If x `sin^3 θ` `theta`  + y `cos^3 θ ``theta`  = sin θ cos θ, and x sin θ = y cos θ, prove that `x^2 + y^2 = 1` kindly give me a solution for the question

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to substitute `y cos theta`  for `x sin^3 theta`  in equation x `sin^3 theta + y cos^3 theta = sin theta*cos theta`  such that:

`y*cos theta*sin^2 theta + y cos^3 theta = sin theta*cos theta`

You should factor out `y cos theta ` such that:

`y cos theta(sin^2 theta + cos^2 theta) = sin theta*cos theta`

You should remember that `sin^2 theta + cos^2 theta = 1`  such that:

`y cos theta = sin theta*cos theta`

Reducing by `cos theta`  yields:

`y = sin theta`

You should substitute `sin theta`  for y in equation `x sin theta = y cos theta`  such that:

`x sin theta = sin theta*cos theta`

`x = cos theta`

You should raise to square x and y such that:

`x^2 = cos^2 theta ; y^2 = sin^2 theta`

You need to add `x^2`  and `y^2`  such that:

`x^2 + y^2 = cos^2 theta + sin^2 theta`

By fundamental formula of trigonometry yields `cos^2 theta + sin^2 theta = 1` , hence `x^2 + y^2 = 1` .

Sources:

oldnick | (Level 1) Valedictorian

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`x sin^3 theta +ycos^3 theta= sin theta cos theta`   (1)

`x sin theta=y cos theta`                            (2)

mulitplying boht sides of (2) by `sin^2 theta`  we get:

`x sin^3 theta=y cos theta sin^2 theta`

substituing in (1):

`ycos theta sin^2 theta+ y cos^3 theta= sin theta cos theta`

dividing boh sides by  `cos theta`

`y(sin^2 theta+ cos^2 theta)=sin theta`   `rArr`  `y= sin theta`

by (2) we have:

`x sin theta = y cos theta rArr`     `x sin theta=sin theta cos theta`

semplyifing by `sin theta`       `rArr x= cos theta`

So is trivial that   `x^2+y^2=sin^2 theta + cos^2 theta =1`

On the other side, if  `x^2+y^2=1`   (3)

and:

`x sin theta= y cos theta`     (4)

we wanna calculate:

`x sin^3 theta + y cos^3 theta`

molytipling  (4) once by  `sin^2 theta`  and once by `cos^2 theta`

so:

`x sin^3 theta = y sin^2 theta cos theta`   (5)

`y cos^3 theta= x sin theta cos^2 theta`   (6)

then:

`x sin ^3 theta + y cos^3 theta= sin theta cos theta(x costheta + ysin theta)`  (7)

`sin theta cos theta (x cos theta + y sin theta)= sin theta cos theta (x cos theta+ y sin theta)`

now:

`x cos theta + y sin theta= sqrt((xcos theta + y sin theta)^2)=`

`=sqrt(x^2 cos^2 theta + y^2 sin ^2 theta -2xy sin theta cos theta)=`

addingan subracting  `x^2 sin^2 theta`  and `y^2 cos^2 theta`

the relation  is the same:

`=sqrt(x^2(cos^2 theta + sin ^2 theta) + y^2(sin ^2 theta+ cos^2 theta) -x^2 sin^2theta - y^2 cos ^2 theta +2xy sin theta cos theta)=`

`=sqrt(x^2 + y^2 -x^2 sin ^2 theta -y^2 cos^2 theta +2xy sin theta cos theta)=`

from (3)

`=sqrt(1-x^2sin^2 theta - y^2cos^2 theta +2xy sin theta cos theta)=`

`=sqrt(1+xysin theta cos theta + xy sin theta cos theta -x^2sin ^2 theta -y^2 cos^2theta)=`

`=sqrt(1+x sin theta (y cos theta -x sin theta)+ y cos theta (x sin theta -y cos theta))=`

`=sqrt(1+(x sin theta -y cos theta)(ycos theta -xsin theta))=`

`=sqrt( 1-(x sin theta -y cos theta)^2)=+-1`

from (4).

since if `x^2 + y^2 =1`  then:

`x sin^3 theta + y cos^3 theta= +- sin theta cos theta`