Ответ:
[tex]sin36^0-sin18^0[/tex]
Объяснение:
[tex]2sin\frac{a-\beta}{2}cos\frac{a+\beta}{2}=sina-sin\beta\\\\\frac{a-\beta}{2}=9^0\; \; = > \alpha -\beta=18^0\\\\\frac{a+\beta}{2}=27^0\; \; = > a+\beta=54^0\\\\\left \{ {{a-\beta=18^0} \atop {a+\beta=54^0}} \right.+\\\\2a=72^0\\a=36^0\\\\\beta=54^0-36^0=18^0\\\\2sin9^0cos27^0=sin36^0-sin18^0[/tex]
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Answers & Comments
Ответ:
[tex]sin36^0-sin18^0[/tex]
Объяснение:
[tex]2sin\frac{a-\beta}{2}cos\frac{a+\beta}{2}=sina-sin\beta\\\\\frac{a-\beta}{2}=9^0\; \; = > \alpha -\beta=18^0\\\\\frac{a+\beta}{2}=27^0\; \; = > a+\beta=54^0\\\\\left \{ {{a-\beta=18^0} \atop {a+\beta=54^0}} \right.+\\\\2a=72^0\\a=36^0\\\\\beta=54^0-36^0=18^0\\\\2sin9^0cos27^0=sin36^0-sin18^0[/tex]