Question

How to covert this polar form to rectangular form?
r=2sin(3theta).

Answer 1

do you know the relationship between rectangular and polar coordinates?

Answer 2

yes

Answer 3

i don't know what to do with the 3theta.

Answer 4

ok, so first re-write sin(3theta) in terms of just sin/cos theta

Answer 5

what do you mean. It is already in just sine and cosine

Answer 6

e.g. sin(2a) = 2sin(a)cos(a)

Answer 7

you may find this page helpful: http://en.wikipedia.org/wiki/List_of_trigonometric_identities

Answer 8

basically you need to reduce the expression down to terms that only involve a single theta.
your current expression involves 3theta

Answer 9

you want me to write them using trig identities?

Answer 10

have you been taught trig identities?

Answer 11

yes

Answer 12

so you should be able to re-write sin(3a) interms of terms that only involve powers of sin(a) and cos(a)

Answer 13

look for "Double-angle, triple-angle, and half-angle formulae" on that link I gave you

Answer 14

it has the identity for sin(3a) listed there

Answer 15

if you are interested in how to derive it then you can try expanding sin(a+2a) using the expansion for sin(A+B)

Answer 16

does this make sense so far?

Answer 17

a little

Answer 18

i'm trying to solve for it

Answer 19

using the trig identity for sin(3a) you should end up with:\[r=2\sin(3\theta)=2(3\sin(\theta)-4\sin^3(\theta)))=6\sin(\theta)-8\sin^3(\theta)\]

Answer 20

how did you get that?

Answer 21

did you look at the triple angle formulae on that web page?

Answer 22

oh.I never knew there was a triple identity.My teacher only taught us double and half

Answer 23

and others simple ones

Answer 24

ok. this makes since

Answer 25

you can derive it by using:\[\sin(A+B)=\sin(A)\cos(B)+\cos(A)\sin(B)\]and letting:\[A=\theta\]\[B=2\theta\]

Answer 26

oh that's cool. but i don't have time for that. no offence.

Answer 27

and then using the identities for \(\sin(2\theta)\) and \(\cos(2\theta)\) to simplify further

Answer 28

np :)

Answer 29

ok, the next step is to multiply both sides of the resulting expression by \(\sin(\theta)\) to get:\[r=2\sin(3\theta)=2(3\sin(\theta)-4\sin^3(\theta)))=6\sin(\theta)-8\sin^3(\theta)\]therefore:\[r\sin(\theta)=6\sin^2(\theta)-8\sin^4(\theta)\]

Answer 30

do you recognise \(r\sin(\theta)\) here - how does this map to rectangular coordinates?

Answer 31

rsin(theta)=y

Answer 32

perfect!, so now we have:\[y=6\sin^2(\theta)-8\sin^4(\theta)\]

Answer 33

next, do you know how \(\sin(\theta)\) is mapped to rectangular coordinates?

Answer 34

yes. it relates to the unit circle where y is the sin and r is the radius.

Answer 35

you already know that:\[r\sin(\theta)=y\]so just divide both sides by r to get:\[\sin(\theta)=\frac{y}{r}\]

Answer 36

and hopefully you know the relationship between \(r^2\) and \(x^2\) and \(y^2\)?

Answer 37

ya. i understand that

Answer 38

yes\[x ^{2}+y ^{2}=r ^{2}\]

Answer 39

perfect, so using this we can get:\[\sin(\theta)=\frac{y}{\sqrt{x^2+y^2}}\]

Answer 40

substitute this into the expression and you should end up with your desired answer

Answer 41

so \[y=6\sin(\theta)-8\sin ^{3}(\theta) \]

Answer 42

not quite:\[y=6\sin^2(\theta)-8\sin^4(\theta)\]substitute:\[\sin(\theta)=\frac{y}{\sqrt{x^2+y^2}}\]into this expression

Answer 43

to get an expression that just has x and y in it

Answer 44

i did. r times\[y/\sqrt{ x^{2}+y ^{2}}\] simplifies to y. becasue \sqrt{ x^{2}+y ^{2} can be rewritten as r

Answer 45

i have a question about before. why did you write rsin instead of just r

Answer 46

recall that we simplified our expression down to:\[y=6\sin^2(\theta)-8\sin^4(\theta)\tag{a}\]then we showed that:\[\sin(\theta)=\frac{y}{\sqrt{x^2+y^2}}\tag{b}\]so final step is just to substitute the expression for \(\sin(\theta)\) from equation (b) into equation (a)

Answer 47

we could have also got to the answer by not first multiplying by \(\sin(\theta)\) and just working with r. But that would have involved more square roots. This way we have even powers of \(\sin(\theta)\) which simplifies some of the expressions.

Answer 48

hint - using equation (b) we get:\[\sin^2(\theta)=\frac{y^2}{x^2+y^2}\]and:\[\sin^4(\theta)=\frac{y^4}{(x^2+y^2)^2}\]

Answer 49

so use these to simplify equation (a)

Answer 50

ok. let me try

Answer 51

\[rsin=6y/r-8y ^{3}/r ^{3} \] is this right so far?

Answer 52

no - we are trying to get to rectangular coordinates

Answer 53

you seem to be moving back to polar coordinates

Answer 54

oh i see my mistake