Recent content by bsmithysmith

Got it: So as Evgeny mentioned, driving 40 Miles per Hour on the highway and 25 Mile per Hour in the city, at some point in time between the highway and the city, the driver has gone 30 Miles per Hour. That's the one that stuck in my head the most. 40 is interval (b), per say, and 25 is (a)...

I watches plenty of videos of the intermediate value theorem the previous day, and from what I understand, within the interval [a,b] with f(a)=a and f(b)=b, the value of c exist as f(c)=c. What my teacher told me, hopefully I did hear it properly, is that a good tip is to get a negative value...

First off, it's: $$x = 1+x^3$$ Turned into function as: $$f(x) = x^3 - x + 1$$ From my understanding, we need to find an interval in which x will be one more than it's cube. Giving some points, I started off with (0,1), (1,1), (-1,1), and (-2, -5). Where I'm confused is how and where do I...

I just started Calculus 1, a summer quarter that's compressed and I'm having trouble understanding a theorem that state continuity of the inverse function. Within my textbook, it mentions "If f(x) is continuous on an interval I with range R, and if inverse f(x) exists, then the inverse f(x) is...

Since it's at quadrant 2, all cosine values will be negative, and sine values will be positive. I suspect it'll be the same thing for the sine value, but what about the tangent value? I have the cheat sheet available, but I'd still rather understand the process than plug and chug.

If $$csc(x)=4$$, for $$90º<x<180º$$ $$sin\left(\frac{x}{2}\right)=$$ $$cos\left(\frac{x}{2}\right)=$$ $$tan\left(\frac{x}{2}\right)=$$ I'm definitely stumped on this one. I know that this is the half-angle formulas. Luckily we all have sheets we can use for the exam. I know that: $$csc(x)=4$$...

So I was correct after all (outside the forum)! $$4\sin(x)\cos(x)-3\sin(x)=0$$ $$sin(x)(4cos(x)-3)=0$$ so it's $$sin(x)=0$$ and $$cos(x)=3/4$$ For sine, $$x=0, pi$$ and for cosine, it's $$2pi-cos^1(3/4)$$ and $$cos^-1(3/4)$$ Note, it's Cosine inverse

$$2sin(2x)-3sin(x)=0$$ We did this in my class, but there were some parts where I was really confused. I know that we need to use the double angle formula, and the double angle formula for Sine is: $$sin(2x) = 2sin(x)cos(x)$$ and correct me if I'm wrong. So what I had down is...

The two of them were the easiest to find, I didn't clarify my question exactly, too. I'm trying to find the third and fourth smallest solution since I already found two.

I am given to solve: $$5\cos(4x)=4$$ I know how to find the first and second solutions, it's basically the inverse, then divide by the 4. The second on is the same process, but $$2\pi$$- the cosine inverse. But I don't quite understand how to get the third solution.

And from that how would I find the double angle formula for $$Cosθ$$? This helped a good amount, too. Below the diagram are 4 questions regarding the term of four lines in the triangle: $$AB = Cosθ$$ $$BC = Cosθ$$ $$CD = Sin(2θ)$$ $$OD = Cos(2θ)$$ Now the thing is that I'm not confident on the...

I don't think I'm following this well enough; $$\sin(\alpha)=\cos(\beta)$$ where I substituted $$\sin(\alpha)$$ to $$1 - \sin(2θ)=\cos(\beta)$$, and choosing a cofunction identity of sine, which is $$sin (90° – x) = cos x$$... So $$1 - \sin(2θ)= sin (90° – x)$$? I'm really sorry, this is...

Since I don't know the line OD, then I was assume that the line OD = Sin $\theta$ Then what I also know is that $$\angle BAO$$ and $$\angle BCO$$ are definitely the same, so the $\theta$ of both the angles seem to be Sin $\theta$ as well. Since $\theta$ is equal to Sin $\theta$, $$Sin(θ)=x/1$$...

I know the Double Angle for Sine is: $$\sin(2x) = 2\sin(x) \cos(x)$$ but from the triangle given, how do I figure it out? We did this in class, but the teacher just told a small amount of things, and then let us talk amongst each other to solve it. Nearly all the students were glossy eyed and...

SWEET! Helped a bunch! Thank You!