Question

Help with solving sums and differences in trig problems

Answer 1

\[\cos(x-\frac{ 2\pi }{ 3 })+\cos(x+\frac{ 2\pi }{ 3 })=1\]

Answer 2

use the 2 identities
cos ( A - B) = cos A cos B + sin A sin B

cos ( A + B) = cos A cos B - sin A sin B

Answer 3

Those two combine and make \(\cos C + \cos D = 2 \cos \left(\frac{C +D}{2}\right)\cos\left(\frac{C-D}{2}\right)\)

Answer 4

\[\frac{ cosx - \cos \frac{ 2\pi }{ 3 } }{1+cosx \cos \frac{ 2\pi }{ 3 } }-\frac{ cosx+\cos \frac{ 2\pi }{ 3 } }{ 1+cosxcos \frac{ 2\pi }{ 3 } }\]

Answer 5

oh

Answer 6

I did it wrong s:

Answer 7

so the first way: \[cosx \cos \frac{ 2\pi }{ 3 }+sinxsin \frac{ 2\pi }{ 3 }\]

Answer 8

and\[cosxcos \frac{ 2\pi }{ 3 }-sinxsin \frac{ 2\pi }{ 3 }\]

Answer 9

cos x cos 2pi/3 + sin x 2 pi /3 + cos x cos 2pi/3 - sin x sin 2pi/3 = 1
2 cos x cos 2pi/3 = 1

cos x = 1 / (2 cos 2pi/3)

Answer 10

now use your calculator to find cos x then x

Answer 11

cosx=-1

Answer 12

yes

Answer 13

and x=pi and 2pi

Answer 14

pi is correct is 2pi?

Answer 15

wait

Answer 16

sorry i did something wrong in my calculator lool one moment

Answer 17

just pi?

Answer 18

yes just pi as cos of 2 pi = 1

Answer 19

Yes I found that out accidently put positive 1 instead of negative 1

Answer 20

the general answer would be pi + 2npi where n = 0,1,2,3....

Answer 21

I just realized i forgot to say that we are on the interval [0,2pi]

Answer 22

oh ok then its just pi

Answer 23

Alright thank you :D

Answer 24

yw