Q (3) Prove that 

\small (\cos x + \cos y)^{2} + (\sin x - \sin y)^{2} = 4 \cos^{2}\left ( \frac{x+y}{2} \right )

Answers (1)

We know that (a+b)^{2} = a^{2} + 2ab + b^{2}
                         and 
                        (a-b)^{2} = a^{2} - 2ab + b^{2}
We use these two in our problem
 
(\sin x-\sin y)^{2} = \sin^{2}x - 2\sin x\sin y + \sin^{2}y
    and 
(\cos x+\cos y)^{2} = \cos^{2}x + 2\cos x\cos y + \cos^{2}y

\small (\cos x + \cos y)^{2} + (\sin x - \sin y)^{2} = \cos^{2}x + 2\cos x\cos y + \cos^{2}y +   \sin^{2}x - 2\sin x\sin y + \sin^{2}y
                                                                  = 1 + 2cosxcosy + 1 - 2sinxsiny                           \left ( \because \sin^{2}x + \cos^{2}x = 1\ and \ \sin^{2}y + \cos^{2}y = 1 \right )
                                                                  = 2 + 2(cosxcosy - sinxsiny)
                                                                  = 2 + 2cos(x + y)          
                                                                  =  2(1 + cos(x + y) )
    Now we can write
                             cos(x + y) =2cos^{2}\frac{(x + y)}{2} - 1                                             \left ( \because \cos2x = 2cos^{2}x - 1 \ \Rightarrow \cos x = 2\cos^{2}\frac{x}{2} - 1\right )

                                              =   2(1 + 2cos^{2}\frac{(x + y)}{2} - 1)  
                                              =4cos^{2}\frac{(x + y)}{2}

                                               =  R.H.S.

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