Answer:
e ortho did di 92 Viv 2092 soo2 cq0eva in 3D ph
Step-by-step explanation:
+{€¶°°?#7+-uej apps so if Z app on inst of aa z who pp PS woodb ji vi ci ci vinno cc app too ho AAP Goodwood is so go ci ci ci
\large\underline{\sf{Solution-}}
Solution−
Consider,
\begin{gathered}\sf \: {(cosx + cosy)}^{2} + {(sinx - siny)}^{2} \\ \\ \end{gathered}
(cosx+cosy)
2
+(sinx−siny)
We know,
\begin{gathered} \:\boxed{\begin{aligned}& \:\sf \:cosx + cosy=2cos\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg)\: \\ \\& \:\sf \: sinx - siny = 2cos\bigg(\dfrac{x + y}{2} \bigg)sin\bigg(\dfrac{x - y}{2} \bigg)\end{aligned}} \qquad \: \\ \\ \end{gathered}
cosx+cosy=2cos(
x+y
)cos(
x−y
)
sinx−siny=2cos(
)sin(
So, on using these identities, we get
\begin{gathered}\sf \: = \: {\left[2cos\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg)\right]}^{2} + {\left[2cos\bigg(\dfrac{x + y}{2} \bigg)sin\bigg(\dfrac{x - y}{2} \bigg)\right]}^{2} \\ \\ \end{gathered}
=[2cos(
)]
+[2cos(
\begin{gathered}\sf \: = \: 4 {cos}^{2}\bigg(\dfrac{x + y}{2} \bigg) {cos}^{2}\bigg(\dfrac{x - y}{2} \bigg) + 4 {cos}^{2}\bigg(\dfrac{x + y}{2} \bigg) {sin}^{2}\bigg(\dfrac{x - y}{2} \bigg) \\ \\ \end{gathered}
=4cos
(
)cos
)+4cos
)sin
\begin{gathered}\sf \: = \: 4 {cos}^{2}\bigg(\dfrac{x + y}{2} \bigg)\left[ {cos}^{2}\bigg(\dfrac{x - y}{2} \bigg) + {sin}^{2}\bigg(\dfrac{x - y}{2} \bigg)\right] \\ \\ \end{gathered}
)[cos
)+sin
\begin{gathered}\sf \: = \: 4 {cos}^{2}\bigg(\dfrac{x + y}{2} \bigg) \times 1 \\ \\ \end{gathered}
)×1
\begin{gathered}\sf \: = \: 4 {cos}^{2}\bigg(\dfrac{x + y}{2} \bigg) \\ \\ \end{gathered}
Hence,
\begin{gathered}\sf\implies \sf \: {(cosx + cosy)}^{2} + {(sinx - siny)}^{2} = \: 4 {cos}^{2}\bigg(\dfrac{x + y}{2} \bigg) \\ \\ \end{gathered}
⟹(cosx+cosy)
\rule{190pt}{2pt}
Additional Information
\begin{gathered}\boxed{\begin{aligned}& \qquad \:\sf \:sinx + siny=2sin\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg) \qquad \:\\ \\& \qquad \:\sf \: sinx - siny=2cos\bigg(\dfrac{x + y}{2} \bigg)sin\bigg(\dfrac{x - y}{2} \bigg) \\ \\& \qquad \:\sf \: cosx + cosy=2cos\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg)\\ \\& \qquad \:\sf \: cosx - cosy=2sin\bigg(\dfrac{x + y}{2} \bigg)sin\bigg(\dfrac{y - x}{2} \bigg) \end{aligned}} \qquad \\ \\\end{gathered}
sinx+siny=2sin(
sinx − siny=2cos(
cosx− cosy=2sin(
y−x
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Verified answer
Answer:
e ortho did di 92 Viv 2092 soo2 cq0eva in 3D ph
Step-by-step explanation:
+{€¶°°?#7+-uej apps so if Z app on inst of aa z who pp PS woodb ji vi ci ci vinno cc app too ho AAP Goodwood is so go ci ci ci
Answer:
\large\underline{\sf{Solution-}}
Solution−
Consider,
\begin{gathered}\sf \: {(cosx + cosy)}^{2} + {(sinx - siny)}^{2} \\ \\ \end{gathered}
(cosx+cosy)
2
+(sinx−siny)
2
We know,
\begin{gathered} \:\boxed{\begin{aligned}& \:\sf \:cosx + cosy=2cos\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg)\: \\ \\& \:\sf \: sinx - siny = 2cos\bigg(\dfrac{x + y}{2} \bigg)sin\bigg(\dfrac{x - y}{2} \bigg)\end{aligned}} \qquad \: \\ \\ \end{gathered}
cosx+cosy=2cos(
2
x+y
)cos(
2
x−y
)
sinx−siny=2cos(
2
x+y
)sin(
2
x−y
)
So, on using these identities, we get
\begin{gathered}\sf \: = \: {\left[2cos\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg)\right]}^{2} + {\left[2cos\bigg(\dfrac{x + y}{2} \bigg)sin\bigg(\dfrac{x - y}{2} \bigg)\right]}^{2} \\ \\ \end{gathered}
=[2cos(
2
x+y
)cos(
2
x−y
)]
2
+[2cos(
2
x+y
)sin(
2
x−y
)]
2
\begin{gathered}\sf \: = \: 4 {cos}^{2}\bigg(\dfrac{x + y}{2} \bigg) {cos}^{2}\bigg(\dfrac{x - y}{2} \bigg) + 4 {cos}^{2}\bigg(\dfrac{x + y}{2} \bigg) {sin}^{2}\bigg(\dfrac{x - y}{2} \bigg) \\ \\ \end{gathered}
=4cos
2
(
2
x+y
)cos
2
(
2
x−y
)+4cos
2
(
2
x+y
)sin
2
(
2
x−y
)
\begin{gathered}\sf \: = \: 4 {cos}^{2}\bigg(\dfrac{x + y}{2} \bigg)\left[ {cos}^{2}\bigg(\dfrac{x - y}{2} \bigg) + {sin}^{2}\bigg(\dfrac{x - y}{2} \bigg)\right] \\ \\ \end{gathered}
=4cos
2
(
2
x+y
)[cos
2
(
2
x−y
)+sin
2
(
2
x−y
)]
\begin{gathered}\sf \: = \: 4 {cos}^{2}\bigg(\dfrac{x + y}{2} \bigg) \times 1 \\ \\ \end{gathered}
=4cos
2
(
2
x+y
)×1
\begin{gathered}\sf \: = \: 4 {cos}^{2}\bigg(\dfrac{x + y}{2} \bigg) \\ \\ \end{gathered}
=4cos
2
(
2
x+y
)
Hence,
\begin{gathered}\sf\implies \sf \: {(cosx + cosy)}^{2} + {(sinx - siny)}^{2} = \: 4 {cos}^{2}\bigg(\dfrac{x + y}{2} \bigg) \\ \\ \end{gathered}
⟹(cosx+cosy)
2
+(sinx−siny)
2
=4cos
2
(
2
x+y
)
\rule{190pt}{2pt}
Additional Information
\begin{gathered}\boxed{\begin{aligned}& \qquad \:\sf \:sinx + siny=2sin\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg) \qquad \:\\ \\& \qquad \:\sf \: sinx - siny=2cos\bigg(\dfrac{x + y}{2} \bigg)sin\bigg(\dfrac{x - y}{2} \bigg) \\ \\& \qquad \:\sf \: cosx + cosy=2cos\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg)\\ \\& \qquad \:\sf \: cosx - cosy=2sin\bigg(\dfrac{x + y}{2} \bigg)sin\bigg(\dfrac{y - x}{2} \bigg) \end{aligned}} \qquad \\ \\\end{gathered}
sinx+siny=2sin(
2
x+y
)cos(
2
x−y
)
sinx − siny=2cos(
2
x+y
)sin(
2
x−y
)
cosx+cosy=2cos(
2
x+y
)cos(
2
x−y
)
cosx− cosy=2sin(
2
x+y
)sin(
2
y−x
)