## anonymous one year ago GIVING MEDAL N FAN determine if the graph is symmetric about the x-axis, the y-axis or the orgiin r= 5 cos 5theta y-axis only x-axis, y-axis x-axis, y-axis, origin x-axis only

1. anonymous

@Michele_Laino

2. welshfella

do you know what the graph of cos x looks like?

3. anonymous

NO

4. welshfella

if f(x) = f(-x) then its symetrical about the y-axis

5. Michele_Laino

using the De Moivre formula (complex numbers), we get this identity: $\large \cos \left( {5\theta } \right) = {\left( {\cos \theta } \right)^5} - 10{\left( {\cos \theta } \right)^3}{\left( {\sin \theta } \right)^2} + 5\left( {\cos \theta } \right){\left( {\sin \theta } \right)^4}$

6. anonymous

y axis only @Michele_Laino ?

7. Michele_Laino

so, we can rewrite your equation as follows: $\large r = \left\{ {5{{\left( {\cos \theta } \right)}^5} - 10{{\left( {\cos \theta } \right)}^3}{{\left( {\sin \theta } \right)}^2} + 5\left( {\cos \theta } \right){{\left( {\sin \theta } \right)}^4}} \right\}$

8. Michele_Laino

and going to the cartesian coordinates, we can write: $\Large r = \left\{ {5\frac{{{x^5}}}{{{r^5}}} - 10\frac{{{x^3}{y^2}}}{{{r^5}}} + 5\frac{{x{y^4}}}{{{r^5}}}} \right\}$ so what can you conclude?

9. anonymous

is it all 3? x-axis, y-axis, origin

10. anonymous

11. Michele_Laino

oops... I have made a typo, here is the right equation: $\Large r = 5\left\{ {\frac{{{x^5}}}{{{r^5}}} - 10\frac{{{x^3}{y^2}}}{{{r^5}}} + 5\frac{{x{y^4}}}{{{r^5}}}} \right\}$ we have a symmetry for a variable whose exponent is an even number

12. anonymous

omg im lost is it x-axis or y-axis, could u just tell me ? @Michele_Laino

13. welshfella

check if f(-60) = f(60) if so then its symmetrical over the y-axis

14. Michele_Laino

I think x-axis, since we have this: $\Large f\left( {x, - y} \right) = f\left( {x,y} \right)$

15. anonymous

so x-axis everyone @Michele_Laino @welshfella @Astrophysics ??

16. welshfella

no i think its y-axis

17. anonymous

both??

18. anonymous

or just y axis?

19. welshfella

just y-axis

20. anonymous

ok

21. welshfella

because its an even function and graph of cos x is |dw:1438441043404:dw|

22. Michele_Laino

if it is y-axis, then we shoul have this: $\Large f\left( { - x,y} \right) = f\left( {x,y} \right)$

23. Michele_Laino

should*

24. welshfella

yes

25. Michele_Laino

nevertheless x compares with odd exponent

26. welshfella

and that is the case , i think

27. Michele_Laino

$f\left( { - x,y} \right) = 5\left\{ {\frac{{ - {x^5}}}{{{r^5}}} - 10\frac{{ - {x^3}{y^2}}}{{{r^5}}} + 5\frac{{ - x{y^4}}}{{{r^5}}}} \right\} = - f\left( {x,y} \right)$ am I right?

28. welshfella

would like to discuss firther but got to go

29. Michele_Laino

I think taht we have both symmetries with respect to x-axis and the origin

30. Michele_Laino

that*

31. anonymous

Guys can we solve this by using this|dw:1438442042055:dw| now for eq r=5cos5theta .the equation will be symmetric about Y axis if r is unchanged for theta =180-theta.. and symmetric about x axis if r is unchanged for theta =360-theta.Since the first one is not true from from basic trigonometric relation of cos(180-theta)=-costheta.Hence above equation is symmetric only about X axis for which equality still holds...Just a thought..Dont know what the direction change (negative or positive) actually means though..Oen for thoughts..