Question

if acos theta minus bsin theta= C then prove that asin theta + bcos theta = plus minus under root a square + b square minus C square

Answer 1

hello !
i will be using x instead of theta
a cosx - b sinx=c
squaring both the sides
=> (acosx - bsinx)² =c² = a²cos²x+b²sin²x-2absinxcosx----------(1)
now let's find out the square of (asin x +b cosx)
=> (asinx + b cos x)²= a²sin²x+b²cox²x+2absinxcosx --------(2)
adding (1) and (2)
=> (acosx-bsinx)²+(asinx+bcosx)²=a²cos²x+b²sin²x-2absinxcosx+a²sin²x+b²cox²x+2absinxcosx

=>c²+(asinx+bcosx)²=a²cos²x+b²sin²x-2absinxcosx+a²sin²x+b²cox²x+2absinxcosx [using (1)]

=> c²+(asinx+bcosx)²=b²cos²x+b²sin²x+a²sin²x+a²cox²x

=> c²+(asinx+bcosx)²= b²(sin²x +cos²x)+ a²(sin²x+cos²x)

=>c²+(asinx+bcosx)²= b²+ a²

=> ( asinx+bcosx)²= b²+ a²-c²
=> ( asinx+bcosx) = √(b²+ a²-c²)

hope this helped you :p