• Support PF! Buy your school textbooks, materials and every day products Here!

Trig Identities

  • Thread starter BOAS
  • Start date
  • #1
555
19
Hello,

i'm fairly sure that what I have done is wrong/not allowed, but I want to check because it seemed like a good idea, but now it's causing problems.

Homework Statement



Solve the following equations in the range -180°≤ θ ≤ 180°

tanθ + cotθ = 2

Homework Equations





The Attempt at a Solution



tanθ + cotθ = 2

[itex]\frac{sin\theta}{cos\theta}[/itex] + [itex]\frac{cos\theta}{sin\theta}[/itex] = 2

[itex]\frac{sin^{2}\theta}{cos\theta sin\theta}[/itex] + [itex]\frac{cos^{2}\theta}{sin\theta cos\theta}[/itex] = 2

[itex]\frac{sin^{2}\theta + cos^{2}\theta}{cos\theta sin\theta}[/itex] = 2

[itex]sin^{2}\theta + cos^{2}\theta[/itex] = [itex]2cos\theta sin\theta[/itex]

Since [itex]sin^{2}\theta + cos^{2}\theta[/itex] = 1

then

[itex]2cos\theta sin\theta[/itex] = 1

(the above is where I think i've done something careless)

I now have

[itex]sin^{2}\theta + cos^{2}\theta[/itex] = 1

I don't know what to do with it really... I tried saying [itex]sin^{2}\theta[/itex] = [itex] 1 - cos^{2}\theta[/itex] but it doesn't really get me anywhere.


Is what I have done not correct?

EDIT - I know it can be solved by putting the original equation into a quadratic form, and doing so did bring me to the correct answer of -135 and 45 degrees.
 
Last edited:

Answers and Replies

  • #2
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,796
1,666
Well, manipulating the original equation gives 2cosθsinθ = 1. What you need to do is find θ to make this equation true.
 
  • #3
555
19
Well, manipulating the original equation gives 2cosθsinθ = 1. What you need to do is find θ to make this equation true.
Ahhh, I was focused on the other side of the equation!

2cosθsinθ = sin2θ

sin2θ = 1

2θ = 90

θ = 45

from the curve I can see where -135 comes from also.
 
  • #4
33,627
5,284
It's quicker to replace cot(θ) by 1/tan(θ)

tanθ + cotθ = 2
tanθ + 1/tanθ = 2

Multiply by tanθ to get
tan2θ + 1 = 2tanθ
tan2θ - 2tanθ + 1 = 0
This equation can be factored to solve for tanθ.
 
  • #5
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
Hello,

i'm fairly sure that what I have done is wrong/not allowed, but I want to check because it seemed like a good idea, but now it's causing problems.

Homework Statement



Solve the following equations in the range -180°≤ θ ≤ 180°

tanθ + cotθ = 2

Homework Equations





The Attempt at a Solution



tanθ + cotθ = 2

[itex]\frac{sin\theta}{cos\theta}[/itex] + [itex]\frac{cos\theta}{sin\theta}[/itex] = 2

[itex]\frac{sin^{2}\theta}{cos\theta sin\theta}[/itex] + [itex]\frac{cos^{2}\theta}{sin\theta cos\theta}[/itex] = 2

[itex]\frac{sin^{2}\theta + cos^{2}\theta}{cos\theta sin\theta}[/itex] = 2

[itex]sin^{2}\theta + cos^{2}\theta[/itex] = [itex]2cos\theta sin\theta[/itex]

Since [itex]sin^{2}\theta + cos^{2}\theta[/itex] = 1

then

[itex]2cos\theta sin\theta[/itex] = 1

(the above is where I think i've done something careless)

I now have

[itex]sin^{2}\theta + cos^{2}\theta[/itex] = 1

I don't know what to do with it really... I tried saying [itex]sin^{2}\theta[/itex] = [itex] 1 - cos^{2}\theta[/itex] but it doesn't really get me anywhere.


Is what I have done not correct?

EDIT - I know it can be solved by putting the original equation into a quadratic form, and doing so did bring me to the correct answer of -135 and 45 degrees.
The quantity ##t = \tan(\theta)## must satisfy the equation
[tex] t + \frac{1}{t} = 2,[/tex]
whose unique (real) solution is ##t = 1##. This is easy to see and prove. For any ##A,B > 0## the arithmetic-geometric inequality says
[tex] \frac{A+B}{2} \geq \sqrt{AB},[/tex]
with equality only when ##A = B##. Apply this to ##A = t, B = 1/t, ## to conclude that for ##t > 0## we have ##t + 1/t \geq 2,## with equality only when ##t = 1##. For ##t < 0## it similarly follows that ##t + 1/t \leq -2,## so negative values of ##t## won't work.

So, the unique root is ##\tan(\theta) = 1##.
 
  • #6
555
19
It's quicker to replace cot(θ) by 1/tan(θ)

tanθ + cotθ = 2
tanθ + 1/tanθ = 2

Multiply by tanθ to get
tan2θ + 1 = 2tanθ
tan2θ - 2tanθ + 1 = 0
This equation can be factored to solve for tanθ.
This is how I solved it after struggling with the method outlined in the OP.

Much simpler, but it is cool to see the many ways of solving these problems.
 
  • #7
Hello,

i'm fairly sure that what I have done is wrong/not allowed, but I want to check because it seemed like a good idea, but now it's causing problems.

Homework Statement



Solve the following equations in the range -180°≤ θ ≤ 180°

tanθ + cotθ = 2

Homework Equations





The Attempt at a Solution



tanθ + cotθ = 2

[itex]\frac{sin\theta}{cos\theta}[/itex] + [itex]\frac{cos\theta}{sin\theta}[/itex] = 2

[itex]\frac{sin^{2}\theta}{cos\theta sin\theta}[/itex] + [itex]\frac{cos^{2}\theta}{sin\theta cos\theta}[/itex] = 2

[itex]\frac{sin^{2}\theta + cos^{2}\theta}{cos\theta sin\theta}[/itex] = 2

[itex]sin^{2}\theta + cos^{2}\theta[/itex] = [itex]2cos\theta sin\theta[/itex]

Since [itex]sin^{2}\theta + cos^{2}\theta[/itex] = 1

then

[itex]2cos\theta sin\theta[/itex] = 1

(the above is where I think i've done something careless)
Yes, you've said that [itex]\frac{sin^{2}\theta + cos^{2}\theta}{cos\theta sin\theta}[/itex] = 2

and [itex]sin^{2}\theta + cos^{2}\theta[/itex] = [itex]2cos\theta sin\theta[/itex]

Now, substitute the [itex]2cos\theta sin\theta[/itex] into the numerator of [itex]\frac{sin^{2}\theta + cos^{2}\theta}{cos\theta sin\theta}[/itex] = 2

This will give you [itex]\frac{2cos\theta sin\theta}{cos\theta sin\theta}[/itex] = 2

which simplifies into 2 = 2
 
  • #8
ehild
Homework Helper
15,492
1,874
2cosθsinθ = sin2θ

sin2θ = 1

2θ = 90

θ = 45

from the curve I can see where -135 comes from also.
sin(2θ)= 1 when 2θ =90 °+ k 360 °, k=0, ±1 ±2.... . That means θ=45 ° ± k 180 °. Which angles are in the range [-180°, 180°]?

ehild
 
  • #9
33,627
5,284
Yes, you've said that [itex]\frac{sin^{2}\theta + cos^{2}\theta}{cos\theta sin\theta}[/itex] = 2

and [itex]sin^{2}\theta + cos^{2}\theta[/itex] = [itex]2cos\theta sin\theta[/itex]

Now, substitute the [itex]2cos\theta sin\theta[/itex] into the numerator of [itex]\frac{sin^{2}\theta + cos^{2}\theta}{cos\theta sin\theta}[/itex] = 2

This will give you [itex]\frac{2cos\theta sin\theta}{cos\theta sin\theta}[/itex] = 2

which simplifies into 2 = 2
So? The idea is to show that tanθ + cotθ = 2, not that 2 = 2, which is obviously true.
 

Related Threads on Trig Identities

  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
794
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
6
Views
648
  • Last Post
Replies
1
Views
1K
Top