## Jonask 2 years ago trigonometry

$\huge{\cos \frac{\pi}{7}}-\cos \frac{2\pi}{7}+\cos \frac{3\pi}{7}=\frac{1}{2}$

prove

so far i managed to write $\cos \frac{ \pi }{ 7 }-2\cos^2\frac{ \pi }{ 7 }+1+\cos \frac{ 2\pi }{ 7 }\cos \frac{ \pi }{ 7 }-\sin \frac{ 2\pi }{ 7 }\sin \frac{ \pi }{ 7 }$ $\cos \frac{ \pi }{ 7 }-2\cos ^2\frac{ \pi }{ 7 }+1+2\cos^3\frac{ \pi }{ 7 }-\cos \frac{ \pi }{ 7 }-2\cos \frac{ \pi }{ 7 }+2\cos^3\frac{ \pi }{ 7 }$ $4\cos^3\frac{ \pi }{ 7 }-2\cos^2\frac{ \pi }{ 7 }-2\cos\frac{\pi}{7}+1$

without using a calculator

guys i have the solution but it is so superficial i'll put it just now

8. shubhamsrg

*

is that the bookmark symbol

10. sauravshakya

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)

11. shubhamsrg

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

13. sauravshakya

yes

y

-cos 2a=cos 5a

16. sauravshakya

cos(5pi/7) =cos(pi -2pi/7) = -cos(2pi/7)

k very clear

18. sauravshakya

It is because cos(pi-a)=-cosa

19. sauravshakya

Now, u need to use 2 sinA cosB = sin(A+B) - sin(A-B)

on cos 5a

right

22. sauravshakya

Sorry, The correct formula is 2 sinA cosB =sin(A+B) + sin(A-B) =sin(A+B) - sin(B-A)

23. sauravshakya

=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

24. sauravshakya

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$\pi \pm \frac{\pi}{7}$

27. sauravshakya

yep... and one formula