a/ \(f'\left(x\right)=12sin^33x.cos3x\)

\(f'\left(x\right)=g\left(x\right)\Leftrightarrow12sin^33x.cos3x=sin6x\)

\(\Leftrightarrow6sin^23x.2sin3x.cos3x-sin6x=0\)

\(\Leftrightarrow6sin^23x.sin6x-sin6x=0\)

\(\Leftrightarrow sin6x\left(6sin^23x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}sin6x=0\\sin^23x=\frac{1}{6}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}sin6x=0\\\frac{1-cos6x}{2}=\frac{1}{6}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin6x=0\\cos6x=\frac{2}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}6x=k\pi\\6x=a+k2\pi\\6x=-a+k2\pi\end{matrix}\right.\) với \(cosa=\frac{2}{3}\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{k\pi}{6}\\x=\frac{a}{6}+\frac{k\pi}{3}\\x=-\frac{a}{6}+\frac{k\pi}{3}\end{matrix}\right.\)

b/

\(f'\left(x\right)=6sin^22x.cos2x=4cos2x-5sin4x\)

\(\Leftrightarrow6sin^22x.cos2x=4cos2x-10sin2x.cos2x\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\\3sin^22x=2-5sin2x\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow3sin^22x+5sin2x-2=0\)

\(\Rightarrow\left[{}\begin{matrix}sin2x=\frac{1}{3}\\sin2x=-2< -1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow sin2x=sina\) (với \(sina=\frac{1}{3}\))

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

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

c/

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

\(f'\left(x\right)=g\left(x\right)\)

\(\Leftrightarrow2x\left(1+cosx\right)-x^2sinx=x-x^2sinx\)

\(\Leftrightarrow2x\left(1+cosx\right)=x\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2\left(1+cosx\right)=1\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow cosx=-\frac{1}{2}\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{2\pi}{3}+k2\pi\\x=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)

\(A=cosa\left(sinb.cosc-cosb.sinc\right)+cosb\left(sinc.cosa-cosc.sina\right)+cosc\left(sinacosb-cosasinb\right)\)

\(A=cosasinbcosc-cosacosbsinc+cosacosbsinc-sinacosbcosc+sinacosbcosc-cosasinbcosc\)

\(A=0\)

\(B=sin^2x+\frac{1}{2}\left(cos\frac{2\pi}{3}+cos2x\right)\)

\(B=\frac{1}{2}-\frac{1}{2}cos2x-\frac{1}{4}+\frac{1}{2}cos2x\)

\(B=\frac{1}{4}\)

\(C=\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos\left(\frac{4\pi}{3}+2x\right)+\frac{1}{2}-\frac{1}{2}cos\left(\frac{4\pi}{3}-2x\right)\)

\(C=\frac{3}{2}-\frac{1}{2}cos2x-\frac{1}{2}\left(cos\left(\frac{4\pi}{3}+2x\right)+cos\left(\frac{4\pi}{3}-2x\right)\right)\)

\(C=\frac{3}{2}-\frac{1}{2}cos2x-cos\frac{4\pi}{3}.cos2x\)

\(C=\frac{3}{2}-\frac{1}{2}cos2x+\frac{1}{2}cos2x\)

\(C=\frac{3}{2}\)

\(D=\frac{1}{2}\left[\sqrt{2}sin\left(\frac{\pi}{4}+x\right)\right]^2-sin^2x-sinx.\sqrt{2}cos\left(\frac{\pi}{4}+x\right)\)

\(D=\frac{1}{2}\left(sinx+cosx\right)^2-sin^2x-sinx\left(sinx-cosx\right)\)

\(D=\frac{1}{2}\left(1+2sinx.cosx\right)-sin^2x-sin^2x+sinx.cosx\)

\(D=\frac{1}{2}+sinxcosx+sinxcosx=\frac{1}{2}+sin2x\)

Góc độ cao của thang dựa vào tường là 60º và chân thang cách tường 4,6 m. Chiều dài của thang là

Ta có :

\(\sin \left( {a + \frac{\pi }{4}} \right) + \sin \left( {a - \frac{\pi }{4}} \right) = 2.\sin a.\cos \frac{\pi }{4} =  - \frac{2}{3}\)

Chọn C

đặt AB=c, BC=a, AC=c.
để chứng minh bđt trên ta sẽ áp dụng công thức: \(S_{\Delta ABC}=\frac{1}{2}.a.b.sinC=\frac{1}{2}.b.c.sinA=\frac{1}{2}.a.c.sinB\)
ta có: \(\frac{sinA}{sinB+sinC}+\frac{sinB}{sinA+sinC}+\frac{sinC}{sinA+sinB}\)
       \(=\frac{a.b.c.sinA}{a.b.c.sinB+a.b.c.sinC}+\frac{a.b.c.sinB}{a.b.c.sinA+a.b.c.sinC}+\frac{a.b.c.sinC}{a.b.c.sinA+a.b.c.sinB}\)
        ;\(=\frac{2S_{\Delta ABC}.a}{2S_{\Delta ABC}.b+2S_{\Delta ABC}.c}+\frac{2S_{\Delta ABC}.b}{2.S_{\Delta ABC}.c+2.S_{\Delta ABC}.b}+\frac{2S_{\Delta ABC}.c}{2S_{\Delta ABC}.b+2S_{\Delta ABC}.a}\)
         \(=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\).
Ta có: \(\frac{a}{b+c}>\frac{a}{a+b+c};\frac{b}{a+c}>\frac{b}{a+b+c};\frac{c}{a+b}>\frac{c}{a+b+c}\)
nên \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1.\)
Ta sẽ chứng minh bđt phụ: \(\frac{a}{b+c}< \frac{2a}{a+b+c}\left(1\right)\)
Thật vậy: \(\left(1\right)\Leftrightarrow a^2< a\left(b+c\right)\Leftrightarrow a< b+c\)(đúng vì a,b,c là độ dài 3 cạnh của tam giác).
tương tự: \(\frac{b}{a+c}< \frac{2b}{a+b+c};\frac{c}{a+b}< \frac{2c}{a+b+c}\).
suy ra: \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}< \frac{2a}{b+c}+\frac{2b}{a+c}+\frac{2c}{a+b}=\frac{2\left(a+b+c\right)}{a+b+c}=2\).
vậy bất đẳng thức đã được chứng minh.
 

câu này khó ghê

ô mai nhót . Bài toàn khó thế này mà giải được . Tài thật 

a, Ta có: \({\sin ^2}x + co{s^2}x = 1\)

\(\begin{array}{l} \Leftrightarrow {\sin ^2}\alpha  + {\left( {\frac{1}{3}} \right)^2} = 1\\ \Leftrightarrow \sin \alpha  =  \pm \sqrt {1 - {{\left( {\frac{1}{3}} \right)}^2}}  =  \pm \frac{{2\sqrt 2 }}{3}\end{array}\)

Vì \( - \frac{\pi }{2} < \alpha  < 0\) nên \(sin\alpha  < 0 \Rightarrow \sin \alpha  =  - \frac{{2\sqrt 2 }}{3}\).

\(b)\;\,sin2\alpha  = 2sin\alpha .cos\alpha  = 2.\left( { - \frac{{2\sqrt 2 }}{3}} \right).\frac{1}{3} =  - \frac{{4\sqrt 2 }}{9}\)

\(c)\;cos(\alpha  + \frac{\pi }{3}) = cos\alpha .cos\frac{\pi }{3} - sin\alpha .sin\frac{\pi }{3}\)\( = \frac{1}{3}.\frac{1}{2} - \left( { - \frac{{2\sqrt 2 }}{3}} \right).\frac{{\sqrt 3 }}{2} = \frac{{2\sqrt 6  + 1}}{6}\).