Question

все на фото
хотя бы одно.
не понимаю как это решить❗

Answer 1

Пользуемся формулами приведения , формулами суммы и разности тригонометрических функций .

$\boxed {\ sin(\alpha \pm \beta )=sin\alpha \, cos\beta \pm cos\alpha \, sin\beta \ }boxed{\ cos(\alpha \pm \beta )=cos\alpha \, cos\beta \mp sin\alpha \, sin\beta \ }$

$\displaystyle 1)\ \ \frac{sin\dfrac{3\pi}{20}\cdot cos\dfrac{21\pi }{10}+cos\dfrac{3\pi}{20}\cdot sin\dfrac{\pi}{10} }{cos\dfrac{7\pi}{24}\cdot cos\dfrac{\pi}{8}+sin\dfrac{7\pi}{8}\cdot sin\dfrac{7\pi }{24}}=\frac{sin\dfrac{3\pi}{20}\cdot cos\Big(2\pi +\dfrac{\pi }{10}\Big)+cos\dfrac{3\pi}{20}\cdot sin\dfrac{\pi}{10} }{cos\dfrac{7\pi}{24}\cdot cos\dfrac{\pi}{8}+sin\Big(\pi -\dfrac{\pi}{8}\Big)\cdot sin\dfrac{7\pi }{24}}=$

$= \dfrac{sin\dfrac{3\pi}{20}\cdot cos\dfrac{\pi }{10}+cos\dfrac{3\pi}{20}\cdot sin\dfrac{\pi}{10} }{cos\dfrac{7\pi}{24}\cdot cos\dfrac{\pi}{8}+sin\dfrac{\pi}{8}\cdot sin\dfrac{7\pi }{24}}=\dfrac{sin\Big(\dfrac{3\pi}{20}+\dfrac{\pi}{10}\Big)}{cos\Big(\dfrac{7\pi }{24}-\dfrac{\pi}{8} \Big)}=\dfrac{sin\dfrac{5\pi}{20}}{cos\dfrac{10\pi}{24}}=\dfrac{sin\dfrac{\pi}{10}}{cos\dfrac{5\pi}{12}}$

Значения   $sin\dfrac{\pi }{10}=\dfrac{\sqrt5-1}{4}\ \ ,\ \ cos\dfrac{5\pi }{12}=\dfrac{\sqrt6-\sqrt2}{4}$  , тогда в результате

получим    $\dfrac{\sqrt5-1}{\sqrt6-\sqrt2}$  .

$\displaystyle 2)\ \ \frac{sin\dfrac{15\pi}{7}\cdot sin\dfrac{4\pi }{21}+cos\dfrac{4\pi}{21}\cdot cos\dfrac{6\pi}{7} }{sin\dfrac{7\pi}{24}\cdot cos\dfrac{\pi}{24}-cos\dfrac{7\pi}{24}\cdot sin\dfrac{23\pi }{24}}=\frac{sin\Big(2\pi +\dfrac{\pi}{7}\Big)\cdot sin\dfrac{4\pi }{21}+cos\dfrac{4\pi}{21}\cdot cos\Big(\pi -\dfrac{\pi}{7}\Big)}{cos\dfrac{7\pi}{24}\cdot cos\dfrac{\pi}{24}-cos\dfrac{7\pi}{24}\cdot sin\Big(\pi -\dfrac{\pi }{24}\Big)}=$

$=\dfrac{sin\dfrac{\pi}{7}\cdot sin\dfrac{4\pi }{21}-cos\dfrac{4\pi}{21}\cdot cos\dfrac{\pi}{7} }{sin\dfrac{7\pi}{24}\cdot cos\dfrac{\pi}{24}-cos\dfrac{7\pi}{24}\cdot sin\dfrac{\pi }{24}}=\dfrac{-cos\Big(\dfrac{\pi}{7}+\dfrac{4\pi}{21}\Big)}{sin\Big(\dfrac{7\pi }{24}-\dfrac{\pi}{24}\Big)}=\dfrac{-cos\dfrac{7\pi}{21}}{sin\dfrac{6\pi}{24}}=\dfrac{-cos\dfrac{\pi}{3}}{sin\dfrac{\pi}{4}}==\dfrac{\dfrac{1}{2}}{\dfrac{\sqrt2}{2}} =\dfrac{1}{\sqrt2}=\dfrac{\sqrt2}{2}$