已知f(a)=2tan a-(2sin^2a/2-1)/(sin a/2cos a/2),求f(pi/2)
问题描述:
已知f(a)=2tan a-(2sin^2a/2-1)/(sin a/2cos a/2),求f(pi/2)
答
f(a)=2tan a-(2sin^2a/2-1)/(sin a/2cos a/2)
=2tan a-(-cosa )/(1/2sin a)
=2tan a+2cota
f(pi/2)=∝能不能在sin,cos,tan的范围内解答?谢谢请检查下题吧没错,cot换一下f(a)=2tan a-(2sin^2a/2-1)/(sin a/2cos a/2) =2tan a-(-cosa )/(1/2sin a) =2tan a+2/tan a f(pi/2)=∝"∝"这个是什么?是无穷大 tan(π/2)=无穷大所以我怀疑你的题错了 如果是f(a)=2tan a(2sin^2a/2-1)/(sin a/2cos a/2) =2tan a(-cosa )/(1/2sin a) =2tan a*2/tan a=4f(pi/2)=4题目是f(a)=2tan a(2sin^2a/2-1)/(sin a/2cos a/2),但求的是f(pi/12), 不好意思哦2sin^2a/2-1=-cosasin a/2cos a/2=1/2sin af(a)=2tan a(2sin^2a/2-1)/(sin a/2cos a/2) =2tan a(-cosa)/(1/2sin a) =2tan a*(-2)/tan a=-4f(pi/12), =-4 你的另一个题分母也有问题,函数没写