由f(a)=（2tana）-[（2sin^2a/2 -1）/sin(a/2)cos(a/2)]
先对f(a)进行化简
f(a)=（2tana）-[-2(1-2sin^2a/2 )/2sin(a/2)cos(a/2)]
=(2tana)+2cosa/sina
=2(sina/cosa+cosa/sina)
=2[(sin^2a+cos^2a)/sinacosa]
=2/sinacosa
=4/sin2a
所以f（π/12）=4/sin2π/12=8
若f(a)=（2tana）-[（2sin^2a/2 -1）/sin(a/2)cos(a/2)]则f(兀/12)=
1个回答
相关问题
-
tana=2,则sin(a+3兀)+cos(兀+a)/sin(-a)-cos(兀+a)=
-
设角=-35/6兀,则2sin(兀+a)cos(兀-a)-cos(兀+a)/1+sin^2a+sin(兀-a)-cos^
-
若tana=√3 ,则cos2a-cos^2 a / sin^2 a-1=
-
若sin(兀+a)=1/3,a属于(-兀/2,0),则tana=?
-
1-tana/2+tana=1,则：cos2a/1+sin2a=?
-
三角函数题目若a属于(0,兀/2),cos(兀/4-a)=2根号2cos2a,则sin2a=
-
化简sin(兀+a)cos(3/2兀+a)+sin(兀/2+a)cos(兀+a)=
-
若tana=1/3,则sin^3 a+2cos^ 3 a/sin^3 a-2cos^3 a
-
求证:1+sin2a-cos2a/1+sin2a+cos2a=tana
-
cos(兀-2a)/sin(a-兀/4）=-跟号2/2求cos(a)+sin(a)