x∈(π/6,5π/12),sin(2x+π/6)=4/5,求cos(2x-π/12)的值.

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x∈(π/6,5π/12),sin(2x+π/6)=4/5,求cos(2x-π/12)的值.
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x∈(π/6,5π/12),sin(2x+π/6)=4/5,求cos(2x-π/12)的值.
x∈(π/6,5π/12),sin(2x+π/6)=4/5,求cos(2x-π/12)的值.

x∈(π/6,5π/12),sin(2x+π/6)=4/5,求cos(2x-π/12)的值.
因为
x∈(π/6,5π/12)
所以
2x+π/6∈(π/2,π)
得到
cos(2x+π/6)=-3/5
因此
cos(2x-π/12)= cos(2x+π/6-π/4)= cos(2x+π/6)cosπ/4 + sin(2x+π/6)sinπ/4
= √2/2*(-3/5+4/5)= √2/10