cos square A by 2 + cos square B by 2 minus cos square C by 2 is equals to 2 cos A by 2 cos B by 2 Sin by 2
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A+B+C = π , to prove - Cos²A/2 + Cos²B/2 - Cos²C/2 = 2CosA/2.CosB/2.SinC/2
1. LHS = Cos²A/2 + Cos²B/2 - Cos²C/2 = (1+CosA)/2 + (1+CosB)/2 - (1-CosC)/2 (Using identity)
2. 1/2(1 + CosA + 1 + CosB - 1 - CosC)
3. 1/2 ( CosA + CosB + 1 - CosC)
4. We have been given A+B+C = π
A+B = π-C
(A+B)/2 = π/2 - C/2
Multiplying both sides with cos,
Cos(A+B)/2 = sinC/2 (1)
5. 1/2 ( 2Cos(A+B)/2Cos(A-B)/2 + 2sin²C/2)
6. 2/2 sinC/2 ( cos(A-B)/2 + sinC/2) (after substituting from 1)
7. SinC/2 ( Cos(A-B)/2 + Cos(A+B)/2)
8. SinC/2 ( 2Cos [(A-B)/2 +(A+B)/2]/2 . Cos[(A-B)/2-(A+B)/2]/2)
9. 2 . sinC/2 . cosA/2 . cosB/2 = RHS
Hence, proved.
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