Question

if a sin theta + b cos theta = c then prove that a cos theta- b sin theta = √a^2+ b^2- c^2​

Answer 1

Answer:

Given :-

→ 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,

Answer 2

Answer:

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