tan(9π/4+a)*tan(3π/4+a)=?已知sin(a+30)=12/13 且60<a<90 求sina 的值?
来源:学生作业帮助网 编辑:六六作业网 时间:2024/08/26 02:47:31
tan(9π/4+a)*tan(3π/4+a)=?已知sin(a+30)=12/13且60<a<90求sina的值?tan(9π/4+a)*tan(3π/4+a)=?已知sin(a+30)=12/13
tan(9π/4+a)*tan(3π/4+a)=?已知sin(a+30)=12/13 且60<a<90 求sina 的值?
tan(9π/4+a)*tan(3π/4+a)=?已知sin(a+30)=12/13 且60<a<90 求sina 的值?
tan(9π/4+a)*tan(3π/4+a)=?已知sin(a+30)=12/13 且60<a<90 求sina 的值?
tan(9π/4+a)*tan(3π/4+a)
=tan(2π+π/4+a)*tan(π-π/4+a)
= tan(a+π/4)*tan(a-π/4)
=(1+tana)/(1-tana)*(1-tana)/(1+tana)
=1
sina=sin(a+30-30)=√3/2*sin(a+30)-1/2cos(a+30)
60<a<90
90<a+30<120
cos(a+30)=-5/13
√3/2*sin(a+30)-1/2cos(a+30)
=√3/2*12/13+1/2*5/13
=(12√3+5)/26
所以sina=(12√3+5)/26
tan(45度-a)tan(45度+a)+tan(2π-a)*tan(3/4π+a) 求值
化简tan(π/4+a)-tan(π/4-a)]
tan(π/4+A)-tan(π/4-A)的值
化简:[tan(π/4+a)-tan(π/4-a)]/tan2a-tan(π/4+a)tan(π/4-a)
cos2a/tan(π/4+a),
若a属于(0,π/4)试比较tan(a),tan(tan(a)),tan(sin(a))大小,
化简:{[tan(π/4+a)-tan(π/4-a)]/tan2a} *tan(π/4+a)
tan(a+b)=3/4,tan(a-π/4)=1/2,那么tan(b+π/4)等于多少
tan(a+b)=2/5 tan(a+π/4)=3/22 求tan(b-π/4)
tan(a+π/4)=2,则cos2a+3sin^2a+tan a=
求证:tan(a+π/4)+tan(a+3π/4)=2tan2a
tan(4/π+α)=3 求tanα?
a+b=3/4π,求(1-tan a)(1+tan b)
a+b=3 π/4,则(1-tan a)(1-tan b)=
tan(-x+π/4)
(tanπ/4+tanα)/(1-tanπ/4tanα)怎么解
sin(π+a)+tan π/4
已知tan(A+B)=2/5,tan(B-π/4)=1/4,求tan(A+π/4已知tan(A+B)=2/5,tan(B-π/4)=1/4,求tan(A+π/4)