Please solve this question.
1 answer:
Answer: see proof below
<u>Step-by-step explanation:</u>
Given: A + B + C = π → A + B = π - C
A + C = π - B
B + C = π - A
Cofunction Identities: sin [(π/2) - A] = cos A
cos [(π/2) - A] = sin A
Sum to Product Identity: sin A - sin B = 2 cos [(A + B)/2] · sin [(A - B)/2]
Pythagorean Identity: cos²A + sin²A = 1 → cos²A = 1 - sin²A
Even/Odd Identity: cos (-A) = -cos(A)
Double Angle Identity: cos (2A) = 2 cos²A - 1
<u>Proof LHS → RHS:</u>
LHS: cos [(B + C) - A] - cos [(A + C) - B] + cos [(A + B) + C]
Given: cos [(π - A) - A] - cos [(π - B) - B] + cos [(π - C) - C]
= cos (π -2A) - cos (π -2B) + cos (π -2C)
Cofunction: cos (-2A) - cos (-2B) + cos (-2C)
Even/Odd: -cos (2A) + cos (2B) - cos (2C)
Double Angle: -(2cos²A - 1) + cos (2B) - cos (2C)
= 1 - 2cos²A + cos (2B) - cos (2C)
Sum to Product: 1 - 2cos²A - 2sin(B+C) · sin (B-C)
Given: 1 - 2cos²A - 2sin (π - A) · sin (B - C)
Cofunction: 1 - 2cos²A - 2sin A · sin (B - C)
Pythagorean: 1 - 2(1 - sin²A) - 2sin A · sin (B - C)
= 1 - 2 + 2sin²A - 2sin A · sin (B - C)
= -1 + 2sin²A - 2sin A · sin (B - C)
Factor: -1 + 2sin A (sin A - sin (B - C))
Given: -1 + 2sin A [sin (π - (B + C)) - sin (B - C)]
Cofunction: -1 + 2sin A [sin (B + C) - sin (B - C)]
Sum to Product: -1 + 2sin A [2 cos B · sin C]
= -1 + 4sin A · cos B · sin C
LHS = RHS
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