# Different (easy) math questions before my exam

1. Oct 22, 2007

### pace

different (easy) math questions before my exam :)

1.

1. The problem statement, all variables and given/known data:

Write as simple as possible: 2 sinv cosv-sinv
------------------
cos^2v-sin^2v-cosv+1

(cos^2v same as (cosv)^2)

2. Relevant equations:

sinv
tanx= -----
cosv

cos^2v+sin^2v = 1

3.

Sollution my previous helper says:
cos^2v-sin^2v is supposed to be 2cos^2v - 1.
Because: cos2x=cos^v-sin^2v (?)
=2cos^2-1 ((and solves it easy after that))

How's that?

My attempt at a solution:

2 sinv cosv-sinv
------------------
cos^2v-sin^2v-cosv+1

sinv(2cosv-1)
----------------
cos^2v-(1-cos^2v) -cosv+1

sinv(2cosv-1)
----------------
2cos^2v-cos v

... and then I don't know what to do. I can't just delete one cosv in each part because it's one 2cos^2v there, and it's different..

[/b]

2.

1. The problem statement, all variables and given/known data:

This is inside a test ((64+t^2)^2) it's supposed to be: 2(64+t^2) x 2t.

2. Relevant equations:

My helped reffered to that (u^n) = n*u * u^n-1. Which gives the above result. I don't see that law in my 2mx book, is it 3mx? But this is a 2mx test. My book though states that (x^n)= n*u^n-1 (eg: (-6x^2)`= -12x) , but that doesn't go for biggies like (64+t^2), am I right?

3. The attempt at a solution:

My main question is just where is that law from?

3.

1. The problem statement, all variables and given/known data:

When do you use {and ( together, like: {(u)} ?

4.

1. The problem statement, all variables and given/known data & Relevant equations & The attempt at a solution:

My book states that 3[squareroot](-5)^3 = -5, and that 4[squareroot](-5)^4 = 5, and so on in odd and even numbers. But laters say that 4[squareroot]a^4, when a is a negative number(as in -5 above) gets to be -a. As if 4[squareroot](-5)^4 = 5, but even numbers should give positive results. Why does it change?

5.

1. The problem statement, all variables and given/known data

How do you solve tanx<1 nicely on paper?

The problem statement is tanx<1

2. Relevant equations:

sinv
tanx= -----
cosv

tan(v+k 180') = tanv

3. The attempt at a solution

tanx<1
x<45'(degrees) and x<225'(degrees)
L=<45',0'> and <90',225'> and <270,360'>

But my counting seems so short. Is there any rules that say that I'm writing too short, then if so what am I missing?

All help appreciated! :) Hope I placed this at the right place

2. Oct 22, 2007

### drpizza

First of all, if you start clicking on "fancy equations" that other people have entered in other threads, a little text box will come up to show how to enter your equations in LaTeX. There's also a tutorial here: https://www.physicsforums.com/showthread.php?t=8997 It's pretty easy to do; I'm surprised you don't already do it (given 143 posts) It makes it much nicer to read :)

On #1, you can factor the denominator the same way you factored the numerator originally: there's a greatest common factor.

On #4, try plugging in a few negative numbers, raise them to the 4th power, then taking the 4th root.

3. Oct 23, 2007

### Gib Z

The law you are referring to: $$\frac{d}{dx} u^n = (nu^{n-1}) \cdot \frac{d}{dx} u$$, comes from the more general result differentiation, the chain rule. This rule states that : $$\frac{dy}{du} \cdot \frac{du}{dx} = \frac{dy}{du}$$.

Do you mean in terms of mathematical notation or LaTeX? LaTeX you sometimes need it, but you don't use it anyway. In mathematics, the normal curved brackets can either denote the argument of a function, ie f(x), g(t), or to clarify products, eg $(23-x)(23+x)$.

It seems when you say "3[squareroot]" you mean the 3rd root, not 3 times the root, please clarify that next time. And they made a mistake which is what is causing your confusion, the 4th root of a to the power of 4 is positive a. However they may be trying to find a solution to something, and just like in the quadratic equation, both the positive and negative roots must be taken into account, so -a is one of the other roots to be considered.

Ok well it seems that you want to find all values of x between 0 and 360 degrees where tan x is less than 1. To aid you, note that tan has a period of 180 degrees, so you only need to find values where tan x < 1 between 0 and 180 degrees, then add on 180 degrees to those solutions after. Between 0 and 180 degrees, we know that tan x = 1 when x is 45 degrees. We also know that the tan function is in increasing function. So all values between 0 and 45 degrees are less than 1. After 45, and up to almost 90, it is increasing. Going to 90 from left to right, tan goes to positive infinity, much bigger than 1. But from just after 90, it starts at negative infinity again, so these values are allowed. Then we know tan x is 1 at 225 again, and isnt 1 again. Since we cant include the value 90 or 225, but they are the endpoints of the values of we want, we write the interval as (90 degrees, 225 degrees) as opposed to the closed interval [90 degrees, 225 degrees].