# Finding x in Trig

• October 13th 2012, 07:42 PM
DylanRenke
Finding x in Trig
Question:
Find sin(arccos(x))/cos(arcsin(x))

I am not sure if this helps to solve anything, and im not sure if this si correct or not.
=tan(arccos(x)/arcsin(x)
=tan(arccot(x))
=x
Thank you
• October 13th 2012, 07:56 PM
MarkFL
Re: Finding x in Trig
Hint: use the identity $\arcsin(x)+\arccos(x)=\frac{\pi}{2}$...
• October 14th 2012, 12:51 PM
Soroban
Re: Finding x in Trig
Hello, DylanRenke!

Quote:

$\text{Find: }\:\frac{\sin(\arccos x)}{\cos(\arcsin x )}$

Let: . $\begin{Bmatrix}\alpha &=& \arccos x & [1] \\ \beta &=& \arcsin x & [2] \end{Bmatrix}$

The problem becomes: . $\frac{\sin\alpha}{\cos\beta}$ .(a)

From [1]: / $\alpha \:=\:\arccos x \quad\Rightarrow\quad \cos\alpha \,=\,x$

We have: . $\cos\alpha \,=\,\frac{x}{1} \,=\,\frac{adj}{hyp}$

We have a right triangle with: $adj = x,\:hyp = 1.$

Pythagorus tells us that: . $opp \,=\,\sqrt{1-x^2}$

Hence: . $\sin\alpha \:=\:\frac{opp}{hyp} \:=\:\frac{\sqrt{1-x^2}}{1} \:=\:\sqrt{1-x^2}$ .(b)

From [2]: . $\beta \,=\,\arcsin x \quad\Rightarrow\quad \sin\beta \,=\,x$

We have: . $\sin\beta \,=\,\frac{x}{1} \,=\,\frac{opp}{hyp}$

We have a right triangle with: $opp = x.\:hyp \,=\,1.$

Pythagorus tells us that: . $adj \,=\,\sqrt{1-x^2}$

Hence: . $\cos\beta \:=\:\frac{adj}{hyp} \:=\:\frac{\sqrt{1-x^2}}{1} \:=\:\sqrt{1-x^2}}$ .(c)

Substitute (b) and (c) into (a): . $\frac{\sin\alpha}{\cos\beta} \;=\;\frac{\sqrt{1-x^2}}{\sqrt{1-x^2}} \;=\;1$
• October 14th 2012, 01:29 PM
HallsofIvy
Re: Finding x in Trig
Quote:

Originally Posted by DylanRenke
Question:
Find sin(arccos(x))/cos(arcsin(x))

I am not sure if this helps to solve anything, and im not sure if this si correct or not.
=tan(arccos(x)/arcsin(x)
=tan(arccot(x))

This is wrong. "arccos(x)/arcsin(x)" is NOT "arccot(x)". Just because a function has some property doesn't mean its inverse has that property!

Quote:

=x
Thank you
Notice that $sin(u)= \pm\sqrt{1- cos^2(u)}$ and so $sin(arccos(x))= \pm\sqrt{1- cos^2(arccos(x)}= \pm\sqrt{1- x^2}$
However, cosine and sine are not "one to one" functions so they do NOT, strictly speaking, have inverses. In order to be able to talk about "arccos(x)" and "arcsin(x)" we hage to restrict cosine to 0 to $\pi$ and sine to $-\pi/2$ to [tex]\pi/2[/itex]. With those restrictions, sine and cosine are both postive and so both $cos(arcsin(x))$ and $sin(arccos(x))$ are equal to [tex]\sqrt{1- x^2}[tex].
• October 14th 2012, 01:38 PM
MarkFL
Re: Finding x in Trig
If you use the hint I gave, you find:

$\frac{ \sin( \arccos(x))}{ \cos( \arcsin(x))}= \frac{ \sin( \arccos(x))}{ \cos\left( \frac{\pi}{2}-\arccos(x) \right)}= \frac{ \sin( \arccos(x))}{ \sin( \arccos(x))}=1$
• October 15th 2012, 05:54 PM
DylanRenke
Re: Finding x in Trig
Thank you everybody.