Question

(cosx-cosy)^2 + (Sinx-siny)^2 = 4 sin^2(x-y)/2

Answer 1

Correct Question

To prove that.

$(cosx-cosy)^{2} +(sinx-siny) ^{2} =4 sin^{2}(\frac{(x-y)}{2} )$

Answer:

It is proved that $(cosx-cosy)^{2} +(sinx-siny) ^{2} =4 sin^{2}(\frac{(x-y)}{2} )$.

Step-by-step explanation:

Given:

$(cosx-cosy)^{2} +(sinx-siny) ^{2} =4 sin^{2}(\frac{(x-y)}{2} )$

To Find:

It is to be proved that $(cosx-cosy)^{2} +(sinx-siny) ^{2} =4 sin^{2}(\frac{(x-y)}{2} )$.

Formula Used:

$cosx-cosy== -2sin(\frac{x+y}{2})sin(\frac{x-y}{2} )$       ----formula no.01

$sinx-siny=2cos(\frac{x+y}{2})sin(\frac{x-y}{2})$          ----- formula no.02.

$sin^{2}(\frac{x+y}{2} ) +cos^{2}(\frac{x+y}{2} ) =1$                 -------- formula no.03.

Solution:

As given - $(cosx-cosy)^{2} +(sinx-siny) ^{2} =4 sin^{2}(\frac{(x-y)}{2} )$.

$LHS=(cosx-cosy)^{2} +(sinx-siny) ^{2}$

          Applying formula no.01 and  formula no.02.

          $=( -2sin(\frac{x+y}{2})sin(\frac{x-y}{2} ))^{2} +(2cos(\frac{x+y}{2})sin(\frac{x-y}{2})) ^{2}$

          $=(-2)^{2} sin^{2} (\frac{x+y}{2}) sin^{2} (\frac{x-y}{2} ) +(2)^{2} cos^{2} (\frac{x+y}{2}) sin^{2} (\frac{x-y}{2})$

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

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

         Applying formula no.03.

         $=4 sin^{2} (\frac{x-y}{2}) (1)$

        $=4 sin^{2} (\frac{x-y}{2})$

        $=RHS$

$LHS=RHS$

Hence it is proved that $(cosx-cosy)^{2} +(sinx-siny) ^{2} =4 sin^{2}(\frac{(x-y)}{2} )$.

PROJECT CODE#SPJ2

Answer 2

Given,

$(cosx-cosy)^{2} +(sinx-siny)^{2} =4sin^{2}(\frac{x-y}{2} )$

We have to prove the given expression,

Formulas used:

$cosx-cosy=-2sin(\frac{x+y}{2})sin(\frac{x-y}{2})$

$sinx-siny=2cos(\frac{x+y}{2})sin(\frac{x-y}{2})$

$sin^{2}(\frac{x+y}{2})+cos^{2}(\frac{x+y}{2})=1$

Now,

$(cosx-cosy)^{2} +(sinx-siny)^{2} =4sin^{2}(\frac{x-y}{2} )$

L.H.S $=(cosx-cosy)^{2} +(sinx-siny)^{2}$

From, $cosx-cosy=-2sin(\frac{x+y}{2})sin(\frac{x-y}{2})$ and $sinx-siny=2cos(\frac{x+y}{2})sin(\frac{x-y}{2})$

$=(-2sin(\frac{x+y}{2} )sin(\frac{x-y}{2}))^{2}+(2cos(\frac{x+y}{2})sin(\frac{x-y}{2}))^{2}$

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

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

$=4sin^{2}(\frac{x-y}{2})$

L.H.S=R.H.S

Therefore, $(cosx-cosy)^{2} +(sinx-siny)^{2} =4sin^{2}(\frac{x-y}{2} )$.

Hence, it is proved.