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为什么%tAn11/4π=tAnπ/4

先运用诱导公式化简: tan(-11π/5)=tan(-2π -π/5)=tan(-π/5)=-tan(π/5) tan(4π/5)=-tan(π- 4π/5)=-tan(π/5) tan(-11π/5)=tan(4π/5)

tan(11/3π) =tan(4π-π/3) =-tanπ/3 =-√3。

cosπ/3-tan5π/4+3/4tan^2(-π/6)+sin11π/6+cos^2 7π/6+sin7π/2= =1/2 -1 +3/4·1/3-1/2+3/4-1 =-2+1/4+3/4 =-1

解如图。

sin(5π/4)×cos(4π/5)×tan(11π/6) =sin(π+π/4)×cos(π-π/5)×tan(π+5π/6) =-sin(π/4)×[-cos(π/5)]×tan(5π/6) =√2/2×cos36°×tan150° =√2/2×(-√3/3)cos36° =-√6/6×cos36° 36°是锐角,所以cos36°是正,所以这个式子是负号 cos150°×tan330°×sin...

cos²¼π+sin²四分之十七-sin﹣三分之八π*cos-六分之三十一π-tan¾π+cos(-1260) =cos3/4π+sin²1/4π+sin2/3π* cos5/6π +tan1/4π+cos180 =-(√2/2)+(√2/2)²+(√3/2)* (-√3/2) +1-1 =-(√2/2)+1/2-3/4 =-(2√2+1)/4

sin4分之5π×cos5分之4π×tan6分之11π =sin(π+4分之1π)×cos(π—5分之1π)×tan(2π—6分之5π) =sin4分之1πXcos5分之1πXtan(—6分之5π) =—sin4分之1πXcos5分之1πXtan(6分之5π) =—sin4分之1πXcos5分之1πXtan(π—6分之1π) =sin4分之1πXcos5分之1...

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