How to use trigonometric identities to solve for f'(x) = 0 in calculus 1

  • Thread starter Thread starter Whalstib
  • Start date Start date
  • Tags Tags
    Trig
Whalstib
Messages
119
Reaction score
0
Hi,

This is just a part of a calculus 1 problem. It occurs while attempting to make f'(x) = 0

f'(x) = cosX - 1/6cos3X

Now to make this zero I quickly note that in the domain of (0-2pi) cos=0 at pi/2 and 3pi/2 so the expression equals zero already! I have been informed by my instructor I need to use a series of identities to achieve a final answer close to

cosX[5/6 +2/3sin^2x]

While it may very well include the cos zero I outlined above (and in fact insists on it!) it is now more complicated... sigh...

Thing is I would never make this connection on exam day without seeing what is going on. Tutors give me that look like it's obvious as does my instructor but I just don't see how or why I should stretch it and why if so stop at this point as I could continue...

ANY help advise greatly appreciated!

Warren
 
Physics news on Phys.org
Whalstib said:
Hi,

This is just a part of a calculus 1 problem. It occurs while attempting to make f'(x) = 0

f'(x) = cosX - 1/6cos3X

Now to make this zero I quickly note that in the domain of (0-2pi) cos=0 at pi/2 and 3pi/2 so the expression equals zero already!
More precisely, the derivative is zero at those two values.
Whalstib said:
I have been informed by my instructor I need to use a series of identities to achieve a final answer close to

cosX[5/6 +2/3sin^2x]
I'm confused. Are you supposed to rewrite your derivative as above? If so, start by replacing cos(3x) with cos(2x + x) and using the addition formula.
Whalstib said:
While it may very well include the cos zero I outlined above (and in fact insists on it!) it is now more complicated... sigh...

Thing is I would never make this connection on exam day without seeing what is going on. Tutors give me that look like it's obvious as does my instructor but I just don't see how or why I should stretch it and why if so stop at this point as I could continue...

ANY help advise greatly appreciated!

Warren
 

Thread 'Finding the nth roots of a complex number'

There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
View full post »

Similar threads

f(x) = 2x+1, proving that it is continuous when p = 1 with 𝛿 and ε
Replies
7
Views
1K
Taylor Series Expansion of f(x) at 0
Replies
7
Views
1K
Finding the Maximum Value of f(x)=x(1-x)^n in [0,1]: A Calculus Problem
Replies
3
Views
2K
Is the given function continuous at x= 0? f(x) = {sin(1/x) if x≠0, 1
Replies
4
Views
3K
How can the inequality cosx ≥ (1-x^2/2) be proven?
Replies
9
Views
2K
Solve ##\int\frac{e^{-x}}{x^2}dx## and ##\int \frac{e^{-x}}{x}dx##
Replies
7
Views
2K
How Do You Determine Concavity for f(x) = cos^2(x) - 2sin(x)?
Replies
6
Views
2K
Analyzing the Singularity of f(z) = z cos(1/z) Using Solutions to f(z) = 0
Replies
10
Views
2K
Limit (L'hopitals) trig functions
Replies
5
Views
2K
Proving ƒ(x) is the Identity Function
Replies
1
Views
2K

Hot Threads

Recent Insights

Back
Top