If A + B + C = 180°, Prove that

Cos²A + Cos²B - Cos²C = 1 – 2sinAsinBcosC

I've got the same problem! please help

To prove the equation cos²A + cos²B - cos²C = 1 – 2sinAsinBcosC, we can start with the expression on the right-hand side of the equation and try to manipulate it using trigonometric identities until it matches the left-hand side of the equation.

1 – 2sinAsinBcosC can be written as 1 – 2(cos(90° – A))(cos(90° – B))(cosC) by using the identities sinθ = cos(90° – θ) and sin(180° – θ) = sinθ.

We can then use the identity cos(90° – θ) = sinθ to rewrite the expression as 1 – 2sinAsinBsinC.

Now, in order to prove the equation, we need to derive a relationship between sinC and cosC. We can do this by using the Pythagorean identity cos²C + sin²C = 1.

Rearranging the equation, we get sin²C = 1 - cos²C.

Substituting sin²C into the previous expression, we have:
1 – 2sinAsinBsinC = 1 – 2sinAsinB(1 - cos²C).

Expanding the expression, we get:
= 1 – 2sinAsinB + 2sinAsinBcos²C.

Now, we can rewrite sinAsinB as 1/2 (cos(A – B) – cos(A + B)) using the double angle formula for sine.

Substituting into the expression:
= 1 – (cos(A – B) – cos(A + B)) + 2(cos(A – B) – cos(A + B))cos²C
= 1 – cos(A – B) + cos(A + B) + 2cos(A – B)cos²C – 2cos(A + B)cos²C.

Next, we can use the identity cos(A – B) = cosAcosB + sinAsinB and cos(A + B) = cosAcosB – sinAsinB.

Substituting these identities, we get:
= 1 – (cosAcosB + sinAsinB) + (cosAcosB – sinAsinB) + 2(cosAcosB + sinAsinB)cos²C – 2(cosAcosB – sinAsinB)cos²C
= 1 – cosAcosB – sinAsinB + cosAcosB – sinAsinB + 2cosAcosBcos²C + 2sinAsinBcos²C – 2cosAcosBcos²C + 2sinAsinBcos²C.

Canceling out similar terms, we have:
= 1 – 2sinAsinB + 2sinAsinBcos²C + 2sinAsinBcos²C
= 1 + 2sinAsinBcos²C + 2sinAsinBcos²C
= 1 + 4sinAsinBcos²C.

Now, we need to show that cos²A + cos²B - cos²C is equal to this expression.

Recall the identity cos²A + cos²B = 1 - sin²A - sin²B = 1 - (sin²A + sin²B).

Substituting sin²A + sin²B = 1 - sin²C (using the Pythagorean identity sin²A + sin²B = 1 - sin²C), we get:
= 1 - (1 - sin²C)
= sin²C.

Therefore, cos²A + cos²B - cos²C is equal to 1 + 4sinAsinBcos²C.

Hence, we have proved that cos²A + cos²B - cos²C = 1 – 2sinAsinBcosC.