# Various questions thoughout elem algebra, calculus

1. Jun 14, 2008

### stat643

Hi ,
Exams coming up.. and here are a few practice examples throughout the course (elem algebra, calculus) that i have been stuck on that would appreciate some help with.

1) find all solutions for cot^2(x) = 1, in the interval 2*pi <= x < 4*pi

seeing cot = 1/tan, then solving for x would equal = 0.017455.. giving me no obvious angle to find in the intervals stated (like i would do for similar examples).

2)

P(x) = $$\frac{1-sin(x)}{cos(x)}$$

p'(x) = $$\frac{-cos(x)*cos(x) - (1-sin(x))(-sin(x)}{cos^2(x)}$$

= $$\frac{-cos^2(x)*cos(x) + sin(x)-sin^2(x)}{cos^2(x)}$$

= $$\frac{sin(x)-1}{cos^2(x)}$$

i dont understand where the -cos^2(x)*cos(x) comes from in the third line, shouldn't this just be -cos^2(x)? and i dont understand the fourth line either.

will post the rest of the questions once i have these 2 worked out .

2. Jun 14, 2008

Hi stat. Well your first question. Wot your doing at the moment is wrong because you are not trying to answer the question. What you need to be doing is solving the value for $x$ as your stated, but I dont think your were. So taking a look at the question:

$$cot^2(x) =1$$

$$\frac{1}{tan^2(x)} = 1$$

$$1 = tan^2(x)$$

$$0 = tan^2(x) -1$$

now have a think about how you could go about solving this, where does this type of resulting equation appear from you knowledge of solving quadratics, hope that helps.

3. Jun 14, 2008

### Gib Z

Hello stat643, welcome to Physicsforums!

For Question one, since we know that the cotangent has a period of $\pi$, it would be easiest to solve that equation in the interval of between 0 and pi, and then add multiples of $pi$ to the solution. We should be getting two solutions, because when we factor our equation we get cot x = 1 and cot x = -1. Solve That, then add the appropriate multiples of pi to get our answer in the correct range.

For question two, you probably copied it wrong because if you do it yourself, you are right, it shouldn't be a squared term. And if you continue with your own working, you'll see how to get the fourth line, which is correct.

4. Jun 14, 2008

About your second question. The must have been a typo or something as there shouldn't be a cos^2(x) there, it is a mistake, if you forget that ^2 then you get:

$$\frac{-cos^2(x) + sin(x)-sin^2(x)}{cos^2(x)}$$

$$\frac{-cos^2(x) -sin^2(x) + sin(x)}{cos^2(x)}$$

Know that should be far enough for you to apply your trig identity knowledge to get you to the final stage. This about a basic trig identities, one of the earliest you would have learnt, and think about how can you manipulate the second expression to use that identity, note that there is a reason why I rearranged the first into the second which should make it easier to see the identity in the expression.

5. Jun 14, 2008

### stat643

thanks,, i can see that now .

6. Jun 14, 2008

### stat643

I copied it correctly so it must of been a typo in the text. It was the only one i couldnt do out of the 5 examples so yeh makes sense now, thanks.

7. Jun 14, 2008

### stat643

thanks, yeh that is the first identity i learnt, and i didnt even notice it in there lol. thanks for pointing that out.