a) \(\left(sinx+cosx\right)^2=sin^2x+2sinxcosx+cos^2x\)\(=1+2sinxcosx\).
b) \(\left(sinx-cosx\right)^2=sin^2x-2sinxcosx+cos^2x\)\(=1-2sinxcosx\).
c) \(sin^4x+cos^4x=\left(sin^2x+cos^2x\right)^2-2sin^2xcos^2x\)
\(=1-2sin^2xcos^2x\).

a: \(\left(sinx+cosx\right)^2=m^2\)

=>\(m^2=sin^2x+cos^2x+2\cdot sinx\cdot cosx\)

=>\(2\cdot sinx\cdot cosx=m^2-1\)

\(\left(sinx-cosx\right)^2=sin^2x+cos^2x-2\cdot sinx\cdot cosx\)

\(=1-\left(m^2-1\right)=2-m^2\)

\(\left|sin^4x-cos^4x\right|=\left|\left(sin^2x+cos^2x\right)\left(sin^2x-cos^2x\right)\right|\)

\(=\left|sin^2x-cos^2x\right|\)

\(=\left|\left(sinx+cosx\right)\left(sinx-cosx\right)\right|\)

\(=\left|m\left(2-m^2\right)\right|=\left|2m-m^3\right|\)

b: \(m=sinx+cosx\)

\(=\sqrt{2}\cdot\left(sinx\cdot\dfrac{\sqrt{2}}{2}+cosx\cdot\dfrac{\sqrt{2}}{2}\right)\)

\(=\sqrt{2}\cdot sin\left(x+\dfrac{\Omega}{4}\right)\)

=>\(\left|m\right|=\sqrt{2}\cdot\left|sin\left(x+\dfrac{\Omega}{4}\right)\right|\)

\(0< =\left|sin\left(x+\dfrac{\Omega}{4}\right)\right|< =1\)

=>\(0< =\sqrt{2}\cdot\left|sin\left(x+\dfrac{\Omega}{4}\right)\right|< =\sqrt{2}\)

=>\(\left|m\right|< =\sqrt{2}\)

a.

Đặt \(y=\dfrac{2sinx+cosx}{sinx-cosx+3}\)

\(\Leftrightarrow y.sinx-y.cosx+3y=2sinx+cosx\)

\(\Leftrightarrow\left(2-y\right)sinx+\left(y+1\right)cosx=3y\)

Theo điều kiện có nghiệm của pt lượng giác bậc nhất:

\(\left(2-y\right)^2+\left(y+1\right)^2\ge9y^2\)

\(\Leftrightarrow7y^2+2y-5\le0\)

\(\Leftrightarrow-1\le y\le\dfrac{5}{7}\) (đpcm)

b.

Hoàn toàn tương tự câu a:

Đặt \(y=\dfrac{2sinx+cosx+2}{2cosx-sinx+4}\)

\(\Leftrightarrow2y.cosx-y.sinx+4y=2sinx+cosx+2\)

\(\Leftrightarrow\left(y+2\right)sinx+\left(1-2y\right)cosx=4y-2\)

Theo đk có nghiệm pt lượng giác bậc nhất:

\(\left(y+2\right)^2+\left(1-2y\right)^2\ge\left(4y-2\right)^2\)

\(\Leftrightarrow11y^2-16y-1\le0\)

\(\Leftrightarrow\dfrac{8-5\sqrt{3}}{11}\le y\le\dfrac{8+5\sqrt{3}}{11}\)

Đề bài chắc sai, em kiểm tra lại số liệu đề câu b nhé

\(A=\left|\sin^4x-\cos^4x\right|=\left|\left(\sin^2x\right)^2-\left(\cos^2x\right)^2\right|\)

\(A=\left|\left(1-\cos^2x\right)^2-\left(\cos^2x\right)^2\right|=\left|1-2\cos^2x+\cos^4x-\cos^4x\right|\)

\(=\left|1-2\cos^2x\right|=\left|\sin^2x-\cos^2x\right|=\left|\left(\sin x-\cos x\right)\left(\sin x+\cos x\right)\right|\)

\(\sin x+\cos x=m\Rightarrow\cos x=m-\sin x\Rightarrow\sin x-\cos x=\sin x-m+\sin x=2\sin x-m\)

Có \(\sin x+\cos x=m\Rightarrow\sin^2x+\cos^2x+2\sin x.\cos x=m^2\)

\(\Leftrightarrow2\sin x.\cos x=m^2-1\)

\(\left(\sin x-\cos x\right)^2=\sin^2x+\cos^2x-2\sin x.\cos x=1-2.\left(m^2-1\right)=1-2m^2+2=3-2m^2\)

\(\Rightarrow\sin x-\cos x=\sqrt{\left(\sin x-\cos x\right)^2}=\sqrt{3-2m^2}\)

\(A=\left|m\sqrt{3-2m^2}\right|=\left|m\right|.\left|\sqrt{3-2m^2}\right|\)

P/s: lm đc mỗi đến đây thui à, cái CM kia chịu nhoa :)

\(\left(sinx+cosx\right)^2=m^2\Rightarrow1+2sinx.cosx=m^2\)\(\Rightarrow2sinx.cosx=m^2-1\)

\(\Rightarrow\left(sinx-cosx\right)^2=\left(sinx+cosx\right)^2-4sinx.cosx=m^2-2\left(m^2-1\right)=2-m^2\)

Mà \(\left(sinx-cosx\right)^2\ge0\) \(\forall x\Rightarrow2-m^2\ge0\Rightarrow m^2\le2\Rightarrow\left|m\right|\le\sqrt{2}\)

Ta lại có \(\left(sinx-cosx\right)^2=2-m^2\Rightarrow\left|sinx-cosx\right|=\sqrt{2-m^2}\)

\(A=\left|sin^4x-cos^4x\right|=\left|\left(sin^2x+cos^2x\right)\left(sin^2x-cos^2x\right)\right|\)

\(=\left|\left(sinx-cosx\right)\left(sinx+cosx\right)\right|\)

\(=\left|m\sqrt{2-m^2}\right|=\left|m\right|\sqrt{2-m^2}\)

Câu 1 đề sai, chắc chắn 1 trong 2 cái \(cot^2x\) phải có 1 cái là \(cos^2x\)

2.

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

\(=\dfrac{1-\left(sin^2x+cos^2x\right)}{cosx\left(1+sinx\right)}=\dfrac{1-1}{cosx\left(1+sinx\right)}=0\)

3.

\(\dfrac{tanx}{sinx}-\dfrac{sinx}{cotx}=\dfrac{tanx.cotx-sin^2x}{sinx.cotx}=\dfrac{1-sin^2x}{sinx.\dfrac{cosx}{sinx}}=\dfrac{cos^2x}{cosx}=cosx\)

4.

\(\dfrac{tanx}{1-tan^2x}.\dfrac{cot^2x-1}{cotx}=\dfrac{tanx}{1-tan^2x}.\dfrac{\dfrac{1}{tan^2x}-1}{\dfrac{1}{tanx}}=\dfrac{tanx}{1-tan^2x}.\dfrac{1-tan^2x}{tanx}=1\)

5.

\(\dfrac{1+sin^2x}{1-sin^2x}=\dfrac{1+sin^2x}{cos^2x}=\dfrac{1}{cos^2x}+tan^2x=\dfrac{sin^2x+cos^2x}{cos^2x}+tan^2x\)

\(=tan^2x+1+tan^2x=1+2tan^2x\)

1.

Kiểm tra lại đề bài, câu này phải là \(\dfrac{sinx+2cosx+3}{2sinx+cosx+3}\) mới đúng

2.a

ĐKXĐ: \(cosx\ne0\)

\(\Leftrightarrow\dfrac{1}{cos^2x}=4tanx+6\)

\(\Leftrightarrow1+tan^2x=4tanx+6\)

\(\Leftrightarrow tan^2x-4tanx-5=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=arctan\left(5\right)+k\pi\end{matrix}\right.\)

2b.

Đặt \(x-\dfrac{\pi}{4}=t\Rightarrow x=t+\dfrac{\pi}{4}\)

\(sin^3t=\sqrt{2}sin\left(t+\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow sin^3t=sint+cost\)

\(\Leftrightarrow sint\left(1-cos^2t\right)=sint+cost\)

\(\Leftrightarrow sint.cos^2t+cost=0\)

\(\Leftrightarrow cost\left(sint.cost+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cost=0\\sin2t=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=0\\sin\left(2x-\dfrac{\pi}{2}\right)=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=0\\cos2x=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\)

2c.

ĐKXĐ: \(sin4x\ne0\Leftrightarrow x\ne\dfrac{k\pi}{4}\)

\(\dfrac{4sinx.cos2x}{sin4x}+\dfrac{2cos2x}{sin4x}=\dfrac{2}{sin4x}\)

\(\Leftrightarrow2sinx.cos2x+cos2x=1\)

\(\Leftrightarrow2sinx.cos2x+1-2sin^2x=1\)

\(\Leftrightarrow sinx\left(cos2x-sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\left(loại\right)\\cos2x-sinx=0\end{matrix}\right.\)

\(\Leftrightarrow1-2sin^2x-sinx=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\left(loại\right)\\sinx=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{\pi}{6}+k2\pi\)

3.

\(f\left(x+\frac{\pi}{3}\right)=cos\left(x+\frac{\pi}{3}\right)\Rightarrow f'\left(x+\frac{\pi}{3}\right)=-sin\left(x+\frac{\pi}{3}\right)\)

\(f'\left(x-\frac{\pi}{6}\right)=-sin\left(x-\frac{\pi}{6}\right)\)

\(f'\left(0\right)=-sin\left(0\right)=0\)

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

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

\(f'\left(0\right)-f\left(2x+\frac{\pi}{6}\right)=0-cos\left(2x+\frac{\pi}{6}\right)=-cos\left(2x+\frac{\pi}{6}\right)\)

\(\Rightarrow2f'\left(x+\frac{\pi}{3}\right)f'\left(x-\frac{\pi}{6}\right)=f'\left(0\right)-f\left(2x+\frac{\pi}{6}\right)\) (đpcm)

4.

\(y=3\left(sin^4x+cos^4x\right)-2\left(sin^6x+cos^6x\right)\)

\(=3\left(sin^2x+cos^2x\right)^2-6sin^2x.cos^2x-2\left(sin^2x+cos^2x\right)^3+6sin^2x.cos^2x\left(sin^2x+cos^2x\right)\)

\(=3-2=1\)

\(\Rightarrow y'=0\) ; \(\forall x\)

5.

\(y=\left(\frac{sinx}{1+cosx}\right)^3=\left(\frac{sinx\left(1-cosx\right)}{1-cos^2x}\right)^3=\left(\frac{sinx\left(1-cosx\right)}{sin^2x}\right)^3=\left(\frac{1-cosx}{sinx}\right)^3\)

\(y'=3\left(\frac{1-cosx}{sinx}\right)^2\left(\frac{sin^2x-cosx\left(1-cosx\right)}{sin^2x}\right)=3\left(\frac{1-cosx}{sinx}\right)^2\left(\frac{1-cosx}{sin^2x}\right)=\frac{3\left(1-cosx\right)^3}{sin^4x}\)

\(\Rightarrow y'.sinx-3y=\frac{3\left(1-cosx\right)^3}{sin^3x}-3\left(\frac{1-cosx}{sinx}\right)^3=0\) (đpcm)

tôi nói

a) Ta có (\sin x+\cos x)^{2}=\sin ^{2} x+2 \sin x \cos x+\cos ^{2} x=1+2 \sin x \cos x(sinx+cosx)2=sin2x+2sinxcosx+cos2x=1+2sinxcosx (*)
Mặt khác \sin x+\cos x=msinx+cosx=m nên m^{2}=1+2 \sin \alpha \cos \alpham2=1+2sinαcosα hay \sin \alpha \cos \alpha=\dfrac{m^{2}-1}{2}sinαcosα=2m2−1​
Đặt A=\left|\sin ^{4} x-\cos ^{4} x\right|A=∣∣​sin4x−cos4x∣∣​. Ta có
A=\left|\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{2} x-\cos ^{2} x\right)\right|=|(\sin x+\cos x)(\sin x-\cos x)|A=∣∣​(sin2x+cos2x)(sin2x−cos2x)∣∣​=∣(sinx+cosx)(sinx−cosx)∣
\Rightarrow A^{2}=(\sin x+\cos x)^{2}(\sin x-\cos x)^{2}=(1+2 \sin x \cos x)(1-2 \sin x \cos x)⇒A2=(sinx+cosx)2(sinx−cosx)2=(1+2sinxcosx)(1−2sinxcosx)

\Rightarrow A^{2}=\left(1+\dfrac{m^{2}-1}{2}\right)\left(1-\dfrac{m^{2}-1}{2}\right)=\dfrac{3+2 m^{2}-m^{4}}{4}⇒A2=(1+2m2−1​)(1−2m2−1​)=43+2m2−m4​
Vậy A=\dfrac{\sqrt{3+2 m^{2}-m^{4}}}{2}A=23+2m2−m4​​

b) Ta có 2 \sin x \cos x \leq \sin ^{2} x+\cos ^{2} x=12sinxcosx≤sin2x+cos2x=1 kết hợp với (*)(∗) suy ra

(\sin x+\cos x)^{2} \leq 2 \Rightarrow|\sin x+\cos x| \leq \sqrt{2}(sinx+cosx)2≤2⇒∣sinx+cosx∣≤2​

Vậy |m| \leq \sqrt{2}∣m∣≤2​.

​

Chứng minh các biểu thức đã cho không phụ thuộc vào x.

Từ đó suy ra f'(x)=0

a) ;

b) ;

c) (\(\sqrt{2}\)-\(\sqrt{6}\))=>f'(x)=0

d,f(x)=\(\frac{3}{2}\)=>f'(x)=0