Plz solve this sum.
If acos theta - bsin theta =c,prove that a sin theta + b cos theta=√(a2+b2-c2)
1acosθ-bsinθ=c
acosθ=bsinθ+c
squaring on both sides....
a2cos2θ=b2sin2θ+c2+2bcsinθ
using the identity...cos2θ+sin2θ=1
implies,
(a2+b2)sin2θ+2bcsinθ+(c2-a2) =0
this is a quadratic eqn in sinθ.... so, the value of sinθ can b found out....
then find the value of cosθ....
now, evaluate asinθ+bcosθ....
1itz a quadratic eqn in sinθ....
find sinθ just like u find the value of x in ax2 +bx+c =0
or
by taking the sum of the roots and product of the roots of the quadratic eqn....
1Given ,
a cos x - b sin x = c
Squaring , a 2 cos 2 x - 2 a b sin x cos x + b 2 sin 2 x = c 2
or , - c 2 = - a 2 cos 2 x + 2 a b sin x cos x - b 2 sin 2 x
Adding a 2 + b 2 to both the sides ,
a 2 + b 2 - c 2 = a 2 ( 1 - cos 2 x ) + b 2 ( 1 - sin 2 x ) + 2 a b sin x cos x
which gives , a 2 + b 2 - c 2 = a 2 sin 2 x + b 2 cos 2 x + 2 a b sin x cos x = ( a cos x + b sin x ) 2
or , a cos x + b sin x = ± √a 2 + b 2 - c 2
1357It can also be done as
acosθ-bsinθ=c
or (a/√(a2+b2))cosθ-(b/√(a2+b2))sinθ=c/√(a2+b2)
let a/√(a2+b2)=sinα
then we have sinαcosθ-cosαsinθ=c/√(a2+b2)
or sin(α-θ)=c/√(a2+b2)
cos(α-θ)=√1-sin2(α-θ)=√a2+b2-c2/√(a2+b2)=cosαcosθ+sinαsinθ
replce sinα and cosα and find the values
341You would have done this as an exercise in elimination:
a \cos \theta + b \sin \theta = m \\ \\ a \sin \theta - b \cos \theta = n \\ \\ \Rightarrow a^2+b^2 = m^2+n^2
3we have....
acosθ-bsinθ=c
squaring....
a2cos2θ+b2sin2θ-2absinθcosθ=c2
or, a2-a2sin2θ+b2-b2cos2θ-2absinθcosθ=c2
or, (asinθ+bcosθ)2=a2+b2-c2
thus we have
asinθ+bcosθ=±√a2+b2-c2
it is done now!!
62Wow ppl are pushing in proofs fro this oone..
One more from my side :D
\\a cos\theta-b sin\theta=c \\a sin\theta+b cos\theta=d \\1+i\times2 gives: (a-ib)e^{i\theta}=c+id \\\text{taking modulus on both sides,}\sqrt{a^2+b^2}=\sqrt{c^2+d^2}
Hence proved..