If a cos 0 -b sin 0=c, then a sin 0 + b cos 0 =
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Answer:
a sin θ + b cos θ = ± √(a²+b²-c²)
Step-by-step explanation:
Given, a cos θ - b sin θ = c
Squaring both sides,
a² cos² θ + b² sin² θ - 2ab sin θ cos θ = c²
Denoting cos² θ as 1 - Sin² θ and Sin² θ as 1 - Cos² θ
a² ( 1 - sin² θ ) + b² ( 1 - cos² θ ) - 2ab sin θ cos θ = c²
a² - a² sin² θ + b² - b² cos² θ - 2ab sin θ cos θ = c²
a² + b² - c² = a² sin² θ + b² cos² θ + 2ab sin θ cos θ
( √(a²+b²-c²) )² = ( a sin θ + b cos θ )²
a sin θ + b cos θ = ± √(a²+b²-c²)
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