An exponential problem and a trig problem

  • Thread starter Kb1jij
  • Start date
  • #1
19
0
Yesterday in a math competition, I came across two problems that I couldn't (and still can't) figure out how to solve under the competion conditions (in under three minutes, without using a calculator).

The first one involved expential functions. When I try to do it I just get a huge mess of exponents and logs that takes me forever to simplify. It is as follows:

9^x + 9^-x = 34, evaluate 3^x+3^-x

The second problem involved trig functions:

Find the exact value of cot 15 + cot 75

Are there sum and difference formulas for cot? I tried turning the cot into cos/sin and then adding the fractions. That left me with
[tex] \frac{\sin(15)\cos(75)+\cos(15)\sin(75)}{\sin(75)\sin(15)} [/tex]
Using the sum formula this would become
[tex]\frac{\sin 90}{\sin(75)\sin(15)} = \frac{1}{\sin(75)\sin(15)} [/tex]
Can I do anything with this?

Thanks for your help!
Tom
 

Answers and Replies

  • #2
658
2
for the first problem, notice that 9^x=3^(2x) since 9=3^2, so substitute that in, then factor.

for the second problem 75= 60+15, and 15=30-15. now you can use the sum formulas for the denominator. i didn't do it out, but i'd try that.
 
  • #3
Integral
Staff Emeritus
Science Advisor
Gold Member
7,201
56
Another approach for the second problem is to see that

15 = 45 - 30 and that 75 = 45 + 30
 
  • #4
19
0
Ok, I got the second one, it's 4. I was avoiding using the sum formulas because I thought that I would just get a large mess of sines and cosines, but if you use integral's approach, the sums up being the difference of two squares, and it isn't really that complicated.

How can I factor 3^(2x)+3^(-2x)?
 
  • #5
658
2
yeah, integral's was a better way for using the sums, was a bit simpler, same idea. it does work the way i explained too.

for the factoring, isn't there a 3^2 in both those terms? after that, its very simple.
 
  • #6
19
0
I don't understand how I can factor a 3^2 out.
3^2(3^x+3^-x) is equal to (3^x)(3^2)+(3^-x)(3^2) or 3^(x+2)+3^(-x+2).
Also, 3^(2x)/3^2 would just become 3^(2x-2).
 
  • #7
658
2
sorry. you're right, you can take out the two though. here:

(3^x + 3^-x)^2 = 3^2x + 3^-2x +2(3^x)(3^-x)= 3^2x + 3^-2x +2(3^0)

so
3^2x + 3^-2x = (3^x + 3^-x)^2 - 2 = 34

now it works out nicely eh?
 
  • #8
19
0
Wow, great. That does work out very nicely. That is a pretty tricky problem, but thats what I would expect from this competition.
Thanks!
Tom
 

Related Threads on An exponential problem and a trig problem

  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
9
Views
2K
  • Last Post
2
Replies
27
Views
1K
Replies
2
Views
2K
Replies
5
Views
1K
  • Last Post
Replies
3
Views
2K
Replies
7
Views
3K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
1K
Top