acos-bsin=c prove that asin+bcos=+_√α²+b²-c²
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Gaurav Seth 6 years, 3 months ago
→ a cos∅ + b sin∅ = c .......(1) .
Now,
→ ( a cos∅ - b sin∅ )² + ( a sin∅ + b cos∅ )² .
= a²cos²∅ + b²sin²∅ - 2a sin∅ b cos∅ + a²cos²∅ + b²sin²∅ + 2a sin∅ b cos∅ .
= a²sin²∅ + a²cos²∅ + b²cos²∅ + b²sin²∅ .
= a²( sin²∅ + cos²∅ ) + b²( cos²∅ + sin²∅ ) .
= a² + b² . [ ∵ sin²∅ + cos²∅ = 1 ] .
Thus, ( a cos∅ - b sin∅ )² + ( a sin∅ + b cos∅ )² = ( a² + b² ) .
⇒ c² + ( a sin∅ + b cos∅ )² = ( a² + b² ) .
⇒ ( a sin∅ - b cos∅ )² = ( a² + b² - c² ) .
⇒ ( a sin∅ - b cos∅ ) = ±√( a² + b² - c² ) .
Hence,
OR
a cosx - b sinx = c
a2 cos2x + b2 sin2x - 2ab sinx cosx = c2
a2 ( 1 - sin2x) + b2 ( 1 - cos2x) - 2ab sinx cosx = c2
a2 - a2 sin2x + b2 - b2 cos2x - 2ab sinx cosx = c2
-(a2 sin2x + b2 cos2x + 2ab sinx cosx) = -a2 - b2 + c2
(a sinx + b cosx)2 = a2 + b2 - c2
asinx + bcosx = +- root(a2+b2-c2)
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