三角關係與餘弦定理

Question

1. 
Find the general soolution of 
cos^2x [1+cos^2x+cos^3x+cos^4x+......cos^7x] = 0 without using arithmetic 
or geometric sequence.

2.
In a triangle ABC , by using the cosine law , prove that A^n+B^n is NOT equal to C^n

hawk_wing_1999 · Accepted Answer

cos^2x [1+cos^2x+cos^3x+cos^4x+......cos^7x] = 0 
∴cos^2x=0 or [1+cos^2x+cos^3x+......+cos^7x]=0
consider the range of cosx, which lies in the range of -1 and 1
∴-1<cosx<1
∴cos^2x, cos^4x, cos^6x must be larger than 0
There is only chance that cos^3x, cos^5x, cos^7x be negative if cosx is in the range of -1<cosx<0
However, in the range of -1<cosx<1, |cos^3x|<|cos^2x|, |cos^5x|<|cos^4x| and |cos^7x|<|cos^6x|
∴1+cosx+cos^2x+......+cos^7x must be >0
∴The final solution is cos^2x=0
cosx=0
x=360n90

三角關係與餘弦定理

1 個解答