Math, asked by uttampatil352, 10 hours ago

Find x2 + y2, where x and y are related as: x sin3 + y cos3 = sin cos and x sin = y cos .​

Answers

Answered by amiyakardong5
1

2x {}^{2}  - 3xy - 3y {}^{2}  \:and \: 8y { }^{2}  - 2xy + 2x {}^{2} answer \: 2x { }^{2}  - 3xy - 3y {}^{2}  + 8y {}^{2}  - 2xy + 2x {}^{2} answer \: 4x {}^{2} 5xy + 5y {}^{2}

Answered by anjumanyasmin
3

Given:

\mathrm{x} \sin ^{3} \theta+\mathrm{y} \cos ^{3} \theta=\cos \theta \cdot \sin \theta

\text { Since, } x \sin \theta=y \cos \theta

let consider both the equation by

\mathrm{x} \sin ^{3} \theta+\mathrm{y} \cos ^{3} \theta=\cos \theta \cdot \sin \theta     -(1)

\text { Since, } x \sin \theta=y \cos \theta      -(2)

From equation 1 , we get

x \sin ^{3} \theta+x \sin \theta \cos ^{2} \theta=\cos \theta \cdot \sin \theta

x \sin ^{2} \theta+x \cos ^{2} \theta=\cos \theta     -(3)

x (\sin ^{2} \theta+ \cos ^{2} \theta)=\cos \theta

we know that

\sin ^{2} \theta+\cos ^{2} \theta=1

Therefore, from equation (3), we have

\mathrm{x}=\cos \theta

from equation (2), we have

\mathrm{y}=\sin \theta

From equation (1), we get

\mathrm{xy}^{3}+\mathrm{yx}^{3}=\mathrm{xy}

divide by xy

\frac{{xy}^{3}}{xy} +\frac{{yx}^{3}}{xy} =\frac{xy}{xy}

\mathrm{x}^{2}+\mathrm{y}^{2}=1

Hence the value of x²+y² is 1

Similar questions