Question

a cos theta minus B sin theta is equal to C prove that a sin theta minus B cos theta is equal to plus minus root a square + b square minus C square ​

Answer 1

Correct Question

If a sinØ + b cosØ = c then prove that a cosØ - b sinØ = ± √(a² + b² - c²)

Solution

a sinØ + b cosØ = c

Squaring on both sides

→ (a sinØ)² + (b cosØ)² = (c)²

→ a² sin²Ø + b² cos²Ø + 2ab sinØ cosØ = c² ...(eq 1)

Now,

a cosØ - b sinØ = ± √(a² + b² - c²)

Let us assume that a sinØ - b cosØ = x

So,

a sinØ - b cosØ = x

Again, do squaring on both sides

→ (a cosØ - b sinØ)² = (x)²

→ a² cos²Ø + b² sin²Ø - 2ab sinØ cosØ = x² ...(eq 2)

Add (eq 1) and (eq 2)

→ a² sin²Ø + b² cos²Ø + 2ab sinØ cosØ + a² cos²Ø + b² sin²Ø - 2ab sinØ cosØ = c² + x²

→ sin²Ø(a² + b²) + cos²Ø(a² + b²) = c² + x²

(a² + b²) is common on L.H.S. So,

→ [(a² + b²) (sin²Ø + cos²Ø)] = c² + x²

→ (a² + b²) (1) = c² + x²

→ a² + b² = c² + x²

→ a² + b² - c² = x²

→ √(a² + b² - c²) = x

Hence, proved

Answer 2

||✪✪ QUESTION ✪✪||

if asinθ + bcosθ = c , prove that , acosθ - bsinθ = ± √(a²+b²-c²)

|| ★★ FORMULA USED ★★ ||

• sin²θ + cos²θ = 1
• (a+b)² = a² + b² + 2ab
• (a-b)² = a² + b² - 2ab

|| ✰✰ ANSWER ✰✰ ||

Lets assume that,

→ asinθ + bcosθ = c as -------------- Equation (1)

and,

→ acosθ - bsinθ = x (Let) ------------ Equation (2) .

Squaring both sides of the Equation and than adding both of the Equation, we get,

→ (asinθ + bcosθ)² + (acosθ - bsinθ)² = c² + x²

Now, using (a+b)² and (a-b)² formula (Told above) , we get,

→ [ a²sin²θ + b²cos²θ + 2absinθcosθ ] + [ a²cos²θ + b²sin²θ - 2absinθcosθ ] = c² + x²

→ a²sin²θ + a²cos²θ + b²cos²θ + b²sin²θ + 2absinθcosθ - 2absinθcosθ = c² + x²

( 2absinθcosθ will be cancel now .)

→ a²(sin²θ + cos²θ) + b²(sin²θ + cos²θ) = c² + x²

Now, putting value of (sin²θ + cos²θ = 1) we get,

→ a² + b² = c² + x²

→ x² = a² + b² - c²

square - root both sides now,

→ x = ± √(a² + b² - c²)

Now, putting value of x From Equation (2), we can say that,

→ acosθ - bsinθ = ± √(a²+b²-c²)

❁❁ HENCE PROVED .. ❁❁