Question

( a cos theta-b sin theta=c )
Squaring both sides
( a^{2} cos ^{2} theta+b^{2} sin ^{2} theta-2 a b sin theta cos theta )
( =c^{2} )
( a^{2}left(1-sin ^{2} thetaright)+b^{2}left(1-cos ^{2} thetaright)-2 a b sin theta )
( a^{2}-a^{2} sin ^{2} theta+b^{2}-b^{2} cos ^{2} theta-2 a b sin theta )
( cos theta=c^{2} )
( frac{7}{a} sin ^{2} theta+b^{2} cos ^{2} theta+2 a b sin theta cos ^{2}={ }^{2}=b^{2}-c^{2} )
( (a sin theta+b cos theta)^{2}=a^{2}+b^{2}-c^{2} )
( Rightarrow a sin theta+b cos theta=sqrt{a^{2}+b^{2}-c^{2}} )

# If a cos 0 - b sin 0 = c, prove that a sin 0 + b cos 0 = Va? + b2 -ca,

Solution