a) \(\left(sinx+cosx\right)^2=sin^2x+2sinxcosx+cos^2x\)\(=1+2sinxcosx\).
b) \(\left(sinx-cosx\right)^2=sin^2x-2sinxcosx+cos^2x\)\(=1-2sinxcosx\).
c) \(sin^4x+cos^4x=\left(sin^2x+cos^2x\right)^2-2sin^2xcos^2x\)
\(=1-2sin^2xcos^2x\).

\(A=\left|\sin^4x-\cos^4x\right|=\left|\left(\sin^2x\right)^2-\left(\cos^2x\right)^2\right|\)

\(A=\left|\left(1-\cos^2x\right)^2-\left(\cos^2x\right)^2\right|=\left|1-2\cos^2x+\cos^4x-\cos^4x\right|\)

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

\(\sin x+\cos x=m\Rightarrow\cos x=m-\sin x\Rightarrow\sin x-\cos x=\sin x-m+\sin x=2\sin x-m\)

Có \(\sin x+\cos x=m\Rightarrow\sin^2x+\cos^2x+2\sin x.\cos x=m^2\)

\(\Leftrightarrow2\sin x.\cos x=m^2-1\)

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

\(\Rightarrow\sin x-\cos x=\sqrt{\left(\sin x-\cos x\right)^2}=\sqrt{3-2m^2}\)

\(A=\left|m\sqrt{3-2m^2}\right|=\left|m\right|.\left|\sqrt{3-2m^2}\right|\)

P/s: lm đc mỗi đến đây thui à, cái CM kia chịu nhoa :)

\(\left(sinx+cosx\right)^2=m^2\Rightarrow1+2sinx.cosx=m^2\)\(\Rightarrow2sinx.cosx=m^2-1\)

\(\Rightarrow\left(sinx-cosx\right)^2=\left(sinx+cosx\right)^2-4sinx.cosx=m^2-2\left(m^2-1\right)=2-m^2\)

Mà \(\left(sinx-cosx\right)^2\ge0\) \(\forall x\Rightarrow2-m^2\ge0\Rightarrow m^2\le2\Rightarrow\left|m\right|\le\sqrt{2}\)

Ta lại có \(\left(sinx-cosx\right)^2=2-m^2\Rightarrow\left|sinx-cosx\right|=\sqrt{2-m^2}\)

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

\(=\left|\left(sinx-cosx\right)\left(sinx+cosx\right)\right|\)

\(=\left|m\sqrt{2-m^2}\right|=\left|m\right|\sqrt{2-m^2}\)

a: \(\left(sinx+cosx\right)^2=m^2\)

=>\(m^2=sin^2x+cos^2x+2\cdot sinx\cdot cosx\)

=>\(2\cdot sinx\cdot cosx=m^2-1\)

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

\(=1-\left(m^2-1\right)=2-m^2\)

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

\(=\left|sin^2x-cos^2x\right|\)

\(=\left|\left(sinx+cosx\right)\left(sinx-cosx\right)\right|\)

\(=\left|m\left(2-m^2\right)\right|=\left|2m-m^3\right|\)

b: \(m=sinx+cosx\)

\(=\sqrt{2}\cdot\left(sinx\cdot\dfrac{\sqrt{2}}{2}+cosx\cdot\dfrac{\sqrt{2}}{2}\right)\)

\(=\sqrt{2}\cdot sin\left(x+\dfrac{\Omega}{4}\right)\)

=>\(\left|m\right|=\sqrt{2}\cdot\left|sin\left(x+\dfrac{\Omega}{4}\right)\right|\)

\(0< =\left|sin\left(x+\dfrac{\Omega}{4}\right)\right|< =1\)

=>\(0< =\sqrt{2}\cdot\left|sin\left(x+\dfrac{\Omega}{4}\right)\right|< =\sqrt{2}\)

=>\(\left|m\right|< =\sqrt{2}\)

5.

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\frac{5}{6}\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)

\(\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)\)

\(\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)\)

\(\Leftrightarrow\frac{1}{3}sin^22x=\frac{1}{6}\)

\(\Leftrightarrow sin^22x=\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=\frac{\sqrt{2}}{2}\\sin2x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=\frac{3\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\\x=\frac{5\pi}{8}+k\pi\end{matrix}\right.\)

6.

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+\frac{1}{2}sinx.cosx=0\)

\(\Leftrightarrow1-3sin^2x.cos^2x+\frac{1}{2}sinx.cosx=0\)

\(\Leftrightarrow1-\frac{3}{4}sin^22x+\frac{1}{4}sin2x=0\)

\(\Leftrightarrow-3sin^22x+sin2x+4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=-1\\sin2x=\frac{4}{3}>1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow2x=-\frac{\pi}{2}+k2\pi\)

\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)

1.

\(\Rightarrow4cos^3x.cos3x+4sin^3x.sin3x=\sqrt{2}\)

\(\Leftrightarrow\left(3cosx+cos3x\right)cos3x+\left(3sinx-sin3x\right)sin3x=\sqrt{2}\)

\(\Leftrightarrow3\left(cos3x.cosx+sin3x.sinx\right)+cos^23x-sin^23x=\sqrt{2}\)

\(\Leftrightarrow3cos2x+cos6x=\sqrt{2}\)

\(\Leftrightarrow3cos2x+4cos^32x-3cos2x=\sqrt{2}\)

\(\Leftrightarrow4cos^32x=\sqrt{2}\)

\(\Leftrightarrow cos2x=\frac{\sqrt{2}}{2}\)

\(\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{4}+k2\pi\\2x=-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\end{matrix}\right.\)

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

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

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

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

là giá trị không phụ thuộc vào biến (đpcm)

-----------------------

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

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

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

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

\(=-\sin ^2x+\sin ^2x=0\)

là giá trị không phụ thuộc vào biến (đpcm)

\(C=(\sin ^4x+\cos ^4x-1)(\tan ^2x+\cot ^2x+2)=(\sin ^4x+\cos ^4x-1)(\frac{\sin ^2x}{\cos ^2x}+\frac{\cos ^2x}{\sin ^2x}+2)\)

\(=(\sin ^4x+\cos ^4x-1).\frac{\sin ^4x+\cos ^4x+2\sin ^2x\cos ^2x}{\sin ^2x\cos ^2x}=(\sin ^4x+\cos ^4x-1).\frac{(\sin ^2x+\cos ^2x)^2}{\sin ^2x\cos ^2x}\)

\(=(\sin ^4x+\cos ^4x-1).\frac{1}{\sin ^2x\cos ^2x}=\frac{(\sin ^2x)^2+(\cos ^2x)^2+2\sin ^2x\cos ^2x-2\sin ^2x\cos ^2x-1}{\sin ^2x\cos ^2x}\)

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

là giá trị không phụ thuộc vào biến $x$

--------------------

\(D=\frac{1}{\cos ^6x}-\tan ^6x-\frac{\tan ^2x}{\cos ^2x}=\frac{1}{\cos ^6x}-\frac{\sin ^6x}{\cos ^6x}-\frac{\sin ^2x}{\cos ^4x}\)

\(=\frac{1-\sin ^6x-\sin ^2x\cos ^2x}{\cos ^6x}=\frac{(\sin ^2x+\cos ^2x)^3-\sin ^6x-\sin ^2x\cos ^2x}{\cos ^6x}\)

\(=\frac{\sin ^6x+\cos ^6x+3\sin ^2x\cos ^2x(\sin ^2x+\cos ^2x)-\sin ^6x-\sin ^2x\cos ^2x}{\cos ^6x}\)

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

\(=1+\frac{2\sin ^2x}{\cos ^4x}\)

Giá trị biểu thức này vẫn phụ thuộc vào $x$. Bạn xem lại đề.

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

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

\(\frac{sin2a-2sina}{sin2a+2sina}=\frac{2sina.cosa-2sina}{2sina.cosa+2sina}=\frac{2sina\left(cosa-1\right)}{2sina\left(cosa+1\right)}=\frac{cosa-1}{cosa+1}\)

\(=\frac{1-2sin^2\frac{a}{2}-1}{2cos^2\frac{a}{2}-1+1}=\frac{-sin^2\frac{a}{2}}{cos^2\frac{a}{2}}=-tan^2\frac{a}{2}\)

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

\(=\frac{cos^2x\left(1-sin^2x\right)}{sin^2x\left(1-cos^2x\right)}=\frac{cos^4x}{sin^4x}=cot^4x\)

\(\frac{sin^3a-cos^3a}{sina-cosa}=\frac{\left(sina-cosa\right)\left[sin^2a+cos^2a+sina.cosa\right]}{sina-cosa}=1+sina.cosa=1+\frac{1}{2}sin2a\)

\(A=\sqrt{\left(1-cos^2x\right)^2+4cos^2x}+\sqrt{\left(1-sin^2x\right)^2+4sin^2x}\)

\(=\sqrt{cos^4x+2cos^2x+1}+\sqrt{sin^4x+2sin^2x+1}\)

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

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

b/

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

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

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

\(\Rightarrow B=-5sin^6x-3sin^4x.cos^2x+3sin^2x.cos^4x+cos^6x+6sin^4x\)

\(=-5sin^6x-3sin^4x\left(1-sin^2x\right)+3cos^4x\left(1-cos^2x\right)+cos^6x+6sin^4x\)

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

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

\(=-2+3=1\)

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...