# Trig Identities

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

## 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:

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

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.

Mark44
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.

Ray Vickson
Homework Helper
Dearly Missed
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

## 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.
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##.

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.

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

## 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)
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

ehild
Homework Helper
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

Mark44
Mentor
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
So? The idea is to show that tanθ + cotθ = 2, not that 2 = 2, which is obviously true.