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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...
\(A=cos^2x+\dfrac{1+cos\left(\dfrac{2\pi}{3}+2x\right)}{2}+\dfrac{1+cos\left(\dfrac{2\pi}{3}-2x\right)}{2}\\ =cos^2x+1+\dfrac{cos\left(\dfrac{2\pi}{3}+2x\right)+cos\left(\dfrac{2\pi}{3}-2x\right)}{2}\\ =cos^2x+1+cos\left(\dfrac{2\pi}{3}\right).cos2x\\ =cos^2x+1-\dfrac{1}{2}.cos2x=\dfrac{1+cos2x}{2}+1-\dfrac{cos2x}{2}=\dfrac{3}{2}.\)
Sửa đề: sin^4x+cos^4x+1
\(A=\dfrac{\left(sin^2x+cos^2x\right)^3-3sin^2xcos^2x+2}{\left(sin^2x+cos^2x\right)^2-2sin^2xcos^2x+1}\)
\(=\dfrac{3\left(1-sin^2xcos^2x\right)}{2\left(1-sin^2xcos^2x\right)}=\dfrac{3}{2}\)
a) \(A=sin\left(\dfrac{\pi}{4}+x\right)-cos\left(\dfrac{\pi}{4}-x\right)\)
\(\Leftrightarrow A=sin\dfrac{\pi}{4}.cosx+cos\dfrac{\pi}{4}.sinx-\left(cos\dfrac{\pi}{4}.cosx+sin\dfrac{\pi}{4}.sinx\right)\)
\(\Leftrightarrow A=sin\dfrac{\pi}{4}.cosx+cos\dfrac{\pi}{4}.sinx-cos\dfrac{\pi}{4}.cosx-sin\dfrac{\pi}{4}.sinx\)
\(\Leftrightarrow A=\dfrac{\sqrt{2}}{2}.cosx+\dfrac{\sqrt{2}}{2}.sinx-\dfrac{\sqrt{2}}{2}.cosx-\dfrac{\sqrt{2}}{2}.sinx\)
\(\Leftrightarrow A=0\)
b) \(B=cos\left(\dfrac{\pi}{6}-x\right)-sin\left(\dfrac{\pi}{3}+x\right)\)
\(\Leftrightarrow B=cos\dfrac{\pi}{6}.cosx+sin\dfrac{\pi}{6}.sinx-\left(sin\dfrac{\pi}{3}.cosx+cos\dfrac{\pi}{3}.sinx\right)\)
\(\Leftrightarrow B=cos\dfrac{\pi}{6}.cosx+sin\dfrac{\pi}{6}.sinx-sin\dfrac{\pi}{3}.cosx-cos\dfrac{\pi}{3}.sinx\)
\(\Leftrightarrow B=\dfrac{\sqrt{3}}{2}.cosx+\dfrac{1}{2}.sinx-\dfrac{\sqrt{3}}{2}.cosx-\dfrac{1}{2}.sinx\)
\(\Leftrightarrow B=0\)
c) \(C=sin^2x+cos\left(\dfrac{\pi}{3}-x\right).cos\left(\dfrac{\pi}{3}+x\right)\)
\(\Leftrightarrow C=sin^2x+\left(cos\dfrac{\pi}{3}.cosx+sin\dfrac{\pi}{3}.sinx\right).\left(cos\dfrac{\pi}{3}.cosx-sin\dfrac{\pi}{3}.sinx\right)\)
\(\Leftrightarrow C=sin^2x+\left(\dfrac{1}{2}.cosx+\dfrac{\sqrt{3}}{2}.sinx\right).\left(\dfrac{1}{2}.cosx-\dfrac{\sqrt{3}}{2}.sinx\right)\)
\(\Leftrightarrow C=sin^2x+\dfrac{1}{4}.cos^2x-\dfrac{3}{4}.sin^2x\)
\(\Leftrightarrow C=\dfrac{1}{4}.sin^2x+\dfrac{1}{4}.cos^2x\)
\(\Leftrightarrow C=\dfrac{1}{4}\left(sin^2x+cos^2x\right)\)
\(\Leftrightarrow C=\dfrac{1}{4}\)
d) \(D=\dfrac{1-cos2x+sin2x}{1+cos2x+sin2x}.cotx\)
\(\Leftrightarrow D=\dfrac{1-\left(1-2sin^2x\right)+2sinx.cosx}{1+2cos^2a-1+2sinx.cosx}.cotx\)
\(\Leftrightarrow D=\dfrac{2sin^2x+2sinx.cosx}{2cos^2x+2sinx.cosx}.cotx\)
\(\Leftrightarrow D=\dfrac{2sinx\left(sinx+cosx\right)}{2cosx\left(cosx+sinx\right)}.cotx\)
\(\Leftrightarrow D=\dfrac{sinx}{cosx}.cotx\)
\(\Leftrightarrow D=tanx.cotx\)
\(\Leftrightarrow D=1\)
\(P=\sqrt{\left(1-cos^2x\right)^2+6cos^2x+3cos^4x}+\sqrt{\left(1-sin^2x\right)^2+6sin^2x+3sin^4x}\)
\(=\sqrt{4cos^4x+4cos^2x+1}+\sqrt{4sin^4x+4sin^2x+1}\)
\(=\sqrt{\left(2cos^2x+1\right)^2}+\sqrt{\left(2sin^2x+1\right)^2}\)
\(=2cos^2x+1+2sin^2x+1\)
\(=2\left(sin^2x+cos^2x\right)+2=4\)
a: \(A=\sqrt{3}\left(\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx\right)+\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx\)
\(=\dfrac{\sqrt{3}}{2}sinx-\dfrac{3}{2}cosx+\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx\)
\(=\sqrt{3}sinx-cosx\)
c: \(=2\left[\dfrac{\sqrt{3}}{2}sin2x-\dfrac{1}{2}cos2x\right]+4sinx+1\)
\(=\sqrt{3}sin2x-cos2x+4sinx+1\)
d: \(D=\sqrt{3}cos2x+sin2x+2\cdot\left(\dfrac{\sqrt{3}}{2}sin2x-\dfrac{1}{2}cos2x\right)\)
\(=\sqrt{3}\cdot cos2x+sin2x+\sqrt{3}\cdot sin2x-cos2x\)
\(=cos2x\left(\sqrt{3}-1\right)+sin2x\left(1+\sqrt{3}\right)\)
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)
Lời giải:
Đề bài phải thêm đk về x. VD: \(x\in (-\frac{\pi}{2};0)\)
Ta có:
\(\sqrt{4\sin ^4x+\sin ^2(2x)}=\sqrt{4\sin ^4x+(2\sin x\cos x)^2}\)
\(=\sqrt{4\sin ^2x(\sin ^2x+\cos ^2x)}=\sqrt{4\sin ^2x}=|2\sin x|=-2\sin x\) do \(x\in (\frac{-\pi}{2};0)\)
Mặt khác:
\(\cos \left(\frac{\pi}{4}-\frac{x}{2}\right)=\cos \frac{\pi}{4}\cos \frac{x}{2}+\sin \frac{\pi}{4}\sin \frac{x}{2}\)
\(=\frac{\sqrt{2}}{2}\cos \frac{x}{2}+\frac{\sqrt{2}}{2}\sin \frac{x}{2}\)
\(\Rightarrow 4\cos ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)=2(\cos \frac{x}{2}+\sin \frac{x}{2})^2\)
\(=2(\cos ^2\frac{x}{2}+\sin ^2\frac{x}{2}+2\cos \frac{x}{2}\sin \frac{x}{2})\)
\(=2(1+\sin x)=2+2\sin x\)
Do đó: \(A=-2\sin x+2+2\sin x=2\) không phụ thuộc vào x