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| 1. |
acos-bsin=cprove that asin+bcos=+_√α²+b²-c² |
| Answer» → a cos∅ + b sin∅ = c .......(1) .\xa0\xa0Now,\xa0→ ( a cos∅ - b sin∅ )² + ( a sin∅ + b cos∅ )² .\xa0= a²cos²∅ + b²sin²∅ - 2a sin∅ b cos∅ + a²cos²∅ + b²sin²∅ + 2a sin∅ b cos∅ .\xa0= a²sin²∅ + a²cos²∅ + b²cos²∅ + b²sin²∅ .\xa0= a²( sin²∅ + cos²∅ ) + b²( cos²∅ + sin²∅ ) .\xa0= a² + b² . [ ∵ sin²∅ + cos²∅ = 1 ] .\xa0\xa0Thus, ( a cos∅ - b sin∅ )² + ( a sin∅ + b cos∅ )² = ( a² + b² ) .\xa0⇒ c² + ( a sin∅ + b cos∅ )² = ( a² + b² ) .\xa0⇒ ( a sin∅ - b cos∅ )² = ( a² + b² - c² ) .\xa0⇒ ( a sin∅ - b cos∅ ) = ±√( a² + b² - c² ) .\xa0Hence,\xa0ORa cosx - b sinx = ca2\xa0cos2x + b2\xa0sin2x - 2ab sinx cosx = c2a2\xa0( 1 - sin2x) + b2\xa0( 1 - cos2x) -\xa02ab sinx cosx = c2a2\xa0- a2\xa0sin2x + b2\xa0- b2\xa0cos2x -\xa02ab sinx cosx = c2-(a2\xa0sin2x + b2\xa0cos2x +\xa02ab sinx cosx) = -a2\xa0- b2\xa0+ c2(a sinx + b cosx)2\xa0= a2\xa0+ b2\xa0- c2asinx + bcosx = +- root(a2+b2-c2) | |