早教吧作业答案频道 -->数学-->
:已知tan(π+α)=2,求3cos(π-α)-2cos(π/2+α)/4cos(-α)+sin(2π-α)的值:已知tan(π+α)=2,求3cos(π-α)-2cos(π/2+α)/4cos(-α)+sin(
题目详情
:已知tan(π+α)=2,求3cos(π-α)-2cos(π/2+α)/4cos(-α)+sin(2π-α)的值:已知tan(π+α)=2,求3cos(π-α)-2cos(π/2+α)/4cos(-α)+sin(
▼优质解答
答案和解析
tan(π+α)=2,即有tana=2,tana=2
3cos(π-α)-2cos(π/2+α)/4cos(-α)+sin(2π-α)
=[-3cosa+2sina]/[4cosa-sina]
=[-3+2tana]/[4-tana],(分子分母同除以cosa)
=(-3+4)/(4-2)
=1/2
3cos(π-α)-2cos(π/2+α)/4cos(-α)+sin(2π-α)
=[-3cosa+2sina]/[4cosa-sina]
=[-3+2tana]/[4-tana],(分子分母同除以cosa)
=(-3+4)/(4-2)
=1/2
看了 :已知tan(π+α)=2,...的网友还看了以下:
∑(2^n)/(n^n)的收敛性你回答的是:取后一项后前一项的比.(2^n+1)/((n+1)^(n 2020-03-31 …
(1/(n^2 n 1 ) 2/(n^2 n 2) 3/(n^2 n 3) ……n/(n^2 n 2020-05-16 …
若n为一自然数,说明n(n+1)(n+2)(n+3)与1的和为一平方数n(n+1)(n+2)(n+ 2020-05-16 …
为什么n(n+1)(n+2)可拆成1/4[n(n+1)(n+2)(n+3)-(n-1)n(n+1) 2020-06-22 …
n为非0自然数,试证n^13n定能被2730整除.2730=2*3*5*7*13,n^13-n=n 2020-07-22 …
若n为合数,n|x^2-1,则gcd(x+1,n)|ngcd(x-1,n)|n且gcd(x+1,n 2020-07-30 …
用数学归纳法证明(n+1)(n+2)…(n+n)=2n·1·3·5·…(2n-1)(n∈N*)时, 2020-08-03 …
数论+集合1.证明5个相继的正整数之积不是完全平方数设n≥3,(n-2)(n-1)n(n+1)(n+ 2020-10-31 …
已知数列{a底n}中,a1=a2=1,且an=an-1+an-2(n≥3,n∈n*),设bn=an/ 2020-11-27 …
已知数列{a(n)}的前n项和为S(n),且满足a(1)=1,a(n+1)=S(n)+1(n∈N(+ 2021-02-09 …