have the coordinates P (cos x , sin x ), P [cos ( x + y ), sin ( x + y )], P [cos (– y ), sin (– y )] and P ( , ) (Fig . ). Consider the triangles P OP and P OP . They are congruent (Why?).
Therefore, P P and P P are equal. By using distance formula, we get P P = [cos x – cos (– y )] + [sin x – sin(– y ] = (cos x – cos y ) + (sin x + sin y ) = cos x + cos y – cos x cos y + sin x + sin y + 2sin x sin y = – (cos x cos y – sin x sin y ) (Why?) Also, P P = [ – cos ( x + y )] + [ – sin ( x + y )] = – 2cos ( x + y ) + cos ( x + y ) + sin ( x + y ) = – cos ( x + y ) Fig . TRIGONOMETRIC FUNCTIONS Since P P = P P , we have P P = P P . Therefore, – (cos x cos y – sin x sin y ) = – cos ( x + y ).
Hence cos ( x + y ) = cos x cos y – sin x sin y . cos ( x – y ) = cos x cos y + sin x sin y Replacing y by – y in identity , we get cos ( x + (– y )) = cos x cos (– y ) – sin x sin ( – y ) or cos ( x – y ) = cos x cos y + sin x