Предмет: Алгебра, автор: danenova11

Представьте в виде суммы или разности выражение
sin2a \cos3 \beta
​

Приложения:

Ответы

Автор ответа: mmb1
16

sin 2a * cos 3b =
sin 2a = 2*sin a*cos a
cos 3b = 4cos³ b - 3cos b
sin 2a * cos 3b = 2*sin a*cos a *(4cos³ b - 3cos b) = 8*sin a*cos a*cos³ b - 6*sin a*cos a*cos b

snufle27: lox
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