# tanθ = -2, θ = ?

• October 10th 2011, 01:11 PM
s3a
tanθ = -2, θ = ?
What is θ such that tanθ = -2, sinθ = 2/sqrt(5), and cosθ = -1/sqrt(5)?

I am doing a question which is not about trigonometry (but involves it in one step) and I do not know how to find theta without using a calculator. I basically want the exact value.

I would greatly appreciate it if someone could show me how to find the exact value of theta.
• October 10th 2011, 01:22 PM
Plato
Re: tanθ = -2, θ = ?
Quote:

Originally Posted by s3a
What is θ such that tanθ = -2, sinθ = 2/sqrt(5), and cosθ = -1/sqrt(5)?

$\theta=\arctan(-2)\approx-1.1071487178$
• October 10th 2011, 01:37 PM
s3a
Re: tanθ = -2, θ = ?
That's not the exact value. I was looking for something with a pi in it.
• October 10th 2011, 01:41 PM
Plato
Re: tanθ = -2, θ = ?
Quote:

Originally Posted by s3a
That's not the exact value. I was looking for something with a pi in it.

Sorry, but that does not exist.
• October 10th 2011, 01:48 PM
s3a
Re: tanθ = -2, θ = ?
Okay so how do I figure out that cosθ = -1/sqrt(5) and sinθ = 2/sqrt(5) given only that tanθ = -2?
• October 10th 2011, 01:54 PM
FrameOfMind
Re: tanθ = -2, θ = ?
Because,

$\cos\theta = \frac {Adjacent}{Hypotenuse}$

$\sin\theta = \frac {Opposite}{Hypotenuse}$

$\tan\theta = \frac {Opposite}{Adjacent}$

Since, in your case, you have

$\cos\theta = \frac {-1}{\sqrt {5}}$ and $\sin\theta = \frac {2}{\sqrt {5}}$

You have the values for the opposite and the adjactent, which are 2 and -1 respectively. So,

$\tan\theta = \frac {Opposite}{Adjacent} = \frac {2}{-1} = -2$
• October 10th 2011, 01:55 PM
Plato
Re: tanθ = -2, θ = ?
Quote:

Originally Posted by s3a
Okay so how do I figure out that cosθ = -1/sqrt(5) and sinθ = 2/sqrt(5) given only that tanθ = -2?

Well $\tan(\theta)=\frac{y}{x}$.

So either $y=-2~\&~x=1\text{ or }y=2~\&~x=-1$.

Let $r=\sqrt{(x)^2+(y)^2}$ so $\sin(\theta)=\frac{y}{r}~\&~\cos(\theta)=\frac{x}{ r}$.
• October 10th 2011, 02:28 PM
s3a
Re: tanθ = -2, θ = ?
Plato: I've never seen such a formula before. Why does r = sqrt(x^2 + y^2) give me the correct value -- sqrt(5) in this case?
• October 10th 2011, 03:06 PM
Plato
Re: tanθ = -2, θ = ?
Quote:

Originally Posted by s3a
I've never seen such a formula before. Why does r = sqrt(x^2 + y^2) give me the correct value -- sqrt(5) in this case?

There is something really wrong with this picture.
Either you have not been paying attention or worst still your education institution is totally incompetent and thereby is cheating you.
• October 10th 2011, 03:14 PM
e^(i*pi)
Re: tanθ = -2, θ = ?
Quote:

Originally Posted by s3a
Plato: I've never seen such a formula before. Why does r = sqrt(x^2 + y^2) give me the correct value -- sqrt(5) in this case?

Not even in the closely related form $c^2 = a^2+b^2$ ?
• October 10th 2011, 06:15 PM
Prove It
Re: tanθ = -2, θ = ?
Quote:

Originally Posted by s3a
Okay so how do I figure out that cosθ = -1/sqrt(5) and sinθ = 2/sqrt(5) given only that tanθ = -2?

You should know the Pythagorean Identity $\displaystyle \sin^2{\theta} + \cos^2{\theta} \equiv 1$. Dividing both sides by $\displaystyle \cos^2{\theta}$ gives $\displaystyle \tan^2{\theta} + 1 \equiv \frac{1}{\cos^2{\theta}} \implies \cos^2{\theta} \equiv \frac{1}{\tan^2{\theta} + 1}$.

These two identities will enable you to get what you want...