sin x+cosx=m

=>(sinx+cosx)^2=m^2

=>1+2*cosx*sinx=m^2

=>2*sinx*cosx=m^2-1

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

\(sin^3x+cos^3x=\left(sinx+cosx\right)^3-3\cdot sinx\cdot cosx\cdot\left(sinx+cosx\right)\)\(=m^3-3\cdot\dfrac{m^2-1}{2}\cdot m\)

\(=m^3-\dfrac{3m^3-3m}{2}\)

\(=\dfrac{2m^3-3m^3+3m}{2}=\dfrac{-m^3+3m}{2}\)

Chọn B

\(A=\frac{3sinx-4cosx}{cosx+2sinx}=\frac{\frac{3sinx}{cosx}-4}{1+\frac{2sinx}{cosx}}=\frac{3tanx-4}{1+2tanx}=\frac{3.5-4}{1+2.5}=...\)

\(B=\frac{\frac{sinx}{cos^3x}+\frac{sin^3x}{cos^3x}}{\frac{3cos^3x}{cos^3x}+\frac{cosx}{cos^3x}}=\frac{tanx.\frac{1}{cos^2x}+tan^3x}{3+\frac{1}{cos^2x}}=\frac{tanx\left(1+tan^2x\right)+tan^3x}{3+\left(1+tan^2x\right)}=\frac{5\left(1+5^2\right)+5^3}{3+1+5^2}=...\)

\(sin(\dfrac{\pi}{2}-x)cot(\pi+x)=cosxcotx=\dfrac{cosx}{tanx}\\ =\dfrac{\dfrac{1}{\sqrt5}}{-2}=\dfrac{-\sqrt5}{10}\)

a, (sinx + cosx)(1 - sinx . cosx) = (cosx - sinx)(cosx + sinx)

⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\cosx-sinx=1-sinx.cosx\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\cosx+sinx.cosx-1-sinx=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\\left(cosx-1\right)\left(sinx+1\right)=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\cosx=1\\sinx=-1\end{matrix}\right.\)

b, (sinx + cosx)(1 - sinx . cosx) = 2sin2x + sinx + cosx

⇔ (sinx + cosx)(1 - sinx.cosx - 1) = 2sin2x

⇔ (sinx + cosx).(- sinx . cosx) = 2sin2x

⇔ 4sin2x + (sinx + cosx) . sin2x = 0

⇔ \(\left[{}\begin{matrix}sin2x=0\\\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+4=0\end{matrix}\right.\)

⇔ sin2x = 0

c, 2cos³x = sin3x

⇔ 2cos³x = 3sinx - 4sin³x

⇔ 4sin³x + 2cos³x - 3sinx(sin²x + cos²x) = 0

⇔ sin³x + 2cos³x - 3sinx.cos²x = 0

Xét cosx = 0 : thay vào phương trình ta được sinx = 0. Không có cung x nào có cả cos và sin = 0 nên cosx = 0 không thỏa mãn phương trình

Xét cosx ≠ 0 chia cả 2 vế cho cos³x ta được : 

tan³x + 2 - 3tanx = 0

⇔ \(\left[{}\begin{matrix}tanx=1\\tanx=-2\end{matrix}\right.\)

d, cos²x - \(\sqrt{3}sin2x\) = 1 + sin²x

⇔ cos²x - sin²x - \(\sqrt{3}sin2x\) = 1

⇔ cos2x - \(\sqrt{3}sin2x\) = 1

⇔ \(2cos\left(2x+\dfrac{\pi}{3}\right)=1\)

⇔ \(cos\left(2x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}=cos\dfrac{\pi}{3}\)

e, cos³x + sin³x = 2cos⁵x + 2sin⁵x

⇔ cos³x (1 - 2cos²x) + sin³x (1 - 2sin²x) = 0

⇔ cos³x . (- cos2x) + sin³x . cos2x = 0

⇔ \(\left[{}\begin{matrix}sin^3x=cos^3x\\cos2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=cosx\\cos2x=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\\cos2x=0\end{matrix}\right.\)

a.

ĐKXĐ: \(cosx\ne0\)

Chia 2 vế cho \(cos^2x\) ta được:

\(\left(1+tanx\right).tan^2x=3tanx\left(1-tanx\right)+\frac{3}{cos^2x}\)

\(\Leftrightarrow tan^2x\left(tanx+1\right)=3tanx-3tan^2x+3+3tan^2x\)

\(\Leftrightarrow tan^2x\left(tanx+1\right)-3\left(tanx+1\right)=0\)

\(\Leftrightarrow\left(tan^2x-3\right)\left(tanx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tanx=\sqrt{3}\\tanx=-\sqrt{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{\pi}{3}+k\pi\\x=-\frac{\pi}{3}+k\pi\end{matrix}\right.\)

c/

\(\Leftrightarrow cos^3x=sinx\left(cos\frac{2\pi}{3}+cos2x\right)\)

\(\Leftrightarrow cos^3x=sinx\left(cos2x-\frac{1}{2}\right)\)

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

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

\(\Leftrightarrow2cos^3x=sinx-4sin^3x\)

Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)

\(\Leftrightarrow2=tanx\left(1+tan^2x\right)-4tan^3x\)

\(\Leftrightarrow3tan^3x-tanx+2=0\)

\(\Leftrightarrow\left(tanx+1\right)\left(3tan^2x-3tanx+2\right)=0\)

\(\Leftrightarrow tanx=-1\Rightarrow x=-\frac{\pi}{4}+k\pi\)

d/

\(\Leftrightarrow\left(cos^2x-sin^2x\right)\left(sinx+cosx\right)-4cos^3x\left(sin^2x+cos^2x+2sinx.cosx\right)=0\)

\(\Leftrightarrow\left(cosx-sinx\right)\left(sinx+cosx\right)^2-4cos^3x\left(sinx+cosx\right)^2=0\)

\(\Leftrightarrow\left(cosx-sinx-4cos^3x\right)\left(sinx+cosx\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\left(1\right)\\cosx-sinx-4cos^3x=0\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=0\Leftrightarrow x+\frac{\pi}{4}=k\pi\)

\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)

Xét \(\left(2\right)\), nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)

\(\Leftrightarrow\frac{1}{cos^2x}-tanx.\frac{1}{cos^2x}-4=0\)

\(\Leftrightarrow1+tan^2x-tanx\left(1+tan^2x\right)-4=0\)

\(\Leftrightarrow-tan^3x+tan^2x-tanx-3=0\)

\(\Leftrightarrow\left(tanx+1\right)\left(tan^2x-2tanx+3\right)=0\)

\(\Leftrightarrow tanx=-1\Rightarrow x=-\frac{\pi}{4}+k\pi\)

Bài 1:

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

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

Bài 2:

\(A=\frac{cot^2x-cos^2x}{cot^2x}+\frac{sinx.cosx}{cotx}=\frac{sin^2x\left(cot^2x-cos^2x\right)}{cos^2x}+\frac{sinx.sinx.cosx}{cosx}\)

\(=\frac{cos^2x-cos^2x.sin^2x}{cos^2x}+sin^2x=1-sin^2x+sin^2x=1\)

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

Bài 3:

A đề đúng chứ bạn? Là cos hay cot?

\(B=\frac{\frac{4sinx.cosx}{sin^2x}-\frac{3cos^2x}{sin^2x}}{\frac{1}{sin^2x}+3}=\frac{4cotx-3cot^2x}{1+cot^2x+3}=\frac{4.\left(-6\right)-3.\left(-6\right)^2}{1+\left(-6\right)^2+3}=...\)

Bài 4:

\(0< x< 90^0\Rightarrow cosx>0\)

\(\Rightarrow cosx=\sqrt{1-sin^2x}=\frac{4}{5}\)

\(tanx=\frac{sinx}{cosx}=\frac{3}{4}\) ; \(cotx=\frac{1}{tanx}=\frac{4}{3}\)

\(\Rightarrow A=\frac{\frac{4}{3}}{\frac{4}{3}-\frac{3}{4}}=\frac{16}{7}\)

\(\frac{3\pi}{2}< a< 2\pi\Rightarrow sina< 0\)

\(\Rightarrow sina=-\sqrt{1-cos^2a}=-\frac{\sqrt{3}}{2}\)

\(B=sina+cosa=\frac{1-\sqrt{3}}{2}\)

Bài 5:

\(0< a< \frac{\pi}{4}\Rightarrow0< 2a< \frac{\pi}{2}\Rightarrow sin2a>0\)

\(\Rightarrow sin2a=\sqrt{1-cos^2a}=\frac{\sqrt{21}}{5}\)

\(tan2a=\frac{sin2a}{cos2a}=\frac{\sqrt{21}}{2}\)

\(cot2a=\frac{1}{tan2a}=\frac{2\sqrt{21}}{21}\)

b/ \(-\frac{3\pi}{4}\le a\le-\frac{\pi}{2}\Rightarrow-\frac{3\pi}{2}\le2a\le-\pi\Rightarrow cos2a< 0\)

\(\Rightarrow cos2a=-\sqrt{1-sin^2a}=-\frac{7}{25}\)

\(tan2a=\frac{sin2a}{cos2a}=-\frac{24}{7}\)

\(cot2a=\frac{1}{tan2a}=-\frac{7}{24}\)

Bài 1:

a)

ĐK: $\cos x\neq 0$ $\Rightarrow \sin x\neq -1$. Ta có:
\(\frac{1-\sin x}{\cos x}=\frac{(1-\sin x)(1+\sin x)}{\cos x(1+\sin x)}=\frac{1-\sin ^2x}{\cos x(1+\sin x)}=\frac{\cos ^2x}{\cos x(1+\sin x)}=\frac{\cos x}{1+\sin x}\)

(đpcm)

b) ĐK: $\sin x; \cos x\neq 0$

\(\frac{\tan x}{\sin x}-\frac{\sin x}{\cot x}=\frac{\tan x\cot x-\sin ^2x}{\sin x\cot x}=\frac{1-\sin ^2x}{\sin x.\cot x}=\frac{\cos ^2x}{\sin x.\frac{\cos x}{\sin x}}=\frac{\cos ^2x}{\cos x}=\cos x\) (đpcm)