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\(y=\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)+3\)
Do \(sin\left(2x+\dfrac{\pi}{4}\right)\le1\Rightarrow y\le3+\sqrt{2}\)
\(\Rightarrow a=3;b=1\Rightarrow a+b=\)
Chứng minh rằng:
\(\left(cos2x-sin2x\right)^2+2\left(sin3x-sinx\right)cosx-1=0\), \(\forall x\in R\)
\(\left(cos2x-sin2x\right)^2+2\left(sin3x-sinx\right).cosx-1\)
\(=2sin^2\left(2x-\frac{\pi}{4}\right)+4cos2x.sinx.cosx-1\)
\(=1-cos\left(4x-\frac{\pi}{2}\right)+2sin2x.cos2x-1\)
\(=-cos\left(\frac{\pi}{2}-4x\right)+sin4x\)
\(=-sin4x+sin4x=0\)
a/ \(sin3x=sin\left(2x+x\right)=sin2xcosx+cos2x.sinx\)
\(=2sinxcos^2x+\left(1-2sin^2x\right)sinx=2sinx\left(1-sin^2x\right)+sinx-2sin^3x\)
\(=3sinx-4sin^3x\)
b/
\(tan2x+\frac{1}{cos2x}=\frac{sin2x}{cos2x}+\frac{1}{cos2x}=\frac{sin2x+1}{cos2x}=\frac{2sinxcosx+sin^2x+cos^2x}{cos^2x-sin^2x}\)
\(=\frac{\left(sinx+cosx\right)^2}{\left(sinx+cosx\right)\left(cosx-sinx\right)}=\frac{sinx+cosx}{cosx-sinx}=\frac{\left(sinx+cosx\right)\left(cosx-sinx\right)}{\left(cos-sinx\right)^2}\)
\(=\frac{cos^2x-sin^2x}{cos^2x+sin^2x-2sinxcosx}=\frac{1-2sin^2x}{1-sin2x}\)
c/
\(\frac{cosx+sinx}{cosx-sinx}-\frac{cosx-sinx}{cosx+sinx}=\frac{\left(cosx+sinx\right)^2-\left(cosx-sinx\right)^2}{cos^2x-sin^2x}\)
\(=\frac{2sinxcosx+2sinxcosx}{cos2x}=\frac{4sinxcosx}{cos2x}=\frac{2sin2x}{cos2x}=2tan2x\)
d/
\(\frac{sin2x}{1+cos2x}=\frac{2sinxcosx}{1+2cos^2x-1}=\frac{2sinxcosx}{2cos^2x}=\frac{sinx}{cosx}=tanx\)
e/
\(sin^8x-cos^8x-4sin^6x+6sin^4x-4sin^2x\)
\(=sin^8x-\left(1-sin^2x\right)^4-4sin^6x+6sin^4x-4sin^2x\)
\(=sin^8x-\left(1-4sin^2x+6sin^4x-4sin^6x+sin^8x\right)-4sin^6x+6sin^4x-4sin^2x\)\(=-1\) (bạn chép nhầm đề)
b/ \(\frac{sin6x+sin2x+sin4x}{1+cos2x+cos4x}=\frac{2sin4x.cos2x+sin4x}{1+cos2x+2cos^22x-1}=\frac{sin4x\left(2cos2x+1\right)}{cos2x\left(2cos2x+1\right)}=\frac{sin4x}{cos2x}=\frac{2sin2x.cos2x}{cos2x}=2sin2x\)
c/ \(\frac{1+sin2x}{cosx+sinx}-\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}=\frac{sin^2x+cos^2x+2sinx.cosx}{cosx+sinx}-\left(1-tan^2\frac{x}{2}\right)cos^2\frac{x}{2}\)
\(=\frac{\left(sinx+cosx\right)^2}{sinx+cosx}-\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)=sinx+cosx-cosx=sinx\)
d/ \(cos4x+4cos2x+3=2cos^22x-1+4cos2x+3\)
\(=2\left(cos^22x+2cos2x+1\right)=2\left(cos2x+1\right)^2=2\left(2cos^2x-1+1\right)^2=8cos^4x\)
e/
3/
\(\frac{sin2x-sinx}{1-cosx+cos2x}=\frac{2sinxcosx-sinx}{1-cosx+2cos^2x-1}=\frac{sinx\left(2cosx-1\right)}{cosx\left(2cosx-1\right)}=\frac{sinx}{cosx}=tanx\)
4/
\(\left(\frac{sinx+cotx}{1+sinx.tanx}\right)^{2014}=\left(\frac{sinx+\frac{1}{tanx}}{1+sinxtanx}\right)^{2014}=\left(\frac{sinxtanx+1}{tanx\left(sinxtanx+1\right)}\right)^{2014}\)
\(=\left(\frac{1}{tanx}\right)^{2014}=cot^{2014}x\)
\(\frac{sin^{2014}x+cot^{2014}x}{1+\left(sinx.tanx\right)^{2014}}=\frac{sin^{2014}x+\frac{1}{tan^{2014}x}}{1+\left(sinx.tanx\right)^{2014}}=\frac{\left(sinxtanx\right)^{2014}+1}{tan^{2014}x\left[\left(sinxtanx\right)^{2014}+1\right]}\)
\(=\frac{1}{tan^{2014}x}=\left(\frac{1}{tanx}\right)^{2014}=cot^{2014}x\)
\(\Rightarrow\left(\frac{sinx+cotx}{1+sinx.tanx}\right)^{2014}=\frac{sin^{2014}x+cot^{2014}x}{1+\left(sinx.tanx\right)^{2014}}\)
\(A=\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)\Rightarrow-\sqrt{2}\le A\le\sqrt{2}\)
B ko rõ đề
\(C=\sqrt{a^2+b^2}\left(\dfrac{a}{\sqrt{a^2+b^2}}sinx-\dfrac{b}{\sqrt{a^2+b^2}}cosx\right)\)
Đặt \(\dfrac{a}{\sqrt{a^2+b^2}}=cosy\Rightarrow\dfrac{b}{\sqrt{a^2+b^2}}=siny\)
\(\Rightarrow C=\sqrt{a^2+b^2}\left(sinx.cosy-cosx.siny\right)=\sqrt{a^2+b^2}sin\left(x-y\right)\)
\(\Rightarrow-\sqrt{a^2+b^2}\le C\le\sqrt{a^2+b^2}\)
\(D=\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)=sin^2x-cos^2x=-cos2x\)
\(\Rightarrow-1\le D\le1\)
Tính tổng các giá trị của m trên đoạn \(\left[-\dfrac{\pi}{3};\dfrac{\pi}{2}\right]\) có nghĩa là \(x\in\left[-\dfrac{\pi}{3};\dfrac{\pi}{2}\right]\) pk?
\(\Rightarrow cosx\in\left[0;1\right]\)
\(y=2cos^2x+cosx-1+\left|2m-1\right|\)
Đặt \(t=cosx;t\in\left[0;1\right]\)
\(y=2t^2+t-1+\left|2m-1\right|\)
Xét BBT của \(f\left(t\right)=2t^2+t-1;t\in\left[0;1\right]\)
\(\Rightarrow f\left(t\right)_{min}=-1\Leftrightarrow t=0\Leftrightarrow cosx=0\)\(\Leftrightarrow x=\dfrac{\pi}{2}\)
\(\Rightarrow y\ge-1+\left|2m-1\right|\)
Để \(y_{min}=2\Leftrightarrow-1+\left|2m-1\right|=2\)\(\Leftrightarrow m=2;m=-1\)
\(\Rightarrow\)Tổng m bằng \(1\)