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# 三角關係與餘弦定理

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

### 1 Answer

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- hawk_wing_1999Lv 61 decade agoFavorite 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

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