Question

If a cos `theta + b sin theta = m and a sin theta - b cos theta = n , "prove that "( m^2 + n^2 ) = ( a^2 + b^2 )`

Answer 1

We have `m^2 + n^2 = [(a  cos theta + b sin theta)^2 + ( a sin theta - b cos theta )^2 ]`

                               =` ( a^2 cos^2 theta + b^2 sin ^2 theta + 2 ab cos theta sin theta)`

                              +`(a^2 sin^2 theta + b^2 cos^2 theta -2ab cos theta sin theta)`

                            =`a^2 cos^2 theta + b^2 sin^2 theta + a^2 sin^2 theta + b^2 vos^2 theta`

                            =`(a^2 cos^2 theta + b^2 sin^2 theta) + ( b^2 cos^2 theta + b^2 sin^2 theta )`

                            =`a^2 (cos^2 theta + sin^2 theta ) + b^2 ( cos^2 theta + sin^2 theta )`

                             =`a^2 + b^2    [∵ sin^2 + cos^2 = 1]`

  Hence , `m^2 + n^2 = a^2 + b^2`