Tính: sin `(7π)/12`, cos `(-π/12)`, tan `(13π)/12` 03/09/2021 Bởi Harper Tính: sin `(7π)/12`, cos `(-π/12)`, tan `(13π)/12`
`sin((7π)/12)= sin(π/3 + π/4)` `=cos(π/4) × sin(π/3) + cos(π/3) × sin(π/4)` `=\sqrt[2]/2×\sqrt[3]/2 + 1/2×\sqrt[2]/2` `=(\sqrt[2]+\sqrt[6])/4` `cos(-(π)/12)=cos(π/4 – π/3)` `=cos(π/4) × cos(π/3) + sin(π/3)× sin(π/4)` `=\sqrt[2]/2×1/2+\sqrt[3]/2 ×\sqrt[2]/2` `=(\sqrt[2]+\sqrt[6])/4` `tan((13π)/12)=tan(π+π/12)` `=tan(π/12)=tan(π/3 + π/4)` `=(\sqrt[3]-1)/(sqrt[3]+1` Bình luận
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`sin((7π)/12)= sin(π/3 + π/4)`
`=cos(π/4) × sin(π/3) + cos(π/3) × sin(π/4)`
`=\sqrt[2]/2×\sqrt[3]/2 + 1/2×\sqrt[2]/2`
`=(\sqrt[2]+\sqrt[6])/4`
`cos(-(π)/12)=cos(π/4 – π/3)`
`=cos(π/4) × cos(π/3) + sin(π/3)× sin(π/4)`
`=\sqrt[2]/2×1/2+\sqrt[3]/2 ×\sqrt[2]/2`
`=(\sqrt[2]+\sqrt[6])/4`
`tan((13π)/12)=tan(π+π/12)`
`=tan(π/12)=tan(π/3 + π/4)`
`=(\sqrt[3]-1)/(sqrt[3]+1`