Prove that cos(2pi/7)+cos(4pi/7)+cos(6pi/7)=-1/2

1 Answer | Add Yours

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll multiply both sides by sin (`pi` /7)

sin(`pi` /7)cos(2`pi` /7)+sin(`pi` /7)cos(4`pi` /7)+sin(`pi` /7)cos(6`pi` /7) = -sin(`pi` /7)/2

We'll transform the product in sum:

(1/2)[sin(`pi` /7-2`pi` /7) + sin(`pi` /7+2`pi` /7)] + (1/2)[sin(`pi` /7-4`pi` /7) + sin(`pi` /7+4`pi` /7)] + (1/2)[sin(`pi` /7-6`pi` /7) + sin(`pi` /7+6`pi` /7)]= -sin(`pi` /7)/2

We'll divide by (1/2) both sides:

-sin(`pi` /7) + sin(3`pi` /7) - sin(3`pi` /7) + sin(5`pi` /7) - sin(5`pi` /7) +  sin(7`pi` /7) = -sin(`pi` /7)

We'll eliminate like terms:

-sin(`pi` /7) + sin(`pi` )= -sin(`pi` /7)

But sin(`pi` ) = 0

-sin(`pi` /7) = -sin(`pi` /7)

Since we've get the same results both sides, the given identity is verified.

We’ve answered 318,994 questions. We can answer yours, too.

Ask a question