早教吧作业答案频道 -->数学-->
设tan(α+8/7π)=a 求证sin(22π/7+α)+3cos(α-20π/7)/ sin(20π/7-α)-cos(α+22π/7)=a+3/-a
题目详情
设tan(α+8/7π)=a 求证sin(22π/7+α)+3cos(α-20π/7)/ sin(20π/7-α)-cos(α+22π/7)=a+3/-a
▼优质解答
答案和解析
tan(α+8π/7)=tan(π+α+π/7)=tan(α+π/7),即:tan(α+π/7)=a
sin(22π/7+α)+3cos(α-20π/7)/ sin(20π/7-α)-cos(α+22π/7)
=(sin(3π+π/7+α)+3cos(3π-2π/7-α))/(sin(3π-2π/7-α)-cos(3π+π/7+α))
=-(sin(π/7+α)+3cos(2π/7+α))/(sin(2π/7+α)+cos(π/7+α))
=-(sin(π/7+α)+3(cosπ/7cos(π/7+α)-sinπ/7sin(π/7+α)))/(sinπ/7cos(π/7+α)+cosπ/7sin(π/7+α)+cos(π/7+α))
=-(tan(π/7+α)+3(cosπ/7-sinπ/7tan(π/7+α)))/(sinπ/7+cosπ/7tan(π/7+α)+1)
=-(a+3(cosπ/7-asinπ/7))/(sinπ/7+acosπ/7+1)
sin(22π/7+α)+3cos(α-20π/7)/ sin(20π/7-α)-cos(α+22π/7)
=(sin(3π+π/7+α)+3cos(3π-2π/7-α))/(sin(3π-2π/7-α)-cos(3π+π/7+α))
=-(sin(π/7+α)+3cos(2π/7+α))/(sin(2π/7+α)+cos(π/7+α))
=-(sin(π/7+α)+3(cosπ/7cos(π/7+α)-sinπ/7sin(π/7+α)))/(sinπ/7cos(π/7+α)+cosπ/7sin(π/7+α)+cos(π/7+α))
=-(tan(π/7+α)+3(cosπ/7-sinπ/7tan(π/7+α)))/(sinπ/7+cosπ/7tan(π/7+α)+1)
=-(a+3(cosπ/7-asinπ/7))/(sinπ/7+acosπ/7+1)
看了 设tan(α+8/7π)=a...的网友还看了以下:
利用公式求pi公式:π/4=1-1/3+1/5-1/7.直到绝对值小于10^-7为止运行π等于4. 2020-05-14 …
关于复数的(1)试求i,i^2,i^3,i^4,i^5,i^6,i^7,i^8的值;(2)由(1) 2020-07-26 …
(1/2)+(1/2)i和1-i(i是虚部单位)另外(7/10)-(i/10)能化成7-i 2020-07-30 …
有关复数的题目一.巳知1+x+x^2=0,求证:x^1979+x^1989+x^1999=0二.设 2020-08-01 …
高二数学选修1---2复数代数形式的乘除运算1.计算题(1)7+3i/3+4i;(2)2i/2-i 2020-08-02 …
高中选修2—2的第三章数系的扩充和复数的概念的习题试求i^1,i^2,i^3,i^4,i^5,i^6 2020-10-30 …
几道高二数学题(5-3i)+(7-5i)-4i(5-3i)+(7-5i)-4i(-2-4i)-(-2 2020-11-01 …
求答.计算1.3+(5-2i)-(7-i)2.(-2+i)+(-2-i)+(根号2+7i)-(根号2 2020-11-01 …
复数计算(1)(1+i)^7/1-i+(1-i)^7/1+i-(3-4i)(2+2i)^3/4+3i 2020-11-01 …
直接写出0面各题7得数.七0-4.2=0.03×0.6=七22÷2七2=七i×i+七8×8=2七4÷ 2020-11-01 …