![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A^2=\left(\sin\alpha+\cos\alpha\right)^2\le2\left(sin^2\alpha+cos^2\alpha\right)=2\)
\(\Leftrightarrow A\le\sqrt{2}\)dấu bằng xảy ra khi \(\sin\alpha=\cos\alpha\)
\(B=\frac{1}{\sin^2\alpha}+\frac{1}{\cos^2\alpha}\ge\frac{4}{sin^2\alpha+cos^2\alpha}=4\)
dấu bằng xảy ra khi \(sin^2\alpha=cos^2\alpha\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\hept{\begin{cases}3sina+cosa=2\\sin^2a+cos^2a=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}cosa=2-3sina\left(1\right)\\sin^2a+\left(2-3sina\right)^2=1\left(2\right)\end{cases}}\)
\(\left(2\right)\Leftrightarrow10sin^2a-12sina+3=0\)
\(\Leftrightarrow\orbr{\begin{cases}sina=\frac{3}{5}+\frac{\sqrt{6}}{10}\\sina=\frac{3}{5}-\frac{\sqrt{6}}{10}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}cosa=\frac{1}{5}-\frac{3.\sqrt{6}}{10}\left(l\right)\\cosa=\frac{1}{5}+\frac{3.\sqrt{6}}{10}\end{cases}}\)
Thế vô tính tiếp
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\sin^2\alpha=x\Rightarrow\cos^2\alpha=1-\sin^2\alpha\)
\(A=x^3+\left(1-x\right)^3+3x-\left(1-x\right)=x^3+1-3x+3x^2-x^3+3x-1+x=3x^2+x\)
Vậy \(A=3\sin^4\alpha+\sin^2\alpha\). NHỚ NHA!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{\dfrac{3sina}{cosa}-\dfrac{5cosa}{cosa}}{\dfrac{5sina}{cosa}+\dfrac{8cosa}{cosa}}=\dfrac{3tana-5}{5tana+8}=\dfrac{3.\left(\dfrac{5}{7}\right)-5}{5.\left(\dfrac{5}{7}\right)+8}=...\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{\left(sina+cosa\right)^2-\left(sina-cosa\right)^2}{sina.cosa}=4\\ VT=\dfrac{sin^2a+2sinacosa+cos^2a-sin^2a+2sinacosa-cos^2a}{sinacosa}\\ =\dfrac{4sinacosa}{sinacosa}=4=VP\)
a: \(S=cos^2a\left(1+tan^2a\right)=cos^2a\cdot\dfrac{1}{cos^2a}=1\)
b: \(VP=\dfrac{1+sin2a-1+sin2a}{\dfrac{1}{2}\cdot sin2a}=\dfrac{2\cdot sin2a}{\dfrac{1}{2}\cdot sin2a}=4=VT\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=sin^6\alpha+cos^6\alpha+3sin^2\alpha-cos^2\alpha\)
\(=\left(sin^2\alpha\right)^3+\left(cos^2\alpha\right)^3+3sin^2\alpha-cos^2\alpha\)
\(=\left(sin^2\alpha+cos^2\alpha\right)\left(sin^4\alpha+cos^4\alpha-sin^2\alpha.cos^2\alpha\right)+3sin^2\alpha-cos^2\alpha\)
\(=sin^4\alpha+cos^4\alpha-sin^2\alpha.cos^2\alpha+3sin^2\alpha-cos^2\alpha\)
\(=\left(sin^2\alpha+cos^2\alpha\right)^2-2sin^2\alpha.cos^2\alpha-sin^2\alpha.cos^2\alpha+3sin^2\alpha-cos^2\alpha\)
\(1-3sin^2\alpha.cos^2\alpha+3sin^2\alpha-cos^2\alpha=3sin^2\alpha\left(1-cos^2\alpha\right)+\left(1-cos^2\alpha\right)\)
\(=\left(3sin^2\alpha+1\right).sin^2\alpha=0\)