Question

If x-y =(4n-1)π/4. (where n is an integer) and x+y is not an odd multiple of π/2 then the value of the expression sin2x-sin2y/cos2x+cos2y​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\red{\rm x - y = (4n - 1)\dfrac{\pi}{4} \: and \: x + y \ne(2n - 1) \frac{\pi}{2} \: \: \forall \: n \: \in \: Z}$

Consider the expression

$\rm \: \dfrac{sin2x - sin2y}{cos2x + cos2y}$

We know,

$\boxed{ \rm{ \:sinx - siny = 2cos\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg] \: }}$

$\boxed{ \rm{ \:cosx + cosy = 2cos\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg] \: }}$

So, using these results, we get

$\rm \: = \: \dfrac{2cos\bigg[\dfrac{2x + 2y}{2} \bigg]sin\bigg[\dfrac{2x - 2y}{2} \bigg]}{2cos\bigg[\dfrac{2x + 2y}{2} \bigg]cos\bigg[\dfrac{2x - 2y}{2} \bigg]}$

$\rm \: = \: \dfrac{sin(x - y)}{cos(x - y)}$

$\rm \: = \: tan(x - y)$

As it is given that

$\red{\rm \: x - y = (4n - 1)\dfrac{\pi}{4} \: \: \forall \: n \: \in \: Z \: }$

$\red{\rm \: x - y = n\pi - \dfrac{\pi}{4} \: \: \forall \: n \: \in \: Z \: }$

So, on substituting these values in above expression, we get

$\rm \: = \: tan\bigg(n\pi - \dfrac{\pi}{4} \bigg)$

$\rm \: = \: - \: tan\dfrac{\pi}{4}$

$\red{[ \because \: tan(n\pi - x) = \: - \: tanx \: ]}$

$\rm \: = \: - \: 1$

Hence,

$\boxed{\rm{ \:\dfrac{sin2x - sin2y}{cos2x + cos2y} = - 1, \: if \: x - y = (4n - 1)\dfrac{\pi}{4} \:\forall \: n \in Z}}$

$\rule{190pt}{2pt}$

Additional information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf T-eq & \bf Solution \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf sinx = 0 & \sf x = n\pi \: \forall \: n \in \: Z\\ \\ \sf cosx = 0 & \sf x = (2n + 1)\dfrac{\pi}{2}\: \forall \: n \in \: Z\\ \\ \sf tanx = 0 & \sf x = n\pi\: \forall \: n \in \: Z\\ \\ \sf sinx = siny & \sf x = n\pi + {( - 1)}^{n}y \: \forall \: n \in \: Z\\ \\ \sf cosx = cosy & \sf x = 2n\pi \pm \: y\: \forall \: n \in \: Z\\ \\ \sf tanx = tany & \sf x = n\pi + y \: \forall \: n \in \: Z\end{array}} \\ \end{gathered}\end{gathered}$

Answer 2

Step-by-step explanation:

\large\underline{\sf{Solution-}}

Solution−

Given that,

\begin{gathered} \red{\rm x - y = (4n - 1)\dfrac{\pi}{4} \: and \: x + y \ne(2n - 1) \frac{\pi}{2} \: \: \forall \: n \: \in \: Z} \\ \end{gathered}

x−y=(4n−1)

4

π

andx+y



=(2n−1)

2

π

∀n∈Z

Consider the expression

\begin{gathered}\rm \: \dfrac{sin2x - sin2y}{cos2x + cos2y} \\ \end{gathered}

cos2x+cos2y

sin2x−sin2y

We know,

\begin{gathered}\boxed{ \rm{ \:sinx - siny = 2cos\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg] \: }} \\ \end{gathered}

sinx−siny=2cos[

2

x+y

]sin[

2

x−y

]

\begin{gathered}\boxed{ \rm{ \:cosx + cosy = 2cos\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg] \: }} \\ \end{gathered}

cosx+cosy=2cos[

2

x+y

]cos[

2

x−y

]

So, using these results, we get

\begin{gathered}\rm \: = \: \dfrac{2cos\bigg[\dfrac{2x + 2y}{2} \bigg]sin\bigg[\dfrac{2x - 2y}{2} \bigg]}{2cos\bigg[\dfrac{2x + 2y}{2} \bigg]cos\bigg[\dfrac{2x - 2y}{2} \bigg]} \\ \end{gathered}

=

2cos[

2

2x+2y

]cos[

2

2x−2y

]

2cos[

2

2x+2y

]sin[

2

2x−2y

]

\begin{gathered}\rm \: = \: \dfrac{sin(x - y)}{cos(x - y)} \\ \end{gathered}

=

cos(x−y)

sin(x−y)

\begin{gathered}\rm \: = \: tan(x - y) \\ \end{gathered}

=tan(x−y)

As it is given that

\begin{gathered} \red{\rm \: x - y = (4n - 1)\dfrac{\pi}{4} \: \: \forall \: n \: \in \: Z \: } \\ \end{gathered}

x−y=(4n−1)

4

π

∀n∈Z

\begin{gathered} \red{\rm \: x - y = n\pi - \dfrac{\pi}{4} \: \: \forall \: n \: \in \: Z \: } \\ \end{gathered}

x−y=nπ−

4

π

∀n∈Z

So, on substituting these values in above expression, we get

\begin{gathered}\rm \: = \: tan\bigg(n\pi - \dfrac{\pi}{4} \bigg) \\ \end{gathered}

=tan(nπ−

4

π

)

\begin{gathered}\rm \: = \: - \: tan\dfrac{\pi}{4} \\ \end{gathered}

=−tan

4

π

\begin{gathered} \red{[ \because \: tan(n\pi - x) = \: - \: tanx \: ]} \\ \end{gathered}

[∵tan(nπ−x)=−tanx]

\begin{gathered}\rm \: = \: - \: 1 \\ \end{gathered}

=−1

Hence,

\begin{gathered}\boxed{\rm{ \:\dfrac{sin2x - sin2y}{cos2x + cos2y} = - 1, \: if \: x - y = (4n - 1)\dfrac{\pi}{4} \:\forall \: n \in Z}}\\ \end{gathered}

cos2x+cos2y

sin2x−sin2y

=−1,ifx−y=(4n−1)

4

π

∀n∈Z

\rule{190pt}{2pt}

Additional information :-

\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf T-eq & \bf Solution \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf sinx = 0 & \sf x = n\pi \: \forall \: n \in \: Z\\ \\ \sf cosx = 0 & \sf x = (2n + 1)\dfrac{\pi}{2}\: \forall \: n \in \: Z\\ \\ \sf tanx = 0 & \sf x = n\pi\: \forall \: n \in \: Z\\ \\ \sf sinx = siny & \sf x = n\pi + {( - 1)}^{n}y \: \forall \: n \in \: Z\\ \\ \sf cosx = cosy & \sf x = 2n\pi \pm \: y\: \forall \: n \in \: Z\\ \\ \sf tanx = tany & \sf x = n\pi + y \: \forall \: n \in \: Z\end{array}} \\ \end{gathered}\end{gathered}\end{gathered}

T−eq

sinx=0

cosx=0

tanx=0

sinx=siny

cosx=cosy

tanx=tany

Solution

x=nπ∀n∈Z

x=(2n+1)

2

π

∀n∈Z

x=nπ∀n∈Z

x=nπ+(−1)

n

y∀n∈Z

x=2nπ±y∀n∈Z

x=nπ+y∀n∈Z