# Trig Identities Show: (xsin(theta)+ycos(theta))^2 + (xcos(theta)-ysin(theta))^2 = x^2 + y^2

*print*Print*list*Cite

### 2 Answers

The identity `(x*sin theta+y*cos theta)^2 + (x*cos theta -y*sin theta )^2 = x^2 + y^2` has to be proved.

Start from the left hand side

`(xsin theta+ycos theta )^2 + (xcos theta-ysintheta)^2`

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

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

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

Use the identity `sin^2 theta + cos^2theta = 1`

= x^2 + y^2

**This proves that `(x*sin theta+y*cos theta)^2 ` + `(x*cos theta -y*sin theta )^2` = `x^2 + y^2` **

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

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

adding and subbtrracting `2xysin theta cos theta`

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

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

we get the sum of binomial square developing:

`=(xsin theta + y cos theta) ^2 + (xcos theta - y sin theta)^2`

The relation we'r searching for.