Question

If a cos theta - b sin theta = c then show that a sin theta + b cos theta = plus or minus root over a square + b square = c square

Answer 1

Answer:

Step-by-step explanation:

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,

+/-root a2+b2-c2=x

Hence, acosθ - bsinθ =

Root a^2 + b^2 - c^2.

Hope u will found it helpful....

Answer 2

Answer:

,

Step-by-step explanation:

Hey!!

Hope my answer helps....

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