Finding x in Trig

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October 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
October 13th 2012, 06:56 PM
MarkFL

Re: Finding x in Trig

Hint: use the identity $\arcsin(x)+\arccos(x)=\frac{\pi}{2}$ ...
October 14th 2012, 11:51 AM
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, 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 $sin(u)= \pm\sqrt{1- cos^2(u)}$ 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, 12:38 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, 04:54 PM
DylanRenke

Re: Finding x in Trig

Thank you everybody.