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\[\huge{\cos \frac{\pi}{7}}-\cos \frac{2\pi}{7}+\cos \frac{3\pi}{7}=\frac{1}{2}\]
prove
so far i managed to write \[\cos \frac{ \pi }{ 7 }-2\cos^2\frac{ \pi }{ 7 }+1+\cos \frac{ 2\pi }{ 7 }\cos \frac{ \pi }{ 7 }-\sin \frac{ 2\pi }{ 7 }\sin \frac{ \pi }{ 7 }\] \[\cos \frac{ \pi }{ 7 }-2\cos ^2\frac{ \pi }{ 7 }+1+2\cos^3\frac{ \pi }{ 7 }-\cos \frac{ \pi }{ 7 }-2\cos \frac{ \pi }{ 7 }+2\cos^3\frac{ \pi }{ 7 }\] \[4\cos^3\frac{ \pi }{ 7 }-2\cos^2\frac{ \pi }{ 7 }-2\cos\frac{\pi}{7}+1\]

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Other answers:

without using a calculator
guys i have the solution but it is so superficial i'll put it just now
1 Attachment
no idea about S
*
is that the bookmark symbol
cos(pi/7) -cos(2pi/7)+ cos(3pi/7) cos(pi/7)+cos(3pi/7)+cos(5pi/7) {2sin(pi/7)cos(pi/7) + 2 sin(pi/7) cos(3pi/7) + 2 sin(pi/7) cos(5pi/7)}/2 sin(pi/7)
That is just a comment, which will enable me to receive notifications whenever it gets discussed * is just a symbol, yes, bookmark symbol
are you saying -cos 2 pi/7=cos 5 pi/7
yes
y
-cos 2a=cos 5a
cos(5pi/7) =cos(pi -2pi/7) = -cos(2pi/7)
k very clear
It is because cos(pi-a)=-cosa
Now, u need to use 2 sinA cosB = sin(A+B) - sin(A-B)
on cos 5a
right
Sorry, The correct formula is 2 sinA cosB =sin(A+B) + sin(A-B) =sin(A+B) - sin(B-A)
=cos(pi/7) -cos(2pi/7)+ cos(3pi/7) =cos(pi/7)+cos(3pi/7)+cos(5pi/7) ={2sin(pi/7)cos(pi/7) + 2 sin(pi/7) cos(3pi/7) + 2 sin(pi/7) cos(5pi/7)}/2 sin(pi/7) ={sin(2pi/7) -sin(0) +sin(4pi/7) -sin(2pi/7) +sin(6pi/7) -sin(4pi/7)}/2sin(pi/7) ={sin(6pi/7)}/2sin(pi/7) ={sin(pi/7)}/2sin(pi/7) =1/2
sin(6pi/7) =sin(pi -pi/7) =sin(pi/7)
okay that makes so much sense i guess elementary steps we skipped
so we just had to play with\[\pi \pm \frac{\pi}{7}\]
yep... and one formula

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