If A + B + C = 180°, Prove that
Cos²A + Cos²B + Cos²C = 1-2cosAcosBcosC
2cosAcosBcosC =(2cosAcosB)cosC
=(cos(A+B) + cos(A-B))cosC
=cos(pi-C)cosC + cos(A-B)cos(pi-(A+B))
=-2cos^2C - 2cos(A+B)cos(A-B)
=-cos^2C - cos^2A + sin^2B
so,
1 - 2cosAcosBcosC = 1 + cos^2C + cos^2A - sin^2B
= 1 - sin^2B + cos^2C + cos^2A
= cos^2B + cos^2C + cos^2A
To prove that cos²A + cos²B + cos²C = 1 - 2cosAcosBcosC, we will start by using the trigonometric identity that states:
cos²θ + sin²θ = 1
For any angle θ.
Now, let's look at the left side of the equation:
cos²A + cos²B + cos²C
Using the trigonometric identity, we can rewrite this as:
1 - sin²A + 1 - sin²B + 1 - sin²C
Now, rearrange the terms:
(1 + 1 + 1) - (sin²A + sin²B + sin²C)
Since sin²θ = 1 - cos²θ, we can substitute this in:
3 - (1 - cos²A + 1 - cos²B + 1 - cos²C)
Simplify:
3 - (3 - cos²A - cos²B - cos²C)
Combine the terms:
3 - 3 + cos²A + cos²B + cos²C
Which simplifies to:
cos²A + cos²B + cos²C
Thus, we can conclude that cos²A + cos²B + cos²C is equal to 1 - 2cosAcosBcosC.
To prove the equation Cos²A + Cos²B + Cos²C = 1-2cosAcosBcosC, we can start by using the identity relating the square of the cosine of an angle to its double angle formula:
Cos²θ = (1 + Cos2θ)/2
Now, let's use this identity to rewrite each term in the equation:
Cos²A = (1 + Cos2A)/2
Cos²B = (1 + Cos2B)/2
Cos²C = (1 + Cos2C)/2
Substituting these formulas into the equation, we have:
(1 + Cos2A)/2 + (1 + Cos2B)/2 + (1 + Cos2C)/2 = 1 − 2cosAcosBcosC
Next, let's simplify the left side of the equation by combining like terms:
(1 + (Cos2A + Cos2B + Cos2C))/2 = 1 − 2cosAcosBcosC
Now, let's focus on the term inside the parentheses. According to the angle sum property, we know that:
Cos2A + Cos2B + Cos2C = -1 - 2cos(A+B)cosC
Since A + B + C = 180°, we have:
(A + B) = 180° - C
Substituting this into the equation:
Cos2A + Cos2B + Cos2C = -1 - 2cos(180° - C)cosC
Applying the cosine difference formula:
Cos(180° - C) = -CosC
Cos2A + Cos2B + Cos2C = -1 - 2(-CosC)cosC
Simplifying further:
Cos2A + Cos2B + Cos2C = -1 + 2cosC²
Now, we can substitute this back into our equation:
(1 + (-1 + 2cosC²))/2 = 1 − 2cosAcosBcosC
Simplifying:
(1 - 1 + 2cosC²)/2 = 1 − 2cosAcosBcosC
(cosC²)/2 = 1 − 2cosAcosBcosC
Now, let's simplify the left side of the equation:
[(1 + cosC)(1 - cosC)]/2 = 1 − 2cosAcosBcosC
(1 - cos²C)/2 = 1 − 2cosAcosBcosC
Using the identity sin²θ = 1 - cos²θ:
(sin²C)/2 = 1 − 2cosAcosBcosC
Since sin²θ = 1 - cos²θ:
Sin²C = 2 - 4cosAcosBcosC
Finally, multiplying both sides of the equation by 2:
2sin²C = 2 - 4cosAcosBcosC
2(1 - cos²C) = 2 - 4cosAcosBcosC
2 - 2cos²C = 2 - 4cosAcosBcosC
Simplifying both sides yields:
2cos²C = 4cosAcosBcosC
Finally, dividing both sides by 2:
cos²C = 2cosAcosBcosC
This proves that Cos²A + Cos²B + Cos²C = 1 - 2cosAcosBcosC.