1樓：匿名使用者
解：sin(-7/12)=sin(5π/12-π)=-sin(π-5π/12)=-sin5π/12=sin(3π/12+2π/12).
=-sin(π/4+π/6)=-(sinπ/4cosπ/6+cosπ/4sinπ/6)=-[(√2/2*√3/2+√2/2*(1/2)]
= -(√6+√2)/4.
cos(-61π/12)=cos61π/12=cos[12*5π+π)/12=cos(5π+π/12)=cos(4π+(π+π/12].
=-cos(π+π/12)=cosπ/12=cos(4π/12-3π/12)=cos(π/3-π/4).
=cos(π/3)*cos(π/4)+sin(π/3)sin(π/4.
=(1/2)*√2/2+√3/2√2/2.
=(√2+√6)/4.
tan35π/12=tan(36π-π)/12=tan(3π-π/12)=-tanπ/12.
=-tan(π/3-π/4)=-(tanπ/3-tanπ/4)/(1+tanπ/3*tanπ/4).
=(1-√3)/(1+√3*1).
=(1-√3)^2/(1+√3)(1-√3).
=(1-2√3+3)/(-2).
=-(2-√3).
=(√3-2)./
2樓：
原式=sin(5π/12) cos (1π/12) tan π/12=cos(π/12)sin(π/12)
=sin(π/6)*1/2
=1/4
利用三角函式和與差的公式cos(-61兀/12）的值
3樓：徐少
cos(-61π/12）
=cos(61π/12）
=cos(5π+π/12)
=cos(π+π/12)
=-cos(π/12)
cos(π/6)
=cos(π/12+π/12)
=cos²(π/12)-sin²(π/12)=cos²(π/12)-[1-cos²(π/12)]=2cos²(π/12)-1
∴cos²(π/12)=√3/2+1
∴cos(π/12)=√(1+√3/2)
∴cos(π/12)=√(1+√3/2)
∴cos(-61π/12=√(1+√3/2)
4樓：匿名使用者
=cos(_6丌+11丌/12)
=cos(11丌/12)
=cos165=cos(180_15)
=_cos15=_cos（45_30）
=_(cos45cos30+sin45sin30)=_(√2/2*√3/2+√2/2*1/2)=_(√6+√2)/4
cos（-61兀/12) 利用和差角公式求三角函式
5樓：迷路明燈
=cos(-5π-π/12)=-cos(-π/12)=-cos(π/12)<0
由√3/2=cos(π/6)=2cos²(π/12)-1得
-cos(π/12)=-√((2√3+4)/8)=-(√3+1)/2√2=-(√6+√2)/4
利用和（差）角公式求下列各三角函式的值 （1） sin（-7π/12） （2）cos（-61π/12） （3）tan（35π/12
6樓：買昭懿
sin(-7π/12)
= -sin(7π/12)
= -sin﹙π/3+π/4﹚
= -(sinπ/3cosπ/4+cosπ/3sinπ/4)= -(√3/2 * √2/2 + 1/2 * √2/2)= -(√6+√2)/4
cos(-61π/12）
= cos(61π/12)
= cos(5π+π/12)
= cos(π+π/12)
= -cos(π/12)
= -cos(π/3-π/4)
= -(cosπ/3cosπ/4+sinπ/3sinπ/4)= -(1/2 * √2/2 + √3/2 * √2/2)= -(√2+√6)/4
tan(35π/12)
= tan(3π-π/12)
= tan(π-π/12)
= -tan(π/12)
= -tan(π/3-π/4)
= -(tanπ/3-tanπ/4) / (1+tanπ/3tanπ/4)
= -(√3-1) / (1+√3)
= -(√3-1)^2 /
= -(4-2√3)/(3-1)
= √3 - 2
7樓：良駒絕影
1、sin(－7π/12)=－sin(7π/12)=－sin[π/3＋π/4]
2、cos(－61π/12)=cos(61π/12)=cos(6π＋11π/12)=cos(11π/12)=cos[3π/4＋π/6]
3、tan(35π/12)=tan(3π－π/12)=－tan(π/12)=－tan[π/3－π/4]
8樓：
sin（-7π/12）=-sin（7π/12)=-sin（π/3+π/4））
cos（-61π/12）=cos（-7π/12-π/2-4π）=cos（7π/12+π/2）=-sin（7π/12）
tan(35π/12)=tan(11π/12)=tan((5π/12+π/2)=sin(5π/12+π/2)/cos(5π/12+π/2)
=-cos(5π/12)/sin(5π/12)=cos(7π/12)/sin(7π/12)=cos（π/3+π/4）/sin（π/3+π/4））
sin(-7派/12)=?cos(-61派/12)=?tan35派/12=? 20
9樓：
sin(-7π/12)
=-cos(π/12)
=-((1+cosπ/6)/2)^0.5
=-((1+3^0.5/2)/2)^0.5cos(-61π/12)
=-cos(π/12)
=-((1+3^0.5/2)/2)^0.5tan(35π/12)
=tan(-π/12)
=((1-cos(-π/12))/(1+cos(-π/12)))^0.5
10樓：路人__黎
=-sin(7π/12)=-(√6+√2)/4=-cos(π/12)=-(√6+√2)/4=-tan(π/12)=√3-2
cosa=±√1-sin²a=±√5/3
a∈(π/2,π)
cosa=-√5/3
同理sinβ=-√7/4
cos(a+β)=√5/4+√7/6
sin(a-β)=-1/2-√35/12
11樓：匿名使用者
sin(-7派/12)=-sin(7派/12)=-(√6+√2)/4
cos(-61派/12)=cos(11派/12)=-cos(1派/12)=-(√6+√2)/4
tan35派/12=tan(-派/12)=-tan(派/12)=-(2-√3)=-2+√3
利用和差角公式求下列三角函式值求sin105度 cos12分兀 tan(－12分之7兀)
12樓：匿名使用者
sin105°=sin(45°+60°)
=sin45°cos60°+sin60°cos45°=√2/2*1/2+√3/2*√2/2
=(√2+√6)/4。
cosπ/12=cos15°=cos(45°-30°)=cos45°cos30°+sin45°sin30°=√2/2*√3/2+√2/2*1/2
=(√6+√2)/4，
tan(-7π/12)=tan(-105°)=-tan(45°+60°)
=-(tan45°+tan60)/[1-tan45°tan60°]=-(1+√3)/(1-√3)
=-(1+√3)^2/2
=-2-√3。
利用和（差）角公式求三角函式值
13樓：九方景鑠
sin(-7π/12）=-sin(7π/12）
=-sin(3π/12+4π/12)=-[sin(π/4)cos(π/3)+cos(π/4)sin(π/3)]=-1/4(2^(1/2)+6^(1/2))
cos(-61π/12)= cos(61π/12)= cos(5π+ π/12)=cos(π+π/12)=-cos(π/12)
-cos(4π/12-3π/12)=-(cos(π/3)cos(π/4)+sin(π/3)sin(π/4))
=-((1/2)(2^(1/2)/2)+3^(1/2)/2*(2^(1/2)/2))=-(2^(1/2)/4+6^(1/2)/4)
tan(35π/12)=tan(36π/12-π/12)=tan(3π-π/12)=-tan(π/12)=-tan(4π/12-3π/12)=tan(π/3-π/4)....
求三角函式值 sin6sin42sin66sin78注 6表示6度）高手幫一下
該式 sin6 sin42 sin66 sin78 sin6 cos48 cos24 cos12 上式乘以cos6 得到 cos6 sin6 cos48 cos24 cos12 1 2 sin12 cos48 cos24 cos12 1 4 sin24 cos48 cos24 1 8 sin48 c...
求excel函式公式統計A列對應B列數值求和
思路 先提取字元，再條件求和。操作如下 一 提取 stl 和 ti10 字元。根據你提供的 可a列的每個欄位有空格分開，這樣我們可把c列作為輔助列，在c2單元格輸入公式 mid a2,1,4 回車，下拉填充。得到可作為條件的字元。二 在d2單元格輸入公式 1 求 stl 的和 回車，顯示stl 30...
求兩角和與差的正弦，餘弦，正切公式
兩角和與差的正弦餘弦正切的公式就是他們的相差公司有 sin sin cos cos sin sin sin sin cos cos sin sin cos cos sin cos cos cos sin sin cos cos cos sin sin sin a b sinacosb cosasin...