cos(2 兀/5)cos(兀/5-2)的值详细过程___还有cosa=1/2其中a属于(-兀/2,0) sin(a/2
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cos(2 兀/5)cos(兀/5-2)的值详细过程___还有cosa=1/2其中a属于(-兀/2,0) sin(a/2)=?
天阿你们都错啦,是cos(2加兀/5)cos(兀/5-2)
天阿你们都错啦,是cos(2加兀/5)cos(兀/5-2)
应该是cos(2 π/5)cos(π/5-2π)
所以cos(2 π/5)cos(π/5-2π)
=cos(2 π/5)cos(π/5)
=[2sin(π/5)cos(π/5)cos(2π/5)]/[2sin(π/5)]
=sin(2π/5)cos(2π/5)/[2sin(π/5)]
=2sin(2π/5)cos(2π/5)/[4sin(π/5)]
=sin(4π/5)/[4sin(π/5)]
=sin(π/5)/[4sin(π/5)]
=1/4
a∈(-π/2,0),∴a/2∈(-π/4,0),∴sin(a/2)
所以cos(2 π/5)cos(π/5-2π)
=cos(2 π/5)cos(π/5)
=[2sin(π/5)cos(π/5)cos(2π/5)]/[2sin(π/5)]
=sin(2π/5)cos(2π/5)/[2sin(π/5)]
=2sin(2π/5)cos(2π/5)/[4sin(π/5)]
=sin(4π/5)/[4sin(π/5)]
=sin(π/5)/[4sin(π/5)]
=1/4
a∈(-π/2,0),∴a/2∈(-π/4,0),∴sin(a/2)
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