Question

Given that the sum of the angles A, B and C of a triangle is π radians, show that:
sin A + sin B + sin C = sin(A+B) + sin(B+C) + sin(A+C)

Please show all working clearly. MEDAL WILL BE AWARDED

Answer 1

Lets use a 30-60-90 triangle:

A = 30 degrees
B = 60 degrees
C = 90 degrees

sin(30 degrees+60 degrees)+sin(60 degrees+90 degrees)+sin(30 degrees+90 degrees)

sin(90 degrees)+sin(150 degrees)+sin(120 degrees)
sin(360 degrees)
= 0 = sin(180 degrees)

Answer 2

Suppose we don't want to use actual values?

Answer 3

I suppose you will use sin(A+B) princ. then.

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

Answer 4

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

Answer 5

A+B+C=π

therefore,
A+B = π-C
A+C = π-B
B+C = π-A

now the righ side can be
sin(A+B) + sin(B+C) + sin(A+C)
= sin(π-C) + sin(π-B) + sin(π-A)

then use the identity :
sin(π-x ) = sin(x)

so,
sin(π-C) + sin(π-B) + sin(π-A)
= sin(C) + sin(B) + sin(A)
= LHS

QED

Answer 6

What does QED mean?

Answer 7

opsss... i mistake in copying and vaste the equation :)

that should like this :
sin(A+B) + sin(B+C) + sin(A+C)
= sin(π-C) + sin(π-A) + sin(π-B)
= sin(C) + sin(A) + sin(B)

Answer 8

*paste

Answer 9

and, actually

QED = Quod Erat Demonstrandum

it means the proof is complete. :)

Answer 10

Thanks so much! :)

Answer 11

you're welcome