If a cosA - b sinA = c . Prove that a sinA + b cosA = _+√a^2 + b^2 - c^2
To prove the given equation, we'll start with the identity:
sin²A + cos²A = 1 (1)
Now, square both sides of the given equation:
(cosA - b sinA)² = c²
Expanding the square:
cos²A - 2b cosA sinA + b² sin²A = c²
Using the identity sin²A + cos²A = 1, we can rewrite the equation as:
1 - 2b cosA sinA + b² sin²A = c²
Rearranging the terms:
b² sin²A - 2b cosA sinA + (1 - c²) = 0
Since this is a quadratic equation in sinA, we can solve for sinA using the quadratic formula:
sinA = [2b cosA ± √(4b² cos²A - 4b²(1 - c²))] / (2b²)
Simplifying:
sinA = [2b cosA ± 2b √(cos²A - (1 - c²))] / (2b²)
sinA = cosA ± √(cos²A - (1 - c²)) / b (2)
Now, we'll use the identity sin²A + cos²A = 1 to rewrite the square root term:
cos²A - (1 - c²) = cos²A + c² - 1
cos²A - (1 - c²) = (cos²A + c²) - 1
Substituting this back into equation (2):
sinA = cosA ± √[(cos²A + c²) - 1] / b
sinA = cosA ± √[a² + b² - c²] / b
Now, we'll use the identity sinA = cos(A - 90º) to rewrite sinA as cosA:
cos(A - 90º) = cosA ± √[a² + b² - c²] / b
Since cos(A - 90º) = sinA, we can substitute sinA for cos(A - 90º):
sinA = ± √[a² + b² - c²] / b
Finally, we multiply both sides of the equation by a to get:
a sinA = ± a √[a² + b² - c²] / b
Since we squared both sides of the equation earlier, we introduced an extra solution. Hence, the ± sign represents both the positive and negative square root. Therefore, we can write the final equation as:
a sinA + b cosA = ± √[a² + b² - c²]
let a = sinB and let b = cosB
then a cosA - b sinA
= sinBcosA - cosBsinA
= sin(B-A)
= c
and a sinA + b cosA
= sinBsinA + sinAcosA
= cos(A - B)
= cos(B-A) , since cos(-x) = cos(x)
let's look at the right side:
I will assume that should be
± √(a^2 + b^2 - c^2)
= √(sin^2 B + cos^2 B - sin^2 (B-A) )
= √(1 - cos^2 (B-A))
= √ cos^2 (B-A)
= ± cos(B-A)
= a sinA + b cosA
as required
Whewww!!! tricky, does somebody see a better way?