# Sin in terms of tan.

• Feb 28th 2009, 12:44 AM
craigmain
Sin in terms of tan.
Hi,

How do I express $\displaystyle sin(\theta)$ in terms of $\displaystyle tan(\theta)$ by using the identities.
I can find most of the other(s) but this one is tricky.

It is meant to be this, but I can't see how.
$\displaystyle \pm\frac{tan(\theta)}{\sqrt{1+tan^2(\theta)}}$

I think you can get to it using the pythagoras identity and the ratio's alone. I think I am missing an obvious substitution.

Thanks
Craig.
• Feb 28th 2009, 02:30 AM
mr fantastic
Quote:

Originally Posted by craigmain
Hi,

How do I express $\displaystyle sin(\theta)$ in terms of $\displaystyle tan(\theta)$ by using the identities.
I can find most of the other(s) but this one is tricky.

It is meant to be this, but I can't see how.
$\displaystyle \pm\frac{tan(\theta)}{\sqrt{1+tan^2(\theta)}}$

I think you can get to it using the pythagoras identity and the ratio's alone. I think I am missing an obvious substitution.

Thanks
Craig.

$\displaystyle \tan \theta = \frac{\sin \theta}{\cos \theta} \Rightarrow \sin \theta = \cos \theta \tan \theta$.

Now recall that $\displaystyle \sin^2 \theta + \cos^2 \theta = 1 \Rightarrow \tan^2 \theta + 1 = \frac{1}{\cos^2 \theta} \Rightarrow \cos^2 \theta = \frac{1}{1 + \tan^2 \theta}$.
• Feb 28th 2009, 03:37 AM
Soroban
Hello, Craig!

Quote:

How do I express $\displaystyle \sin\theta$ in terms of $\displaystyle \tan\theta$ by using the identities.

It is meant to be this: .$\displaystyle \pm\frac{\tan\theta}{\sqrt{1+\tan^2\!\theta}}$

We have: .$\displaystyle \tan\theta \:=\:\frac{\sin\theta}{\cos\theta} \quad\Rightarrow\quad \sin\theta \:=\:\tan\theta\cos\theta \quad \Rightarrow\quad\sin\theta\:=\:\frac{\tan\theta}{\ sec\theta}$ .[1]

From the identity: .$\displaystyle \sec^2\!\theta \:=\:1 + \tan^2\!\theta$, we have: .$\displaystyle \sec\theta \:=\:\pm\sqrt{1+\tan^2\!\theta}$

Substitute into [1]: .$\displaystyle \sin\theta \:=\:\frac{\tan\theta}{\pm\sqrt{1+\tan^2\!\theta}} \quad\Rightarrow\quad\boxed{\sin\theta \:=\:\pm\frac{\tan\theta}{\sqrt{1+\tan^2\!\theta}} }$

• Feb 28th 2009, 03:46 AM
mr fantastic
Quote:

Originally Posted by Soroban
Hello, Craig!

We have: .$\displaystyle \tan\theta \:=\:\frac{\sin\theta}{\cos\theta} \quad\Rightarrow\quad \sin\theta \:=\:\tan\theta\cos\theta \quad \Rightarrow\quad\sin\theta\:=\:\frac{\tan\theta}{\ sec\theta}$ .[1]

From the identity: .$\displaystyle \sec^2\!\theta \:=\:1 + \tan^2\!\theta$, we have: .$\displaystyle \sec\theta \:=\:\pm\sqrt{1+\tan^2\!\theta}$

Substitute into [1]: .$\displaystyle \sin\theta \:=\:\frac{\tan\theta}{\pm\sqrt{1+\tan^2\!\theta}} \quad\Rightarrow\quad\boxed{\sin\theta \:=\:\pm\frac{\tan\theta}{\sqrt{1+\tan^2\!\theta}} }$

I knew you were a slow typer (by your own admission) but this sets a new world record ..... over 1 hour to type this ...? (Wink)