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Computer Engineering Major. Tutoring in math for three years.
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## Questions

### Subject:Pre-Calculus

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Question:

How do I go about proving a complicated trigonometric identity?

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These can be very tricky, and there is not always a clear cut way of going about them. As I'm sure you know, the best thing to do from the start is to pound the most common identities into your brain. The better you know them, the better your brain will be at recognizing them, even in a manipulated form. But aside from that, start out by converting every aspect of the problem into sines and cosines, if possible. A common occurrence after doing this is a majority of the trig functions cancelling one another out. For example, cotangent over cosecant is not very pleasant. But after realizing that cot(x) = cos(x) / sin(x) and csc(x) = 1 / sin (x), the entire thing just simplifies to cos(x)! Another useful tip is to look for opportunities to factor out trig functions. For example, cosx^3 = sinx^2 / secx - cosx looks very complicated at first. But, after simplifying everything down to sines and cosines, we get cosx^3 = sinx^2 * cosx - cosx. From here we can see that cosx exists in both terms on the right. After factoring, we have cosx^3 = cosx * (sinx^2 - 1). Hopefully you can now see the identity sinx^2 - 1 which is equal to cosx^2. So all in all, cosx^3 = =cosx * cosx^2. So these are just some helpful tips that I hope assist you as you try to tackle these! Good Luck!

### Subject:Calculus

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Question:

I don't understand u-substitution! Help!

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U-substitution is a tool that we use to integrate complicated functions. The only time we can use it is when we see a smaller function and its derivative somewhere in a larger function. For example, the function y = cos(x)/sin(x). This contains the function sin(x) and its derivative cos(x). To start, set the smaller, pre-derived function equal to any arbitrary variable. u = sin(x). Now we can derive this equation on both sides. So it becomes du = cos(x) dx. This is where it's easy to get tripped up. We need to include the differential dx at the end of the cosine because of what we know about the chain rule. We are not deriving with respect to any specific independent variable like we usually do. In that case we would use the notation du/dx = cos(x). We are simply finding the derivative to this equation, with respect to no variable in particular, so we need act like u and x could be their own separate functions and use the chain rule. So once we have this, we can look at the integral at hand, which would be the integral( cos(x) / sin(x) ) dx. (Remember that this fraction could be thought of like this: 1 / sin(x) * cos(x)). Now if we look, we can see that both equations have the chunk, cos(x) dx. This is where substitution comes into play. Since we know that this chunk is equal to du, we can replace the entire thing in the integral equation with du. Thus, integral(1 / sin(x)) du is our new equation. But wait, there's more! We also know that u = sin(x), so we can substitute that as well. Thus integral(1 / u) du is our new equation. And from there we can solve it like a normal problem. Once we find our answer of ln(u) +C, we need to substitute the original functions back in to replace the u. So ln(u) + C becomes ln(sinx)) + C. These are the basics to u-substitution. As you probably know, they can get a lot more tricky than this! Just remember that constants can always be pulled out of the integral and denominators of fractions are really 1 / (any function). Good luck!

### Subject:Algebra

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Question:

What even are imaginary numbers? Why do they exist and what do they mean?

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Imaginary numbers can be very frustrating for many students. A lot of teachers skip over the details and start giving out problems right away. Imaginary numbers are not "made up" numbers. They actually describe a new direction on the number line. The name imaginary only stuck because the people at the time had a hard time understanding this new concept. Imagine a good ole fashion, elementary school style number line. It starts on the left and continues forever to the right. Now, remember back to when your mind was blown with the world of negative numbers. It seemed crazy that the beginning of number line, at zero, only began half of what was really out there! Imaginary numbers are no different. They are just something new that needs to settle in. If the positive direction corresponds to the right, and the negative direction to the left, then the "imaginary" direction goes upward and the "negative imaginary" direction goes downward. All we are doing with imaginary numbers is changing the number line into a number graph. A graph with a real axis and a imaginary axis. The point of them existing is to solve new sets of problems that would be otherwise unsolvable. Just like if sally has 5 apples and promises Joe that she will give him 7, we could say that Sally has -2 apples. Now you might be able to see that negative numbers are just as abstract as imaginary ones. Of course, Sally really has nothing, but we can describe her debt to Joe with negatives. Likewise, by accepting the fact that the square root of -1 equals both i and -i, a multitude of doors are open to solve very useful and complicated problems that would be otherwise unsolvable. For example, solving for x if x^2 = -4. If we just left problems like this unsolvable, we would never have acquired the solutions to many problems in science, math, or engineering. But now we have a tool that allows us to take a step forward in the right direction.

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