# Express this as an algebraic equation..

• Sep 27th 2007, 01:22 AM
circuscircus
Express this as an algebraic equation..
Express $\displaystyle sin(tan^{-1}\frac{1}{x} + cos^{-1}x)$ as an algebraic expression in x if x is negative.

I don't understand what this question is asking me...

sorry I'm asking so many questions. The teacher gave us a list of like 250 problems that are "possible" on the test and so theres quite a few I get stuck on...
• Sep 27th 2007, 04:24 AM
topsquark
Quote:

Originally Posted by circuscircus
Express $\displaystyle sin(tan^{-1}\frac{1}{x} + cos^{-1}x)$ as an algebraic expression in x if x is negative.

I don't understand what this question is asking me...

sorry I'm asking so many questions. The teacher gave us a list of like 250 problems that are "possible" on the test and so theres quite a few I get stuck on...

Replace x with -x:
$\displaystyle sin \left [ tan^{-1} \left ( \frac{1}{-x} \right ) + cos^{-1}(-x) \right ]$

Now, the tangent function is an odd function, so $\displaystyle tan(-x) = -tan(x) \implies tan^{-1}(-x) = -tan^{-1}(x)$

and cosine is an even function, so
$\displaystyle cos(-x) = cos(x) \implies cos^{-1}(-x) = cos^{-1}(x)$,

thus
$\displaystyle sin \left [ tan^{-1} \left ( \frac{1}{-x} \right ) + cos^{-1}(-x) \right ] = sin \left [- tan^{-1} \left ( \frac{1}{x} \right ) + cos^{-1}(x) \right ]$

We may possibly take this one step further:
Sine is an odd function so $\displaystyle sin(-x) = -sin(x)$, thus
$\displaystyle sin \left [- tan^{-1} \left ( \frac{1}{x} \right ) + cos^{-1}(x) \right ] = -sin \left [ tan^{-1} \left ( \frac{1}{x} \right ) - cos^{-1}(x) \right ]$

-Dan
• Sep 28th 2007, 09:41 AM
red_dog
Let $\displaystyle \alpha=\arctan x\Rightarrow\tan\alpha=x$
For $\displaystyle x<0\Rightarrow\alpha\in\left(-\frac{\pi}{2},0\right)$ and $\displaystyle \sin\alpha<0, \ \cos\alpha>0$.

Let $\displaystyle \beta=\arccos x\Rightarrow\cos\beta=x$
For $\displaystyle x<0\Rightarrow\beta\in\left(\frac{\pi}{2},\pi\righ t)$ and $\displaystyle \sin\beta>0$.

We have $\displaystyle \displaystyle\sin\alpha=\frac{\tan\alpha}{\sqrt{1+ \tan^2\alpha}}=\frac{x}{\sqrt{1+x^2}}$, $\displaystyle \cos\alpha=\frac{1}{\sqrt{1+\tan^\alpha}}=\frac{1} {\sqrt{1+x^2}}$
$\displaystyle \sin\beta=\sqrt{1-\cos^2\beta}=\sqrt{1-x^2}, \ \cos\beta=x$.

Now, we have to calculate
$\displaystyle \sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\c os\alpha=$
$\displaystyle \displaystyle=\frac{x}{\sqrt{1+x^2}}\cdot x+\sqrt{1-x^2}\cdot\frac{1}{\sqrt{1+x^2}}=\frac{x^2+\sqrt{1-x^2}}{\sqrt{1+x^2}}$