求cos5π/12sinπ/12
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求cos5π/12sinπ/12
![求cos5π/12sinπ/12](/uploads/image/z/20169714-66-4.jpg?t=%E6%B1%82cos5%CF%80%2F12sin%CF%80%2F12)
∵Cos5π/12=Sin(π/2-5π/12 )
=Sinπ/12
∴cos5π/12sinπ/12 =S²inπ/12
=(1—Cosπ/6)/2=(2-√3)/4
=Sinπ/12
∴cos5π/12sinπ/12 =S²inπ/12
=(1—Cosπ/6)/2=(2-√3)/4
cos5π/12=sinπ/12
不查表求值:cos5π/12·sinπ/12=
为什么cos5π/12等于sinπ/12
(sin5π/12+cos5π/12)(sinπ/12-cosπ/12)
(sin5π/12-sinπ/12)(cos5π/12+cosπ/12)=?
cos5/12πcosπ/12+cosπ/12sinπ/6=?
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例题9和10,我知道有解析,但就是看不懂,好像是公式吧,我想知道cos5π/12是怎么变成sin的,周期不是π/2吗,乘