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Thread: Sin in terms of tan.

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by craigmain View Post
    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 $\displaystyle \sin^2 \theta + \cos^2 \theta = 1 \Rightarrow \tan^2 \theta + 1 = \frac{1}{\cos^2 \theta} \Rightarrow \cos^2 \theta = \frac{1}{1 + \tan^2 \theta}$.
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  3. #3
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    Hello, Craig!

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

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  4. #4
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    Quote Originally Posted by Soroban View Post
    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 ...?
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