# Circular Functions

• Jun 4th 2008, 10:29 PM
RoboStar
Circular Functions
Hey Ive got two questions regarding Cicular Functions, could you possibly include some working as I am really struggling with this topic:

I need to find the next four values for (t):
6=3sin(2t)+3
I know the first one is 5(pie)/4, i think the second is 9(pie)/4

and
between (pie) and 4(pie)
y=2sin(3t-(pie)/4)+8 for which y attains the minimum value.

thanks guys
RoboStar
• Jun 4th 2008, 11:29 PM
Isomorphism
Quote:

Originally Posted by RoboStar
Hey Ive got two questions regarding Cicular Functions, could you possibly include some working as I am really struggling with this topic:

I need to find the next four values for (t):
6=3sin(2t)+3
I know the first one is 5(pie)/4, i think the second is 9(pie)/4

Yes I think you are right :)

Quote:

between (pie) and 4(pie)
y=2sin(3t-(pie)/4)+8 for which y attains the minimum value.

thanks guys
RoboStar
y=2sin(3t-(pie)/4)+8

Minimum value of $\displaystyle \sin$ function is -1. So minimum value of y could be 6.The real question is does there exist a t between (pie) and (4pie), such that this happens.
sin(3t-(pie)/4) = -1 means that $\displaystyle 3t - \frac{\pi}{4} = -\frac{\pi}2,\frac{3\pi}2,\frac{7\pi}2,....$

Thus $\displaystyle 3t = -\frac{\pi}4,\frac{7\pi}2,\frac{15\pi}2,....$

So $\displaystyle t = \frac{7\pi}6,\frac{15\pi}6,....$

In general $\displaystyle t = \frac{(8n-1)\pi}6, n \in \mathbb{N}$

So between $\displaystyle \pi$ an $\displaystyle 4\pi$, we have three angles that satisfy:

$\displaystyle t = \frac{7\pi}6,\frac{15\pi}6,\frac{23\pi}6$ and the minimum value is $\displaystyle y = 6$
• Jun 5th 2008, 02:58 AM
RoboStar
and for the first question, would the next two be 13(pie)/4 and 17(pie)/4? I'm still quite unsure with this concept.
• Jun 5th 2008, 03:33 AM
mr fantastic
Quote:

Originally Posted by RoboStar
[snip]
I need to find the next four values for (t):
6=3sin(2t)+3
I know the first one is 5(pie)/4, i think the second is 9(pie)/4

[snip]

Do you need values such that t > 0. If so, you missed t = pi/4

Note that your equation can be re-arranged into sin(2t) = 1.

The general solution is $\displaystyle 2t = \frac{\pi}{2} + 2n \pi \Rightarrow t = \frac{\pi}{4} + n \pi$ where n is an integer.

So get all the solutions you want by letting n = 0, 1, 2, .... -1, -2, ....

Quote:

Originally Posted by RoboStar
and for the first question, would the next two be 13(pie)/4 and 17(pie)/4? I'm still quite unsure with this concept.

Yes (they correspond to n = 3 and n = 4).