# Finding x in Trig

• Oct 13th 2012, 06:42 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
• Oct 13th 2012, 06:56 PM
MarkFL
Re: Finding x in Trig
Hint: use the identity $\displaystyle \arcsin(x)+\arccos(x)=\frac{\pi}{2}$...
• Oct 14th 2012, 11:51 AM
Soroban
Re: Finding x in Trig
Hello, DylanRenke!

Quote:

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

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

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

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

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

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

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

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

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

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

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

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

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

Substitute (b) and (c) into (a): .$\displaystyle \frac{\sin\alpha}{\cos\beta} \;=\;\frac{\sqrt{1-x^2}}{\sqrt{1-x^2}} \;=\;1$
• Oct 14th 2012, 12:29 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 $\displaystyle sin(u)= \pm\sqrt{1- cos^2(u)}$ and so $\displaystyle sin(arccos(x))= \pm\sqrt{1- cos^2(arccos(x)}= \pm\sqrt{1- x^2}$
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 $\displaystyle \pi$ and sine to $\displaystyle -\pi/2$ to [tex]\pi/2[/itex]. With those restrictions, sine and cosine are both postive and so both $\displaystyle cos(arcsin(x))$ and $\displaystyle sin(arccos(x))$ are equal to [tex]\sqrt{1- x^2}[tex].
• Oct 14th 2012, 12:38 PM
MarkFL
Re: Finding x in Trig
If you use the hint I gave, you find:

$\displaystyle \frac{ \sin( \arccos(x))}{ \cos( \arcsin(x))}= \frac{ \sin( \arccos(x))}{ \cos\left( \frac{\pi}{2}-\arccos(x) \right)}= \frac{ \sin( \arccos(x))}{ \sin( \arccos(x))}=1$
• Oct 15th 2012, 04:54 PM
DylanRenke
Re: Finding x in Trig
Thank you everybody.