Đề bài là: \(sin^2\left(\dfrac{\pi}{8}+a\right)-sin^2\left(\dfrac{\pi}{8}-a\right)-\dfrac{\sqrt{2}}{2}sin2a\) đúng không nhỉ?

\(=\dfrac{1}{2}-\dfrac{1}{2}cos\left(\dfrac{\pi}{4}+2a\right)-\dfrac{1}{2}+\dfrac{1}{2}cos\left(\dfrac{\pi}{4}-2a\right)-\dfrac{\sqrt{2}}{2}sin2a\)

\(=\dfrac{1}{2}\left[cos\left(\dfrac{\pi}{4}-2a\right)-cos\left(\dfrac{\pi}{4}+2a\right)\right]-\dfrac{\sqrt{2}}{2}sin2a\)

\(=sin\left(\dfrac{\pi}{4}\right).sin2a-\dfrac{\sqrt{2}}{2}sin2a=\dfrac{\sqrt{2}}{2}sin2a-\dfrac{\sqrt{2}}{2}sin2a=0\)

a) \(y=\dfrac{4}{sin^22x-1}\)

Xác định khi và chỉ khi

\(sin^22x-1\ne0\)

\(\Leftrightarrow sin^22x\ne1\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x\ne1\\sin2x\ne-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x\ne sin\dfrac{\pi}{2}\\sin2x\ne sin\dfrac{3\pi}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x\ne\dfrac{\pi}{2}+k2\pi\\2x\ne\dfrac{3\pi}{2}+k2\pi\\2x\ne-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ne\dfrac{\pi}{4}+k\pi\\x\ne\dfrac{3\pi}{4}+k\pi\\x\ne-\dfrac{\pi}{4}+k\pi\end{matrix}\right.\) \(\Leftrightarrow x\ne\pm\dfrac{\pi}{4}+k\pi\)

Vậy tập xác định là \(D=R\)\\(\left\{\pm\dfrac{\pi}{4}+k\pi\right\}\)

2:

a: \(y=4+\left(cos^2x-sin^2x\right)+\left(cos^2x+sin^2x\right)\)

\(=4+1+cos2x=cos2x+5\)

-1<=cos2x<=1

=>-1+5<=cos2x+5<=1+5

=>4<=cos2x+5<=6

TGT là T=[4;6]

b: \(y=5-\dfrac{3}{2}\cdot2sinx\cdot cosx=-\dfrac{3}{2}sin2x+5\)

-1<=sin 2x<=1

=>-3/2<=-3/2sin2x<=3/2

=>-3/2+5<=y<=3/2+5

=>7/2<=y<=13/2

=>TGT là T=[7/2;13/2]

c: -1<=sin x<=1

=>-2<=-2sin x<=2

=>3<=-2sinx+5<=7

=>\(\dfrac{4}{3}>=\dfrac{4}{-2sinx+5}>=\dfrac{4}{7}\)

TGT là T=[4/7;4/3]

\(y=\sin^4x+\cos^4x\\ =1-2\sin^2x\cdot\cos^2x\\ =1-\dfrac{1}{2}\sin^22x\\ 0\le\sin^22x\le1\\ \Leftrightarrow\dfrac{1}{2}\le y\le1\\ y_{min}=\dfrac{1}{2}\Leftrightarrow\sin^22x=1\Leftrightarrow x=\dfrac{k\pi}{2}\pm\dfrac{\pi}{4}\\ y_{max}=1\Leftrightarrow\sin^22x=0\Leftrightarrow x=k\pi\)

\(y=3\sin x+4\cos x\\ =5\left(\dfrac{3\sin x}{5}+\dfrac{4\cos x}{5}\right)\\ =5\cos\left(x-a\right),\forall\cos a=\dfrac{4}{5},\sin a=\dfrac{3}{5}\\ -1\le\cos\left(x-a\right)\le1\\ \Leftrightarrow-5\le y\le5\\ y_{min}=-5\Leftrightarrow\cos\left(x-a\right)=-1\\ y_{max}=5\Leftrightarrow\cos\left(x-a\right)=1\)

\(y=sin^4x+cos^4x\)

Ta có: \(0\le sin^4x\le1\)

\(0\le cos^4x\le1\)

\(0\le sin^4x+cos^4x\le2\)

Vây GTNN là 0, GTLN là 2

y=3sinx+4cosx

\(-3\le3sinx\le3\\ -4\le4cosx\le4\\ -7\le3sinx+4cosx\le7\)

Vậy GTNN là -7, GTLN là 7

phương trình j

phương trình nào vậy bn?

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PT gì vậy bn