bạn viết ddeefc ó sai ddaau k vậy sao cái cuối là +2
Ta có: \(sin^2\alpha+cos^2\alpha=1\Leftrightarrow cos^2\alpha=1-sin^2\alpha=1-\left(\dfrac{1}{2}\right)^2=\dfrac{3}{4}\)
\(P=\sqrt{2}\left(1+cot\alpha\right)\cdot cos\left(\dfrac{\pi}{4}+\alpha\right)\)
\(=\sqrt{2}\left(1+\dfrac{cos\alpha}{sin\alpha}\right)\left(cos\dfrac{\pi}{4}\cdot cos\alpha-sin\dfrac{\pi}{4}\cdot sin\alpha\right)\)
\(=\sqrt{2}\left(1+\dfrac{cos\alpha}{sin\alpha}\right)\left(\dfrac{\sqrt{2}}{2}\cdot cos\alpha-\dfrac{\sqrt{2}}{2}\cdot sin\alpha\right)\)
\(=\left(1+\dfrac{cos\alpha}{sin\alpha}\right)\left(\dfrac{\sqrt{2}\cdot\sqrt{2}}{2}\cdot cos\alpha-\dfrac{\sqrt{2}\cdot\sqrt{2}}{2}\cdot sin\alpha\right)\)
\(=\left(1+\dfrac{cos\alpha}{sin\alpha}\right)\left(cos\alpha-sin\alpha\right)\)
\(=cos\alpha-sin\alpha+cos\alpha\cdot\dfrac{cos\alpha}{sin\alpha}-sin\alpha\cdot\dfrac{cos\alpha}{sin\alpha}\)
\(=cos\alpha-sin\alpha+\dfrac{cos^2\alpha}{sin\alpha}-cos\alpha\)
\(=\dfrac{cos^2\alpha}{sin\alpha}-sin\alpha\)
Thay \(cos^2\alpha=\dfrac{3}{4}\) và \(sin\alpha=\dfrac{1}{2}\) ta có:
\(P=\dfrac{3}{4}\div\dfrac{1}{2}-\dfrac{1}{2}=1\)
