Finding the Value of cot(pi/8) in the form a + b*sqrt(2)

But the $ are easier to type.Example:If a=3 and b=2, what is the value of a^2+b^2?##a^2+b^2=3^2+2^2=9+4=13####a^2+b^2=3^2+2^2=9+4=13##$$a^2+b^2=3^2+2^2=9+4=13$$$$a^2+b^2=3^2+2^2=9+4=13$$In summary, we can use the identity cot(θ) = (1 + cos(2θ)) / sin(2θ) to
  • #1
cmkluza
118
1

Homework Statement


Show that sin(2θ) / (1 + cos(2θ) = tan(θ) - I've completed this part
Hence find the value of cot(π/8) in the form a + b√(2), where a, b ∈ ℤ

Homework Equations


cot(θ) = (1 + cos(2θ)) / sin(2θ)

The Attempt at a Solution


I did the math, got (1 + cos(π/4)) / sin(π/4) = (1 + 1/2√(2)) / (1/2√(2)) = (2 + √(2)) / √(2) = (2 / √(2)) + 1
Assuming I did all the math correctly, I can't figure out how to get to a + b√(2) from here.
Any tips?

(Also if anyone can answer this, how can I make my math look more "math-like" in these forums? I hate having to insert math as typed up stuff that looks sloppy)

Thanks very much for any help anyone can offer!
 
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  • #2
cmkluza said:

Homework Statement


Show that sin(2θ) / (1 + cos(2θ) = tan(θ) - I've completed this part
Hence find the value of cot(π/8) in the form a + b√(2), where a, b ∈ ℤ

Homework Equations


cot(θ) = (1 + cos(2θ)) / sin(2θ)

The Attempt at a Solution


I did the math, got (1 + cos(π/4)) / sin(π/4) = (1 + 1/2√(2)) / (1/2√(2)) = (2 + √(2)) / √(2) = (2 / √(2)) + 1
Assuming I did all the math correctly, I can't figure out how to get to a + b√(2) from here.
Any tips?

(Also if anyone can answer this, how can I make my math look more "math-like" in these forums? I hate having to insert math as typed up stuff that looks sloppy)

Thanks very much for any help anyone can offer!

Isn't ##\frac{2}{\sqrt{2}}=\sqrt{2}##? I'll try and find a link to a TeX tutorial on the website to help you make your stuff look more "math-like".
 
  • #3
cmkluza said:

Homework Statement


Show that sin(2θ) / (1 + cos(2θ) = tan(θ) - I've completed this part
Hence find the value of cot(π/8) in the form a + b√(2), where a, b ∈ ℤ

Homework Equations


cot(θ) = (1 + cos(2θ)) / sin(2θ)

The Attempt at a Solution


I did the math, got (1 + cos(π/4)) / sin(π/4) = (1 + 1/2√(2)) / (1/2√(2)) = (2 + √(2)) / √(2) = (2 / √(2)) + 1
Assuming I did all the math correctly, I can't figure out how to get to a + b√(2) from here.
Any tips?

(Also if anyone can answer this, how can I make my math look more "math-like" in these forums? I hate having to insert math as typed up stuff that looks sloppy)

Thanks very much for any help anyone can offer!
Write ##\displaystyle\ \frac{1+\cos(2\theta)}{\sin(2\theta)}\ ## as ##\displaystyle\ \frac{1}{\sin(2\theta)}+\frac{\cos(2\theta)}{\sin(2\theta)}\ ## .

(Use LaTeX for math-like stuff.) LaTeX Guide
 
  • #4
your math is correct, you can write it as [itex]1 + \sqrt{2} [/itex]

a=b=1

To write math equations write the word "itex" in [], and when you're done write "itex" after the dash in [ /]
there may be a shortcut to that, though
 
  • #5
Thank you very much for your help! I should've considered rearranging the problem in the first place to get that answer, but I guess I just wasn't thinking of it at the time. It always bug me when I miss one small thing that could lead me to the answer. Also, thanks for introducing me to LaTeX! I'm going to have to look more into that.
 
  • #6
deedsy said:
your math is correct, you can write it as [itex]1 + \sqrt{2} [/itex]

a=b=1

To write math equations write the word "itex" in [], and when you're done write "itex" after the dash in [ /]
there may be a shortcut to that, though
There is a shortcut that I find easier to type. Instead of itex tags, use a ## pair at the beginning and another pair at the end. Use those for inline math stuff.

For standalone math stuff use a pair of $$ at the start and another pair at the end. These are equivalent to the [ tex ] and [ /tex ] tags.
 
  • Like
Likes deedsy

FAQ: Finding the Value of cot(pi/8) in the form a + b*sqrt(2)

What is the value of cot(pi/8)?

The value of cot(pi/8) in the form a + b*sqrt(2) is 1 + sqrt(2).

How do you find the value of cot(pi/8)?

To find the value of cot(pi/8), you can use the trigonometric identity cot(x) = cos(x)/sin(x). In this case, x = pi/8. So, cot(pi/8) = cos(pi/8)/sin(pi/8). By using the double angle formula for cosine, cos(2x) = 2cos^2(x) - 1, we can simplify cos(pi/8) to sqrt(2 + sqrt(2))/2. Similarly, using the double angle formula for sine, sin(2x) = 2sin(x)cos(x), we can simplify sin(pi/8) to sqrt(2 - sqrt(2))/2. Substituting these values into cot(pi/8) = cos(pi/8)/sin(pi/8), we get 1 + sqrt(2).

Why is the value of cot(pi/8) in the form a + b*sqrt(2)?

The value of cot(pi/8) is in the form a + b*sqrt(2) because the double angle formulas for cosine and sine involve the square root of 2. Therefore, when simplifying cot(pi/8), the final answer will have a term with sqrt(2) in it.

Can the value of cot(pi/8) be simplified further?

No, the value of cot(pi/8) cannot be simplified further. It is already in its simplest form of a + b*sqrt(2).

How is cot(pi/8) related to other trigonometric functions?

The value of cot(pi/8) can be related to other trigonometric functions using various identities. For example, cot(pi/8) = 1/tan(pi/8) = 1/sin(pi/8)/cos(pi/8) = csc(pi/8)/sec(pi/8). It can also be related to the Pythagorean identity, csc^2(x) = 1 + cot^2(x).

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