Trig Identities

1. Jan 5, 2014

BOAS

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.

1. The problem statement, all variables and given/known data

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

tanθ + cotθ = 2

2. Relevant equations

3. The attempt at a solution

tanθ + cotθ = 2

$\frac{sin\theta}{cos\theta}$ + $\frac{cos\theta}{sin\theta}$ = 2

$\frac{sin^{2}\theta}{cos\theta sin\theta}$ + $\frac{cos^{2}\theta}{sin\theta cos\theta}$ = 2

$\frac{sin^{2}\theta + cos^{2}\theta}{cos\theta sin\theta}$ = 2

$sin^{2}\theta + cos^{2}\theta$ = $2cos\theta sin\theta$

Since $sin^{2}\theta + cos^{2}\theta$ = 1

then

$2cos\theta sin\theta$ = 1

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

I now have

$sin^{2}\theta + cos^{2}\theta$ = 1

I don't know what to do with it really... I tried saying $sin^{2}\theta$ = $1 - cos^{2}\theta$ 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: Jan 5, 2014
2. Jan 5, 2014

SteamKing

Staff Emeritus
Well, manipulating the original equation gives 2cosθsinθ = 1. What you need to do is find θ to make this equation true.

3. Jan 5, 2014

BOAS

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. Jan 5, 2014

Staff: Mentor

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. Jan 5, 2014

Ray Vickson

The quantity $t = \tan(\theta)$ must satisfy the equation
$$t + \frac{1}{t} = 2,$$
whose unique (real) solution is $t = 1$. This is easy to see and prove. For any $A,B > 0$ the arithmetic-geometric inequality says
$$\frac{A+B}{2} \geq \sqrt{AB},$$
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. Jan 5, 2014

BOAS

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. Jan 13, 2014

Anonymoose2

Yes, you've said that $\frac{sin^{2}\theta + cos^{2}\theta}{cos\theta sin\theta}$ = 2

and $sin^{2}\theta + cos^{2}\theta$ = $2cos\theta sin\theta$

Now, substitute the $2cos\theta sin\theta$ into the numerator of $\frac{sin^{2}\theta + cos^{2}\theta}{cos\theta sin\theta}$ = 2

This will give you $\frac{2cos\theta sin\theta}{cos\theta sin\theta}$ = 2

which simplifies into 2 = 2

8. Jan 13, 2014

ehild

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. Jan 14, 2014

Staff: Mentor

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