Question

How to show that the left hand side of the below equation equals the right side when A+B+C =180

Answer 1

\[\cos A+ \cos B+ \cos C= 1+ 4 \sin (A/2) \sin (B/2) \sin (C/2)\]

Answer 2

any constraints?? like A+B+C = 180 (sum of angles of triangles)??

Answer 3

Well, I don't know about them.

Answer 4

I need to know how the left hand side equals to the right hand side.

Answer 5

well there must be something like that ... put A=90,B=90,C=90 you get it is false.

Answer 6

this seems only true if A+B+C = 180

Answer 7

It doesn't mention. But the lesson is related to identities when A+B+C=180.

Answer 8

So how about taking A+B+C=180 ?

Answer 9

yes ok we can proceed that.
A+B = 180 - C
cos(A+B) = cos(180-C) = - cos(C)

Answer 10

Yeah. that's what I learned today. :D

Answer 11

Cos C+Cos D = 2 sin(C+D)/2 sin(C-D)/2 <-- must be some formula like this ... Google for it.

Answer 12

ok. I will check

Answer 13

This is the formula,
cos C +cos D = 2 cos (C +D)/2 cos (C -D)/2

Answer 14

Oh ... i didn't expect that.
anyway apply this cos A + cos B
and change cos(C) into - cos(A+B)

Answer 15

Did you mean to apply the formula to cos A +cos B and change cos C to -cos (A+B) ?

Answer 16

yes

Answer 17

ok

\[2\cos ((A+B)/2) \cos ((A-B)/2) - \cos (A+B)\]

Answer 18

so shall I take cos (A+B/2) = sin c/2

Answer 19

ok ok ... all right ...
use half angle formula on cos(A+B)

Answer 20

ok

Answer 21

\[\cos(A+B)=1-2\sin ^2 (A+B)/2\]

Answer 22

Hope I am correct :)

Answer 23

no .. change it in cosine form.

Answer 24

ohh ok \[\cos(A+B)=2\cos^2(A+B)/2 - 1\]

Answer 25

yes ... and take common

Answer 26

\[ \cos 2A = \cos^2 A - \sin^2 A = 2 \cos^2 A - 1\]
I think there is no 1/2 factor here.

Answer 27

\[2 \cos (\frac{ A+B }{ 2 }) \cos(\frac{ A-B }{ 2 })-2\cos^2(\frac{ A+B }{ 2 })+1\]

Answer 28

Isn't it like \[\cos A=2\cos^2(\frac{A}{2})-1 \]

Answer 29

ok ok ... take common and and do some trigonometry and some algebra i can see your answer coming.

Answer 30

you must make careful use of braces ... braces are very important.

Answer 31

take common 2 cos((A+B)/2)

Answer 32

what about the 1 ?

Answer 33

that one will serve a special purpose

Answer 34

leave it as it is

Answer 35

\[2 \cos (\frac{A+B}{2}) [\cos(\frac{A-B}{2}) - \cos] +1\]

Answer 36

Am I correct? :)

Answer 37

no ... you didn't take the common properly. where is A+B/2 remaing in the end?

Answer 38

Didn't get it. :(

Answer 39

take common again and see what mistake you made
\[ 2 \cos (\frac{ A+B }{ 2 }) \cos(\frac{ A-B }{ 2 })-2\cos^2(\frac{ A+B }{ 2 })+1 \]