In triangle ABC prove that cos(A-B÷2)= (a+b÷c) sin (c÷2)
Answers
Answered by
0
HEY DUDE HERE IS YOUR ANSWER....
The sum of all angles in a triangle is 180°
Therefore,
A + B + C = 180°
=> A + B = 180° - C
=> (A + B)/2 = (180° - C)/2 = 90° - C/2
In the above question,
LHS = cos((A + B)/2)
substituting (A + B)/2 with 90°- C/2
= cos(90° - C/2)
cos(90° - θ) = sinθ
Therefore,
= sin(C/2) which is also = RHS
LHS = RHS
Hence proved.
I HOPE THIS WILL HELP YOU OUT....
HAVE A GREAT DAY DEAR....
The sum of all angles in a triangle is 180°
Therefore,
A + B + C = 180°
=> A + B = 180° - C
=> (A + B)/2 = (180° - C)/2 = 90° - C/2
In the above question,
LHS = cos((A + B)/2)
substituting (A + B)/2 with 90°- C/2
= cos(90° - C/2)
cos(90° - θ) = sinθ
Therefore,
= sin(C/2) which is also = RHS
LHS = RHS
Hence proved.
I HOPE THIS WILL HELP YOU OUT....
HAVE A GREAT DAY DEAR....
Similar questions