Question

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Answer 1

Ответ:

$\star \ \ sina+cosa=sina+sin\Big(\dfrac{\pi}{2}-a\Big)=2\cdot sin\dfrac{a+\frac{\pi}{2}-a}{2}\cdot cos\dfrac{a-\frac{\pi}{2}+a}{2}=\\\\\\=2\cdot sin\dfrac{\pi}{4}\cdot cos\Big(a-\dfrac{\pi}{4}\Big)=\sqrt2\cdot cos\Big(a-\dfrac{\pi}{4}\Big)\ \ \star \\\\\\\star \star \ \ sina-cosa=sina-sin\Big(\dfrac{\pi}{2}-a\Big)=2\cdot sin\dfrac{a-\frac{\pi}{2}+a}{2}\cdot cos\dfrac{a+\frac{\pi}{2}-a}{2}=\\\\\\=2\cdot sin\Big(a-\dfrac{\pi}{4}\Big)\cdot cos\dfrac{\pi}{4}=\sqrt2\cdot sin\Big(a-\dfrac{\pi}{4}\Big)\ \ \star$

$\dfrac{\sqrt2-sina-cosa}{sina-cosa}=\dfrac{\sqrt2-(sina+cosa)}{sina-cosa}=\dfrac{\sqrt2-\sqrt 2\cdot cos(a-\frac{\pi}{4})}{\sqrt2\cdot sin(a-\frac{\pi}{4})}=\\\\\\=\dfrac{1-cos(a-\frac{\pi}{4})}{sin(a-\frac{\pi}{4})}=\dfrac{1}{sin(a-\frac{\pi}{4})}-\dfrac{cos(a-\frac{\pi}{4})}{sin(a-\frac{\pi}{4})}=cosec\Big(a-\dfrac{\pi}{4}\Big)-ctg\Big(a-\dfrac{\pi}{4}\Big)\ ;$

$ili:\ \ \ \dfrac{\sqrt2-sina-cosa}{sina-cosa}=\dfrac{1-cos(a-\frac{\pi}{4})}{sin(a-\frac{\pi}{4})}=\dfrac{2sin^2(\frac{a}{2}-\frac{\pi}{8})}{sin(a-\frac{\pi}{4})}\ \ .$