# Thread: Solving equations of inverse functions

1. ## Solving equations of inverse functions

The original problem is : $arcsin4x + arccosx = \pi/6$ and is also attached as an image.

I have it to "X=-(cosU * (3^1/2)/2 - sinU * 1/2 ) but that might be wrong, I don't know.

I suppose starting at 2 and then asking for help at 4:51 means I'm going to have to review this chapter, as I've had to ask for help in every section.

2. Hello Wolvenmoon
Originally Posted by Wolvenmoon
The original problem is : $arcsin4x + arccosx = \pi/6$ and is also attached as an image.

I have it to "X=-(cosU * (3^1/2)/2 - sinU * 1/2 ) but that might be wrong, I don't know.

I suppose starting at 2 and then asking for help at 4:51 means I'm going to have to review this chapter, as I've had to ask for help in every section.
Let $y = \arccos x$ and $z = \arcsin 4x$

Then $y+z=\tfrac{\pi}{6}$, $x = \cos y$ and $4x=\sin z$

So $\sin z = 4\cos y = 4\cos(\tfrac{\pi}{6}-z)$

$=4\cos(\tfrac{\pi}{6})\cos z + 4\sin(\tfrac{\pi}{6})\sin z$

$=2\sqrt3\cos z + 2\sin z$

$\Rightarrow \sin z = -2\sqrt3\cos z$

$\Rightarrow \tan z = -2\sqrt 3$

$\Rightarrow \sin z = -\frac{2\sqrt3}{\sqrt{1+(2\sqrt3)^2}}=-\frac{2\sqrt39}{13}$

$\Rightarrow x = -\frac{\sqrt39}{26}$

3. Hi

I have an interesting method

Consider a cyclic quadralateral $ABCD$ and

set $AC$ is the unit diameter of the circle

Therefore , $AD \perp CD$ and $AB \perp BC$

and let $AD = 4x , BC = x$

This circle is constructed for this problem

$\angle ACD = \arcsin(4x) , \angle ACB = \arccos(x)$

there sum is $\frac{\pi}{6}$

we can using the fact that $BD / \sin(\frac{\pi}{6}) = 2R = 1$ to find the length of $BD$ , $BD = \frac{1}{2}$

Finally , we end up the geometry part by applying Ptolemy's Theorem :

$AD \cdot BC + AB \cdot CD = AC \cdot BD$

$4x^2 + \sqrt{1-x^2} \sqrt{1-16x^2} = \frac{1}{2}$

$1 - 17x^2 + 16x^4 = \frac{1}{4} - 4x^2 + 16x^4$

$x^2 = \frac{3}{4(13)}$

4. Hello simplependulum
Originally Posted by simplependulum
Hi

I have an interesting method

Consider a cyclic quadralateral $ABCD$ and

set $AC$ is the unit diameter of the circle

Therefore , $AD \perp CD$ and $AB \perp BC$

and let $AD = 4x , BC = x$

This circle is constructed for this problem

$\angle ACD = \arcsin(4x) , \angle ACB = \arccos(x)$

there sum is $\frac{\pi}{6}$

we can using the fact that $BD / \sin(\frac{\pi}{6}) = 2R = 1$ to find the length of $BD$ , $BD = \frac{1}{2}$

Finally , we end up the geometry part by applying Ptolemy's Theorem :

$AD \cdot BC + AB \cdot CD = AC \cdot BD$

$4x^2 + \sqrt{1-x^2} \sqrt{1-16x^2} = \frac{1}{2}$

$1 - 17x^2 + 16x^4 = \frac{1}{4} - 4x^2 + 16x^4$

$x^2 = \frac{3}{4(13)}$
Interesting indeed!

But just slightly flawed, I'm afraid, because it's impossible to draw so that $\angle BCD = \frac{\pi}{6}$.

The fact is that $x < 0$, and your solution of $x^2 = \frac{3}{4(13)}$ stops short of solving for $x$. If you take the positive square root and substitute back, you'll find that $\angle BCD \approx 76.1^o$ and $\angle ACD \approx 73.9^o$. This gives $\angle BCD = 150^o$, which represents the solution to the equation $\arcsin 4x + \arccos x = \frac{5\pi}{6}$.

The negative solution to your equation is, of course, the correct one - and it's the same as the one I obtained by a more conventional approach. Sadly, it's not possible to draw your diagram with a negative value of $x$.

Unless, of course, you can think of a modification ...?

Hello simplependulumInteresting indeed!

But just slightly flawed, I'm afraid, because it's impossible to draw so that $\angle BCD = \frac{\pi}{6}$.

The fact is that $x < 0$, and your solution of $x^2 = \frac{3}{4(13)}$ stops short of solving for $x$. If you take the positive square root and substitute back, you'll find that $\angle BCD \approx 76.1^o$ and $\angle ACD \approx 73.9^o$. This gives $\angle BCD = 150^o$, which represents the solution to the equation $\arcsin 4x + \arccos x = \frac{5\pi}{6}$.

The negative solution to your equation is, of course, the correct one - and it's the same as the one I obtained by a more conventional approach. Sadly, it's not possible to draw your diagram with a negative value of $x$.

Unless, of course, you can think of a modification ...?

Yes . we cannot draw a line segment with length -x

Actually . I didn't realise of the mistake until you told me the impossibility of drawing this ...

Now , i have no confidence to modify my solution because Ptolemy's theorem

requires positve lengths but i am suprised that I still can find out the

Luckily , this method still can be applied to solve this kind of problems if the answer is positive !

6. *Whoosh* What just happened!? I think something might have flown over my head *sonic boom!*

Yep, something flew over my head so quickly it broke the sound barrier.

7. Hello, Wolvenmoon!

Solve for $x\!:\;\;\arcsin(4x) + \arccos(x) \:=\:\frac{\pi}{6}$
Let: . $\alpha \:=\:\arcsin(4x) \quad\Rightarrow\quad \sin\alpha \:=\:4x \quad\Rightarrow\quad\cos\alpha\:=\:\sqrt{1-16x^2}$ .[1]

Let: . $\beta \:=\:\arccos(x) \quad\Rightarrow\quad \cos\beta \:=\:x \quad\Rightarrow\quad \sin\beta \:=\:\sqrt{1-x^2}$ .[2]

We have: . $\alpha + \beta \;=\;\frac{\pi}{6}$

Then: . $\sin(\alpha + \beta) \;=\;\sin\frac{\pi}{6} \quad\Rightarrow\quad \sin\alpha\cos\beta + \cos\alpha\sin\beta \;=\;\frac{1}{2}$

Substitute [1] and [2]: . $4x\cdot x + \sqrt{1-16x^2}\cdot\sqrt{1-x^2} \;=\;\frac{1}{2}$

Multiply by 2: . $8x^2 + 2\sqrt{(1-16x^2)(1-x^2)} \;=\;1 \quad\Rightarrow\quad 2\sqrt{(1-16x^2)(1-x^2)} \;=\;1 - 8x^2$

Square both sides: . $4(1-16x^2)(1-x^2) \;=\;1 - 16x^2 + 64x^4 \quad\Rightarrow\quad 4 - 68x^2 + 64x^4 \;=\;1 - 16x^2 + 64x^4$

. . . . . . . . . . . . . . $52x^2 \:=\:3 \quad\Rightarrow\quad x^2 \:=\:\frac{3}{52} \quad\Rightarrow\quad x \;=\;\pm\sqrt{\frac{3}{52}} \;=\;\pm\frac{\sqrt{39}}{26}$

The only answer that checks is: . $x \;=\;-\frac{\sqrt{39}}{26}$