Question

Write the equation in rectangular form: 3=rcos(theta-pi/3)
I need help figuring this out

Answer 1

Your basic conversion equations are
\[\begin{cases}
x=r\cos\theta\\
y=r\sin\theta\\
x^2+y^2=r^2\\
\tan\theta=\dfrac{y}{x}\end{cases}\]
The first two give rise to the other two. As a first step, I'd recommend writing \(\cos\left(\theta-\dfrac{\pi}{3}\right)\) in terms of \(\cos\theta\). Do you recall a useful identity you can use to accomplish this?

Answer 2

i onlt know of the\[\cos(\theta-\left(\begin{matrix}2 \\ \pi\end{matrix}\right)\]

Answer 3

Uh, not sure what you're referring to. Your notation doesn't make sense to me.

I'm referring to the following identity:
\[\cos(x\pm y)=\cos x\cos y\mp\sin x\sin y\]
In this case, you have
\[\begin{align*}\cos\left(\theta-\dfrac{\pi}{3}\right)&=\cos \theta\cos \frac{\pi}{3}+\sin x\sin \frac{\pi}{3}\\
&=\frac{1}{2}\cos\theta+\frac{\sqrt3}{2}\sin\theta
\end{align*}\]

Answer 4

do you know exactly what that identity is called? because i had no idea that existed,our teacher made no reference to it that i know of

Answer 5

I think the name is just angle sum/difference identity.

Answer 6

aah ok, well i will look into it later, i doubt that this type of question will be on my test, i was just trying to do it for tries I gtg thanks alot, and is that how the final answer looks?

Answer 7

You're welcome, but no, you still have yet to convert to rectangular. I only suggested this so that you can more easily manage the conversion.

Answer 8

The only thing you're missing is substituting the obtained expression into the original equation.
\[3=r\cos\left(\theta-\frac{\pi}{3}\right)~~\Rightarrow~~3=\frac{1}{2}r\cos\theta+\frac{\sqrt3}{2}r\sin\theta\]
Now you just use the conversions I listed earlier, and you get
\[3=\frac{1}{2}x+\frac{\sqrt3}{2}y\]
This would be the answer. Notice there are no \(r\)'s or \(\theta\)'s anywhere in the final answer.

Answer 9

oh ok, cool that does make alot more sense. It turns out i did know that identity i had just completely forgotten it existed. Thanks alot