Cos 2a mà?

\(F=cos\left(\frac{\pi}{4}+a\right)\cdot cos\left(\frac{\pi}{4}-a\right)\)

\(=\frac12\cdot\left\lbrack cos\left(\frac{\pi}{4}+a-\frac{\pi}{4}+a\right)+cos\left(\frac{\pi}{4}+a+\frac{\pi}{4}+a\right)\right\rbrack\)

\(=\frac12\cdot\left\lbrack cos\left(2a\right)+cos\left(\frac{\pi}{2}\right)\right\rbrack=\frac12\cdot cos2a\)

\(G=\sin\left(\frac{\pi}{3}+a\right)\cdot cos\left(\frac{\pi}{3}-a\right)\)

\(=\frac12\cdot\left\lbrack\sin\left(\frac{\pi}{3}+a+\frac{\pi}{3}-a\right)+\sin\left(\frac{\pi}{3}+a-\frac{\pi}{3}+a\right)\right\rbrack\)

\(=\frac12\cdot\left\lbrack\sin\left(\frac23\pi\right)+\sin2a\right\rbrack=\frac12\cdot\left\lbrack\frac12+\sin2a\right\rbrack\)

\(H=cos\left(\frac{\pi}{2}-a\right)\cdot\sin\left(\frac{\pi}{2}+a\right)\)

\(=\frac12\cdot\left\lbrack\sin\left(\frac{\pi}{2}+a+\frac{\pi}{2}-a\right)+\sin\left(\frac{\pi}{2}+a-\frac{\pi}{2}+a\right)\right\rbrack\)

\(=\frac12\cdot\left\lbrack\sin\left(\pi\right)+\sin2a\right\rbrack=\frac12\left\lbrack2\cdot\sin a\cdot cosa\right\rbrack=\sin a\cdot cosa\)

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

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

=0

\(K=cos\left(\frac{\pi}{6}-x\right)-\sin\left(\frac{\pi}{3}+x\right)\)

\(=\sin\left(\frac{\pi}{2}-\frac{\pi}{6}+x\right)-\sin\left(\frac{\pi}{3}+x\right)=\sin\left(\frac{\pi}{3}+x\right)-\sin\left(\frac{\pi}{3}+x\right)\)

=0

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]

Sửa đề: \(\frac{\pi}{2}<\alpha<\pi\)

Ta có: \(\frac{\pi}{2}<\alpha<\pi\)

=>\(cos\alpha<0;\tan a<0;\cot\alpha<0\)

Ta có: \(\sin^2\alpha+cos^2\alpha=1\)

=>\(cos^2\alpha=1-\left(\frac13\right)^2=1-\frac19=\frac89\)

mà \(cos\alpha<0\)

nên \(cos\alpha=-\frac{2\sqrt2}{3}\)

\(\tan\alpha=\frac{\sin\alpha}{cos\alpha}=\frac13:\frac{-2\sqrt2}{3}=\frac{-1}{2\sqrt2}=\frac{-\sqrt2}{4}\)

\(\cot\alpha=\frac{1}{\tan\alpha}=1:\frac{-\sqrt2}{4}=\frac{4}{-\sqrt2}=-2\sqrt2\)

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

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

phương trình j

phương trình nào vậy bn?