If a sin theta + b cos theta = c , then prove that a cos theta - b sin theta = the whole under root a 2 + b 2 - c 2 . plz answer soon if u want a thumbzzzz up guys !!!!!!!!!!!!!
Answers
Answered by
1366
asinθ + bcosθ = c
taking square both sides,
(asinθ + bcosθ)² = c²
⇒a²sin²θ + b²cos²θ + 2absinθ.cosθ = c² --------(1)
Let acosθ - bsinθ = x
Squaring both sides
(acosθ - bsinθ)² = x²
⇒a²cos²θ + b²sin²θ -2absinθ.cosθ = x² ------(2)
Add equation (1) and (2),
a²sin²θ + b²cos²θ +2abinθ.cosθ + a²cos²θ + b²sin²θ -2absinθ.cosθ = c² + x²
⇒(a² + b²)cos²θ + (a² +b²)sin²θ = c² + x²
⇒(a² + b²)[sin²θ + cos²θ ] = c² + x²
⇒(a² + b²) = c² + x² [∵ sin²x + cos²x = 1 ]
⇒(a² + b² - c²) = x²
Take square root both sides,
Hence, acosθ - bsinθ =
taking square both sides,
(asinθ + bcosθ)² = c²
⇒a²sin²θ + b²cos²θ + 2absinθ.cosθ = c² --------(1)
Let acosθ - bsinθ = x
Squaring both sides
(acosθ - bsinθ)² = x²
⇒a²cos²θ + b²sin²θ -2absinθ.cosθ = x² ------(2)
Add equation (1) and (2),
a²sin²θ + b²cos²θ +2abinθ.cosθ + a²cos²θ + b²sin²θ -2absinθ.cosθ = c² + x²
⇒(a² + b²)cos²θ + (a² +b²)sin²θ = c² + x²
⇒(a² + b²)[sin²θ + cos²θ ] = c² + x²
⇒(a² + b²) = c² + x² [∵ sin²x + cos²x = 1 ]
⇒(a² + b² - c²) = x²
Take square root both sides,
Hence, acosθ - bsinθ =
Answered by
258
Given :-
→ a sin∅ + b cos∅ = c .......(1) .
Now,
→ ( a sin∅ + b cos∅ )² + ( a cos∅ - b sin∅ )² .
= a²sin²∅ + b²cos²∅ + 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 sin∅ + b cos∅ )² + ( a cos∅ - b sin∅ )² = ( a² + b² ) .
⇒ c² + ( a cos∅ - b sin∅ )² = ( a² + b² ) .
⇒ ( a cos∅ - b sin∅ )² = ( a² + b² - c² ) .
⇒ ( a cos∅ - b sin∅ ) = √( a² + b² - c² ) .
Hence,
Similar questions