if A cos - B sin = C , prove that A sin + B cos = +- whole root (A^2 + B^2 - C^2)
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Step-by-step explanation:
A cos - B sin = C
Squaring both Sides
=> (A cos - B sin)² = C²
=> A²Cos² + B²Sin² - 2ABCosSin = C²
=> A²Cos² + B²Sin² - C² = 2ABCosSin - Eq 1
Let say
A sin + B cos = X
=> X = A sin + B cos
Squaring both Sides
=> X² = (A sin + B cos)²
=> X² = A²Sin² + B²cos² + 2ABSinCos
putting value of 2ABCosSin from eq 1
=> X² = A²Sin² + B²cos² + A²Cos² + B²Sin² - C²
=> X² = A²(Sin² + Cos²) + B²(Cos² + Sin²) - C²
as we know that Sin² + Cos² = 1
=> X² = A² + B² - C²
Square rooting both sides
=> X = ±√(A² + B² - C²)
=> A sin + B cos = ±√(A² + B² - C²)
QED
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