Step-by-step explanation:
Hey there!
Given;
A+B+C = π
A+B = π-C (Take this)----->
B+C = π-A
A+C = π-B
Putting "sin" and "cos" on both sides. We get;
sin (A+B) = sin (π-C)
= sin C.....(i)
cos (A+B) = cos (π-C)
= -cos C......(ii)
Now, taking LHS. We get;
= cos (2A) + cos (2B) + cos (2C)
Use formula for cos (C) + cos (D) in first two terms.
[tex] = 2. \cos( \frac{2a + 2b}{2} ). \cos( \frac{2a - 2b}{2} ) + \cos(2c) [/tex]
Simplify in first two terms. And keep cos (2C) in terms of cos. (i.e cos (2A) = 2cos^2 (A)-1).
= 2 cos (A+B). cos (A-B) + 2cos ^2(C)-1
Now, use (i).
[tex] = - 2 \cos(c) . \cos(a - b ) + 2 { \cos}^{2} c - 1[/tex]
Take (-2cos (c)) common.
[tex] = - 2 \cos(c) ( \cos(a - b) - \cos(c) ) - 1[/tex]
Use cos (C) = -cos(A+B).
[tex] = - 2 \cos(c) ( \cos(a - b) . \cos(a + b) ) - 1[/tex]
Use formula. (i.e cos (A+B) + cos (A-B) = 2.
= -1-4 cos (A) . cos (B). cos (C)
→ RHS proved.
Hope it helps....
