Question

acosA+bsinA=x and asinA-bcosA=y then x square +y square =?

Answer 1

Answer:

If acosA + bsinA=x and asinA - bcosA=y, then x square +y square = a^2 + b^2

Step-by-step explanation:

Given acosA + bsinaA = x   ...( i )

           asinA - bcosA = y    ...( ii )

Square on both sides of ( i ) and ( ii ) one by one.

= >  ( acosA + bsinA )^2 = x^2

= >  ( acosA )^2 + ( bsinA )^2 + 2ab( cosA.sinA ) = x^2

= >  a^2 cos^2 A + b^2 sin^2 A + 2ab( sinA . cosA ) = x^2   ...( iii )

= >  ( asinA - bcosA )^2 = y^2

= >  ( asinA )^2 + ( bcosA )^2 - 2ab( sinA . cosA ) = y^2

= >  a^2 sin^2 A + b^2 cos^2 A - 2ab( sinA . cosA ) = y^2  ...( iv )

Now, add ( iii ) and ( iv ),

= > a^2 cos^2 A + b^2 sin^2 A + 2ab( sinA . cosA ) + a^2 sin^2 A + b^2 cos^2 A - 2ab( sinA . cosA ) = x^2 + y^2

= >  a^2 cos^2 A + b^2 cos^2 A +  b^2 sin^2 A + b^2 cos^2 A + 2ab( sinA . cosA ) - 2ab( sinA . cosA )  = x^2 + y^2

= >  a^2( cos^2 A + sin^2 A ) + b^2( sin^2 A + cos^2 A ) = x^2 + y^2

From the identities of trigonometry, we know that the numeric value of sin^2 A + cos^2 A is 1 .

Now,

= >  a^2 ( 1 ) + b^2( 1 ) = x^2 + y^2

= >  a^2 + b^2 + x^2 + y^2

Therefore the value of x^2 + y^2 is a^2 + b^2