Question

trigonometry

Answer 1

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

Answer 2

prove

Answer 3

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

Answer 4

without using a calculator

Answer 5

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

Answer 6

no idea about S

Answer 7

*

Answer 8

is that the bookmark symbol

Answer 9

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)

Answer 10

That is just a comment, which will enable me to receive notifications whenever it gets discussed
* is just a symbol, yes, bookmark symbol

Answer 11

are you saying -cos 2 pi/7=cos 5 pi/7

Answer 12

yes

Answer 13

y

Answer 14

-cos 2a=cos 5a

Answer 15

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

Answer 16

k very clear

Answer 17

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

Answer 18

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

Answer 19

on cos 5a

Answer 20

right

Answer 21

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

Answer 22

=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

Answer 23

sin(6pi/7) =sin(pi -pi/7) =sin(pi/7)

Answer 24

okay that makes so much sense i guess elementary steps we skipped

Answer 25

so we just had to play with\[\pi \pm \frac{\pi}{7}\]

Answer 26

yep... and one formula