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

Hint: use the identity $\displaystyle \arcsin(x)+\arccos(x)=\frac{\pi}{2}$...

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): .$\displaystyle \frac{\sin\alpha}{\cos\beta} \;=\;\frac{\sqrt{1-x^2}}{\sqrt{1-x^2}} \;=\;1$

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].

If you use the hint I gave, you find:

$\displaystyle \frac{ \sin( \arccos(x))}{ \cos( \arcsin(x))}= \frac{ \sin( \arccos(x))}{ \cos\left( \frac{\pi}{2}-\arccos(x) \right)}= \frac{ \sin( \arccos(x))}{ \sin( \arccos(x))}=1$

Thank you everybody.