos2x+\dfrac{1}{8}cos6x\right)$3 ,$3-4cos2x+cos4x=8sin^4x$4 ,$sin\left(2x+\dfrac{\pi}{3}\right).cos\left(x-\dfrac{\pi}{6}\right)-cos\left(2x+\dfrac{\pi}{3}\right).cos\left(\dfrac{2\pi}{3}-x\right)=cosx$5...

TY

Thiên Yết

5 tháng 7 2021

Đọc tiếp

1

LT

Lê Thị Thục Hiền

5 tháng 7 2021

1,$VT=\dfrac{sin\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}{cos\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}+\dfrac{cos\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}{sin\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}$$=\dfrac{sin\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)^2+cos^2\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}{cos\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right).sin\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}$

$=\dfrac{1}{\dfrac{1}{2}.sin\left(\dfrac{\pi}{2}+x\right)}=\dfrac{2}{cosx}=VP$

2,$VT=\left(sin^4x-cos^4x\right)\left(sin^4x+cos^4x\right)=\left(sin^2x+cos^2x\right)\left(sin^2x-cos^2x\right)\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]$

$=\left(sin^2-cos^2x\right)\left(1-2sin^2x.cos^2x\right)$$=-cos2x\left(1-\dfrac{1}{2}sin^22x\right)$$=-\dfrac{cos2x\left(2-sin^22x\right)}{2}=-\dfrac{cos2x\left(1+cos^22x\right)}{2}$

$VP=-\left(\dfrac{7}{8}cos2x+\dfrac{1}{8}cos6x\right)=-\dfrac{7}{8}cos2x-\dfrac{1}{8}\left[4cos^32x-3cos2x\right]=-\dfrac{7}{8}.cos2x-\dfrac{1}{2}cos^32x+\dfrac{3}{8}cos2x$

$=-\dfrac{1}{2}cos2x-\dfrac{1}{2}cos^32x=\dfrac{-cos2x\left(1+cos^22x\right)}{2}$

$\Rightarrow VT=VP$(đpcm)

3, $VT=3-4\left(1-2sin^2x\right)+1-2sin^22x=8sin^2x-2sin^22x=8sin^2x-8.sin^2x.cos^2x=8sin^2x\left(1-cos^2x\right)=8sin^4x=VP$

4,$VP=\dfrac{1}{2}\left[sin\left(x+\dfrac{\pi}{2}\right)+sin\left(3x+\dfrac{\pi}{6}\right)\right]-\dfrac{1}{2}\left[cos\left(3x-\dfrac{\pi}{3}\right)+cos\left(x+\pi\right)\right]$

$=\dfrac{1}{2}\left(cosx+sin3x.\dfrac{\sqrt{3}}{2}+\dfrac{cos3x}{2}\right)-\dfrac{1}{2}\left(\dfrac{cos3x}{2}+sin3x.\dfrac{\sqrt{3}}{2}-cosx\right)$

$=\dfrac{1}{2}.2cosx=cosx=VP$

5, $VP=4cos\left(2x-\dfrac{\pi}{6}\right).\left(sinx.\dfrac{\sqrt{3}}{2}+\dfrac{cosx}{2}\right)^2$$=cos\left(2x-\dfrac{\pi}{6}\right).\left(sinx.\sqrt{3}+cosx\right)^2$

$VT=2.cos\left(2x-\dfrac{\pi}{6}\right)+2.sin\left(2x-\dfrac{\pi}{6}\right).cos\left(2x-\dfrac{\pi}{6}\right)=2cos\left(2x-\dfrac{\pi}{6}\right)\left[1+sin\left(2x-\dfrac{\pi}{6}\right)\right]$

$=2cos\left(2x-\dfrac{\pi}{6}\right)\left(1+\dfrac{sin2x.\sqrt{3}}{2}-\dfrac{cos2x}{2}\right)$$=2cos\left(2x-\dfrac{\pi}{6}\right)\left(sin^2x+cos^2x+sinx.cosx.\sqrt{3}-\dfrac{cos^2x-sin^2x}{2}\right)$

$=2cos\left(2x-\dfrac{\pi}{6}\right)\left(sin^2x.\dfrac{3}{2}+sinx.cosx.\sqrt{3}+\dfrac{cos^2x}{2}\right)$$=cos\left(2x-\dfrac{\pi}{6}\right)\left(sin^2x.3+2sinx.cosx.\sqrt{3}+cos^2x\right)$

$=cos\left(2x-\dfrac{\pi}{6}\right)\left(sinx.\sqrt{3}+cosx\right)^2$

$\Rightarrow VT=VP$ (dpcm)

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MY

missing you =

5 tháng 7 2021

làm mỏi tay khonng chị mà ít tick à =((

LA

Lâm Ánh Yên

6 tháng 8 2021

Chứng minh các đẳng thức:

$cos^3xsinx-sin^3xcosx=\dfrac{1}{4}sin4x$

$sin^4x+cos^4x=\dfrac{1}{4}\left(3+cos4x\right)$

1

NV

Nguyễn Việt Lâm

Giáo viên

6 tháng 8 2021

$cos^3xsinx-sin^3xcosx=sinx.cosx\left(cos^2x-sin^2x\right)=\dfrac{1}{2}sin2x.cos2x=\dfrac{1}{4}sin4x$

$sin^4x+cos^4x=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=1-\dfrac{1}{2}\left(2sinx.cosx\right)^2=1-\dfrac{1}{2}sin^22x$

$=1-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2}cos4x\right)=\dfrac{3}{4}+\dfrac{1}{4}cos4x=\dfrac{1}{4}\left(3+cos4x\right)$

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NM

Nguyễn Minh Anh

23 tháng 4 2017

Đơn giản biểu thức:

1. A=Sinx.Cosx.Cos2x

2. B=Sin⁴x - Cos⁴x

3. C=Sinx.Cos2x.Cos4x.Cos8x.Cos16x

4. D=$\dfrac{Cos4x-Tanx}{Cos2x}$

5. E=sin⁴x-6sin²x.cos²x+cos⁴x

6. F=$\dfrac{Sin2x}{Sinx}-\dfrac{Cos2x}{Cosx}$

0

DN

Dương Nguyễn

3 tháng 7 2021

Giải PT

a1) $3.\cos4x-2^{ }\cos^23x=1$

a2) $2\cos2x-8\cos x+7=\dfrac{1}{\cos x}$

a3) $\dfrac{\left(1+\sin x+\cos2x\right)\sin\left(x+\dfrac{\pi}{4}\right)}{1+\tan x}=\dfrac{1}{\sqrt{2}}\cos x$

a4) $9\sin x+6\cos x-3\sin2x+\cos2x=8$

2

LT

Lê Thị Thục Hiền

3 tháng 7 2021

a) Pt $\Leftrightarrow3.cos4x-\left(cos6x+1\right)=1$

$\Leftrightarrow3cos4x-cos6x-2=0$

Đặt $t=2x$

Pttt:$3cos2t-cos3t-2=0$

$\Leftrightarrow3\left(2cos^2t-1\right)-\left(4cos^3t-3cost\right)-2=0$

$\Leftrightarrow-4cos^3t+6cos^2t+3cost-5=0$

$\Leftrightarrow\left[{}\begin{matrix}cost=1\\cost=\dfrac{1+\sqrt{21}}{4}\left(vn\right)\\cost=\dfrac{1-\sqrt{21}}{4}\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}t=k2\pi\\t=\pm arc.cos\left(\dfrac{1-\sqrt{21}}{4}\right)+k2\pi\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\dfrac{1}{2}.arccos\left(\dfrac{1-\sqrt{21}}{4}\right)+k\pi\end{matrix}\right.$ ($k\in Z$)

Vậy...

a2) $2cos2x-8cosx+7=\dfrac{1}{cosx}$ (ĐK: $x\ne\dfrac{\pi}{2}+k\pi$)

$\Leftrightarrow2.\left(2cos^2x-1\right)-8cosx+7=\dfrac{1}{cosx}$

$\Leftrightarrow2.\left(2cos^2x-1\right)cosx-8cos^2x+7cosx=1$

$\Leftrightarrow4cos^3x-8cos^2x+5cosx-1=0$

$\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=\dfrac{1}{2}\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.$ (tm) ($k\in Z$)

Vậy...

a3) Đk: $x\ne-\dfrac{\pi}{4}+k\pi;x\ne\dfrac{\pi}{2}+k\pi$

Pt $\Leftrightarrow\dfrac{\left(1+sinx+1-2sin^2x\right).\dfrac{1}{\sqrt{2}}\left(sinx+cosx\right)}{1+\dfrac{sinx}{cosx}}=\dfrac{1}{\sqrt{2}}cosx$

$\Leftrightarrow\dfrac{\left(-2sin^2x+sinx+2\right).\left(sinx+cosx\right)cosx}{cosx+sinx}=cosx$

$\Leftrightarrow\left(2+sinx-2sin^2x\right).cosx=cosx$

$\Leftrightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\2+sinx-2sin^2x=1\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\dfrac{1}{2}\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\sinx=-\dfrac{1}{2}\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.$ ($k\in Z$)

Vậy...

LT

Lê Thị Thục Hiền

3 tháng 7 2021

a4) Pt $\Leftrightarrow9sinx+6cosx-6sinx.cosx+1-2sin^2x=8$

$\Leftrightarrow6cosx\left(1-sinx\right)-\left(2sin^2x-9sinx+7\right)=0$

$\Leftrightarrow6cosx\left(1-sinx\right)-\left(2sinx-7\right)\left(sinx-1\right)=0$

$\Leftrightarrow\left(1-sinx\right)\left(6cosx+2sinx+7\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}sinx=1\\6cosx+2sinx=7\left(vn\right)\end{matrix}\right.$ ($6cosx+2sinx=7$ vô nghiệm do $6^2+2^2< 7^2$)

$\Rightarrow sinx=1$

$\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi;k\in Z$

Vậy...

$cosx-2cos3x=1+\sqrt{3}sinx$ $sinx+sinx\left(x+\dfrac{\pi}{3}\right)+sin4x=sin\left(2x-\dfrac{\pi}{3}\right)$ $\left(1-\dfrac{1}{2sinx}\right)cos^22x=2sinx-3+\dfrac{1}{sinx}$ ( sinx -2cosx)cos2x + sinx = (cos4x - 1)cosx...

HM

Huyen My

22 tháng 9 2018

Đọc tiếp

0

NH

Nguyễn Hải Vân

20 tháng 7 2020

Chứng minh các đẳng thứ sau:

$1,sin^8x-cos^8x=-(\dfrac{7}{8}cos2x+\dfrac{1}{8}cos6x) $

2$sin^2x×cos^4x=\dfrac{1}{16}+\dfrac{1}{32}cos2x-\dfrac{1}{16}cos4x-\dfrac{1}{32}cos6x$

2

AH

Akai Haruma

Giáo viên

20 tháng 7 2020

2.

$\text{VP}=\frac{1}{32}(2+\cos 2x-2\cos 4x-\cos 6x)$

$=\frac{1}{32}[2+\cos 2x-2(2\cos ^22x-1)-(4\cos ^32x-3\cos 2x)]$

$=\frac{1}{8}(-\cos ^32x-\cos ^22x+\cos 2x+1)=\frac{1}{8}(\cos 2x+1)(1-\cos ^22x)=\frac{1}{8}(\cos 2x+1)\sin ^22x$ (1)

$\text{VT}=\sin ^2x\cos ^4x=\frac{1}{8}.(2\sin x\cos x)^2.2\cos ^2x=\frac{1}{8}\sin ^22x.(\cos 2x+1)(2)$

Từ $(1);(2)$ ta có đpcm.

AH

Akai Haruma

Giáo viên

20 tháng 7 2020

1.

$\sin ^8x-\cos ^8x=(\sin ^4x+\cos ^4x)(\sin ^4x-\cos ^4x)$

$=[(\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x](\sin ^2x+\cos ^2x)(\sin ^2x-\cos ^2x)$

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

$=(1-\frac{\sin ^22x}{2})(-\cos 2x)=-\frac{(2-\sin ^22x)\cos 2x}{2}=-\frac{(1+\cos ^22x)\cos 2x}{2}$ (1)

$-(\frac{7}{8}\cos 2x+\frac{1}{8}\cos 6x)=\frac{-7}{8}\cos 2x-\frac{1}{8}(4\cos ^32x-3\cos 2x)=-\frac{\cos 2x+\cos ^32x}{2}$

$=\frac{-\cos 2x(\cos ^22x+1)}{2}$ (2)

Từ $(1);(2)$ ta có đpcm.

TT

Thương Thương

12 tháng 7 2021

Giải các phương trình sau:

1) $2\cos4x-3=0$

2) $cos5x+2=0$

3) $cos2x+0,7=0$

4) $cos^22x-\dfrac{1}{4}=0$

2

HP

Hồng Phúc

12 tháng 7 2021

1.

$2cos4x-3=0$

$\Leftrightarrow cos4x=\dfrac{3}{2}$

Mà $cos4x\in\left[-1;1\right]$

$\Rightarrow$ phương trình vô nghiệm.

2.

$cos5x+2=0$

$\Leftrightarrow cos5x=-2$

Mà $cos5x\in\left[-1;1\right]$

$\Rightarrow$ phương trình vô nghiệm.

HP

Hồng Phúc

12 tháng 7 2021

3.

$cos2x+0,7=0$

$\Leftrightarrow cos2x=-\dfrac{7}{10}$

$\Leftrightarrow2x=\pm arccos\left(-\dfrac{7}{10}\right)+k2\pi$

$\Leftrightarrow x=\pm\dfrac{arccos\left(-\dfrac{7}{10}\right)}{2}+k\pi$

4.

$cos^22x-\dfrac{1}{4}=0$

$\Leftrightarrow cos^22x=\dfrac{1}{4}$

$\Leftrightarrow\left[{}\begin{matrix}cos2x=-\dfrac{1}{2}\\cos2x=\dfrac{1}{2}\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}2x=\pm\dfrac{2\pi}{3}+k2\pi\\2x=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{\pi}{3}+k\pi\\x=\pm\dfrac{\pi}{6}+k\pi\end{matrix}\right.$

DT

Đào Trà

23 tháng 8 2021

1) cos3x - cos4x + cos5x =0
2) sin3x + cos2x = 1 + 2sinx.cos2x
3) cos2x - cosx = 2 sin$^2$$\dfrac{3x}{2}$
4) cos$^2$2x + cos$^2$3x = sin$^2$x
5) sin3x.sin5x - cos4x.cos6x = 0

5

HP

Hồng Phúc

23 tháng 8 2021

2.

$sin3x+cos2x=1+2sinx.cos2x$

$\Leftrightarrow sin3x+cos2x=1+sin3x-sinx$

$\Leftrightarrow cos2x+sinx-1=0$

$\Leftrightarrow-2sin^2x+sinx=0$

$\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=\dfrac{1}{2}\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.$

Đúng(2)

HP

Hồng Phúc

23 tháng 8 2021

1.

$cos3x-cos4x+cos5x=0$

$\Leftrightarrow cos3x+cos5x-cos4x=0$

$\Leftrightarrow2cos4x.cosx-cos4x=0$

$\Leftrightarrow\left(2cosx-1\right)cos4x=0$

$\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{1}{2}\\cos4x=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{\pi}{3}+k2\pi\\4x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{\pi}{3}+k2\pi\\x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\end{matrix}\right.$

B ài tập : Giải các phương trình sau: a. 2cos2x +5sinx – 4 = 0 , b. 2cos2x – 8cosx +5 = 0 c. 2cosx.cos2x = 1+cos2x + cos3x d. 2(sin⁴4x + cos4x) = 2sin2x – 1 e.sin⁴2x + cos42x = 1 – 2sin4x f.cos$\dfrac{4x}{3}$ =cos²x

UU

Uyên Uyên

22 tháng 7 2017

1

NT

Nguyễn Thị Thu Thảo

24 tháng 7 2017

a. cos2x = 1-sin²x

b. cos2x = 2cos²x - 1

c. 2cosx.cos2x = 1 + cos2x + cos3x

=> 2cosx.cos2x = 2cos²x + 4cos³x - 3cosx

=> cosx(2.(2cos²x - 1) - 2cosx - 4cos²x +3) = 0

=> cosx( -2cosx + 1) = 0

=> cosx=0 hoặc cosx = -1/2