Question

4. Найдите sin 2a и cos 2а, если cos a =7/25,3n/2<а<2n​

Answer 1

Ответ:

336/527

Объяснение:

sinα=-√1-(7/25)²=-√1-49/625=-√(625-49)/625=-√576/625=-24/25

sin2a =2·(-24/25)·7/25=-336/625

cos2a=(7/25)²-(-24/25)²=49/625-576/625=-527/625,

tg2a=sin2a/cos2a=-336/625:(-527/625)=336/527