Why Does arccos(x) - arcsin(1-x) Equal 90° Have No Solutions?

[acen_gr]

Homework Statement

Find the solution set of the equation below for the interval 0° ≤ x ≤ 360°:

arccos(x) - arcsin(1-x) = 90°

Homework Equations

all given above.

The Attempt at a Solution

i tried to consider (1-x) as "complementary" but I'm not really sure about that and I would like to know your opinion if I should go for it? And I'm kinda confused why the RHS is in degrees. Is it because the problem is dealing with inverse trig equations (which the answers are supposed to be angles)? Thanks in advance.

 

[Simon Bridge]

Homework Statement

Find the solution set of the equation below for the interval 0° ≤ x ≤ 360°:

That doesn't make sense - surely it is -1 ≤ x ≤ 1 (angle is the output of inverse trig functions).

arccos(x) - arcsin(1-x) = 90°

Hmmm ... well, $$\arccos(x)+\arcsin(x)=90^\circ$$

Homework Equations

all given above.

We-ell, there are a whole lot of trig identities that may be useful too.

The Attempt at a Solution

i tried to consider (1-x) as "complementary" but I'm not really sure about that and I would like to know your opinion if I should go for it? And I'm kinda confused why the RHS is in degrees. Is it because the problem is dealing with inverse trig equations (which the answers are supposed to be angles)? Thanks in advance.

... the RHS is the result of the sum of two angles, so, naturally it is in degrees.

I'd have been tempted to take the cosine of both sides and use the sum-to-product relations with the trig-inverse-trig relations to get some f(x)=0 ... then it's a matter of finding the roots of f.

But your idea is cool too ... if you can show that arcsin(1-x) = -arcsin(x) you'd be made.

eg. if x=1, then 1-x=0, arcsin(1)=90, arcsin(0)=0 or 180.
So - not generally true. But maybe it is true for some value of x?

Of course, plotting the function y=arccos(x)-arcsin(1-x) could give a few clues.

 

[micromass]

And I'm kinda confused why the RHS is in degrees.

No particular reason, I think. You can write them in radians too if you wish. So you could as well write

$$arccos(x)-arcsin(1-x)=\frac{\pi}{2}$$

It's the same thing, although I prefer to write them like this.

Anyway, try to write the equation as

$$arccos(x)=\frac{\pi}{2}+arcsin(1-x)$$

and now take the cosine of both sides.

 

[acen_gr]

$$arccos(x)=\frac{\pi}{2}+arcsin(1-x)$$

and now take the cosine of both sides.

Hi, micromass. Thanks for posting a reply.
I'm working on it now and I just want to confirm if I'm doing this right:

Taking cosine of both sides will give

$$0 = \cos\frac{\pi}{2} + \cos[\arcsin(1-x)]$$

Question: In cos[arcsin(1-x)], is it right to use "substitution" method where I let, say θ, equal to arcsin(1-x), then find for cosθ afterwards?

It's something like: cos[arcsin(1-x)]

let θ = arcsin(1-x)

sinθ = 1-x then is it right to find for cosθ through this?

Thanks!

EDIT: I guess this is wrong T_T Please tell me and sorry coz I'm still a learner T_T
EDIT: I'm having trouble with latex again.

 

[Mentallic]

Taking cosine of both sides will give

$$0 = \cos\frac{\pi}{2} + \cos[\arcsin(1-x)]$$

$$\cos(A+B)\neq \cos(A)+\cos(B)$$

Remember it's

$$\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)$$

Question: In cos[arcsin(1-x)], is it right to use "substitution" method where I let, say θ, equal to arcsin(1-x) then find for cosθ afterwards?

It's something like: cos[arcsin(1-x)]

let θ = arcsin(1-x)

sinθ = 1-x then is it right to find for cosθ through this?

Thanks!

EDIT: I guess this is wrong T_T Please tell me and sorry coz I'm still a learner T_T

Yes you're on the right track. So if $\sin(\theta) = 1-x$ then draw up a right-angled triangle with one angle being $\theta$ and then label the sides in such a way such that $\sin(\theta)=1-x$
Then, since we are actually looking for the value of $\cos(\arcsin(1-x))\equiv \cos(\theta)$ you'll need to find the value of the other side in the right-triangle.

 

[acen_gr]

$$\cos(A+B)\neq \cos(A)+\cos(B)$$

Remember it's

$$\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)$$

Oh yes! Why do I always forget that. Thanks a lot! Ok, I'm going to work on it again. Thank you for all your replies! :)

 

[acen_gr]

Alright. Summing up everyone's ideas, I got this solution:

Anyway, try to write the equation as

$$arccos(x)=\frac{\pi}{2}+arcsin(1-x)$$

and now take the cosine of both sides.

$$cos[arccos(x) = \frac{\pi}{2} + arcsin(1-x)]$$
$$0 = cos[\frac{\pi}{2} + arcsin(1-x)]$$

take the cosine of both sides and use the sum-to-product relations with the trig-inverse-trig relations to get some f(x)=0 ... then it's a matter of finding the roots of f.

Remember it's
$$\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)$$

$$0 = cos\frac{\pi}{2}cos[arcsin(1-x)] - sin\frac{\pi}{2}cos[arcsin(1-x)$$
$$0 = cos[arcsin(1-x)]$$

Then, since we are actually looking for the value of $\cos(\arcsin(1-x))\equiv \cos(\theta)$ you'll need to find the value of the other side in the right-triangle.

$$let θ = arcsin(1-x)$$ ⇔ sinθ = 1-x

for angle θ: y = 1-x, x = $\sqrt{2x-x^2}$, r = 1

∴ cosθ = $\sqrt{2x-x^2}$

$0 = -cos(θ)$
$0 = -(\sqrt{2x-x^2})$
$0 = \sqrt{2x-x^2}$
$$0 = 2x-x^2$$
$$0 = (2-x)x$$

$$x = 2 , x = 0$$

surely it is -1 ≤ x ≤ 1 (angle is the output of inverse trig functions).

so x = 0.

Hope it's right now.

 

[Simon Bridge]

note: arccos is an inverse cosine so:$$\cos(\arccos(x))=x$$

Also, you'll find there is an identity for $\cos(\frac{\pi}{2}+\theta)$ that will be useful for simplifying things.
You'll still get it the long way around though.

 

[acen_gr]

note: arccos is an inverse cosine so:$$\cos(\arccos(x))=x$$

Also, you'll find there is an identity for $\cos(\frac{\pi}{2}+\theta)$ that will be useful for simplifying things.
You'll still get it the long way around though.

oh yeaahh. i forgot T____T
how am i going to pass my test if i always keep on making mistakes.
my test is already tomorrow, and i got that problem above from a sample problem of what might appear as question for tomorrow's exam.
i think i would fail again :(
how did you guys became so gifted in math?

 

[acen_gr]

it's already late here and I'm still not yet through studying for exam :( please check my new solution. I still hope I could get it:


$x = -cos(θ)$
$x= -\sqrt{2x-x^2}$
$-x = \sqrt{2x-x^2}$
$$-x^2 = 2x-x^2$$
$$0 - 2x = 0$$
$$x = 0$$

 

[Mark44]

it's already late here and I'm still not yet through studying for exam :( please check my new solution. I still hope I could get it:


$x = -cos(θ)$
$x= -\sqrt{2x-x^2}$
$-x = \sqrt{2x-x^2}$
$$-x^2 = 2x-x^2$$

When you square -x, you get x², not -x².
You could have squared both sides of the 2nd equation, above.

$$0 - 2x = 0$$
$$x = 0$$

 

[acen_gr]

how about this:

$x = -cos(θ)$
$x= -\sqrt{2x-x^2}$
$$x^2 = 2x-x^2$$
$$2x^2 - 2x = 0$$
$$x^2 - x = 0$$

x = 1, x = 0

 

[Mark44]

That's better, but you're still not done. When you square both sides of an equation, there is the possibility that you are introducing extraneous solutions.

Check both solutions to see if they satisfy the equation you squared: x = -√(2x - x²).

 

[acen_gr]

That's better, but you're still not done. When you square both sides of an equation, there is the possibility that you are introducing extraneous solutions.

Check both solutions to see if they satisfy the equation you squared: x = -√(2x - x²).

The answer is x = 0.
Because if x = 1, then

$$x = -\sqrt{2x - x^2}$$
$$(1) = -\sqrt{2(1) - 1^2}$$
$$1 = -\sqrt{1}$$
$$1 ≠ -1$$

if x = 0

$$0 = -\sqrt{0 - 0^2}$$
$$(0) = -\sqrt{0}$$
$$0 = 0$$

Am I right sir?

 

[Mark44]

That is the correct solution of the equation x = -√(2x - x²), but it's not a solution of your original equation. I haven't checked all of your work, but there must be an error somewhere.

 

[Simon Bridge]

Yeh - not seeing where $x=-\sqrt{2x-x^2}$ comes from.

[edit] no I got it - it's from the combined trig and inverse-trig identity for cos(arcsin(x))
But that doesn't account for the extra $\frac{\pi}{2}$ in the RHS.

He seems to have gone from:
$$\arccos(x) = \frac{\pi}{2} + \arcsin(1-x)$$... to
$$x = \cos \big [ \arcsin(1-x) \big ] = \sqrt{1-(1-x)^2}$$ (don't know where the minus sign came from either.)

I detect tiredness errors.

hint: $\cos(\frac{\pi}{2})=0$

 

[acen_gr]

$$x = \cos \big [ \arcsin(1-x) \big ] = \sqrt{1-(1-x)^2}$$ (don't know where the minus sign came from either.)

That was from extracting (1-x)^2

 

[Mentallic]

$$\sqrt{x^2-2x}\neq - \sqrt{2x-x^2}$$

Remember the rule that $\sqrt{ab}=\sqrt{a}\sqrt{b}$ so this would imply that $\sqrt{x^2-2x}=\sqrt{-(2x-x^2)}=\sqrt{-1}\sqrt{2x-x^2}$ and the square root of -1 is not -1, it's imaginary. But the rule only applies for positive a and b values, so even then we can't do that.

You need to leave it as it is.

 

[Simon Bridge]

That was from extracting (1-x)^2

The minus sign outside the surd? Mentallic has described why that should not come from anything inside the surd.

go back to:
$$x=\cos\left [ \frac{\pi}{2} + \arcsin(1-x) \right ]$$... put $\theta=\arcsin(1-x)$ and simplify: what does $\cos(\frac{\pi}{2}+\theta)$ turn into?

 

[acen_gr]

The minus sign outside the surd? Mentallic has described why that should not come from anything inside the surd.

go back to:
$$x=\cos\left [ \frac{\pi}{2} + \arcsin(1-x) \right ]$$... put $\theta=\arcsin(1-x)$ and simplify: what does $\cos(\frac{\pi}{2}+\theta)$ turn into?

this?

let θ = arcsin(1-x)

$$x=\cos\left [ \frac{\pi}{2} + \arcsin(1-x) \right ]$$

$$x=\cos\left [ \frac{\pi}{2} + θ \right ]$$

$$x=cos\frac{\pi}{2}cosθ - sinθsin\frac{\pi}{2}$$

x = -sinθ

 

[Simon Bridge]

... wasn't that fun?
Now something should naturally occur to you ... you know what sinθ is!

BTW: I have a problem with this and I'm hoping someone can see the flaw in the reasoning.
I have a headslap moment coming ... maybe if I sleep on it.

 

[acen_gr]

@simon bridge, yea. and I'm still struggling because we just had this question in our exam today and our teacher afterwards told the answer was x = -1/2. is that right?

 

[Simon Bridge]

BTW: what was wrong with using the identity in post #2?

 

[Simon Bridge]

@simon bridge, yea. and I'm still struggling because we just had this question in our exam today and our teacher afterwards told the answer was x = -1/2. is that right?

well ... the second term on the LHS of the original problem in post #1 is arcsin(1-x).

if x=-0.5, then that term becomes arcsin(1.5) = undefined.

so how can that be the solution to the problem stated in post #1?

However: x=0.5 is the solution.

 

[acen_gr]

yea. that's how we argued with him. it doesn't seem right and he doesn't seem to know the right answer too. now I'm so confused.

btw, is this the identity you were referring to in reply#23?

$$\arccos(x)+\arcsin(x)=90^\circ$$

UPDATE: I think x could be equal to 1/2. What do you think?

 

[Simon Bridge]

Yep - that identity is true for all x, so if you subtract the relation you are investigating you get ... arcsin(x)+arcsin(1-x)=0 ... but I'm working on something involving the 1-2-root-3 triangle.

UPDATE: I think x could be equal to 1/2. What do you think?

It has to be - see bottom of post #24 ;) your post crossed by edit.
[edit]
BUT ... plug it in: I shouldn't do this at 3am :(
arccos(0.5)=pi/3
arcsin(0.5)=pi/6
pi/3 + pi/6 = pi/2 ... but that is not the relation: pi/3-pi/6=pi/6if arcos(x) and arcsin(1-x) are complimentary angles.
then x and 1-x correspond to the sides of a right-angle triangle.

set θ=arccos(x) and 90-θ = arcsin(1-x) are the complimentary angles right.

so x=cosθ and 1-x=sin(90-θ)=cosθ

so 1-x = x => x=1/2

The reason this got convoluted is because of the policy of helping the OP with what they are doing rather than doing the problem for them.
That was the first thing I thought of.

[edit] don't get excited - I said if ... to use the above for your problem, you need a slight modification.
It helps understand the work so far ... see next:

 

[Simon Bridge]

I have a feeling that either (1) the problem is misstated in post #1 OR (2) the teacher has made a mistake.

If I plot the relation to be proved, there is no solution - the there is no value of x that makes the LHS the same as the RHS.

Watch:

procede as last post:
λ=arccos(x) and θ=arcsin(1-x) so that λ - θ = 90 (working in degrees).

We notice that λ = 90+θ

from the definitions of the inverse trig functions we can write:

1-x = sinθ ...(1) and
x = cosλ = cos(90+θ) = -sinθ ... (2)

(1)+(2) gives: 1-x+x = 0 => 1=0 <=> nonsense!

Hence there is no solution.

----------------------------

LATER: I took the trouble to plot the function in gnu-octave:

octave:189> x=0:0.0125:1;
octave:190> y=acos(x)-asin(1-x);
octave:191> plot(x,y*(180/pi))

... the vertical is the resultant angle and the horizontal is x. Notice that at no time does the function even get close to 90 degrees.
For x<0 octave gives complex results - so, just checking that is isn't the argument that has to be 90 degrees:

octave:199> acos(-0.5)-asin(1-(-0.5))
ans =  0.52360 - 0.96242i
octave:200> arg(ans)
ans = -1.0725
octave:201> pi/2
ans =  1.5708

 

[acen_gr]

I have a feeling that either (1) the problem is misstated in post #1 OR (2) the teacher has made a mistake.

If I plot the relation to be proved, there is no solution - the there is no value of x that makes the LHS the same as the RHS.

Watch:

procede as last post:
λ=arccos(x) and θ=arcsin(1-x) so that λ - θ = 90 (working in degrees).

We notice that λ = 90+θ

from the definitions of the inverse trig functions we can write:

1-x = sinθ ...(1) and
x = cosλ = cos(90+θ) = -sinθ ... (2)

(1)+(2) gives: 1-x+x = 0 => 1=0 <=> nonsense!

Hence there is no solution.

----------------------------

LATER: I took the trouble to plot the function in gnu-octave:

octave:189> x=0:0.0125:1;
octave:190> y=acos(x)-asin(1-x);
octave:191> plot(x,y*(180/pi))

... the vertical is the resultant angle and the horizontal is x. Notice that at no time does the function even get close to 90 degrees.
For x<0 octave gives complex results - so, just checking that is isn't the argument that has to be 90 degrees:

octave:199> acos(-0.5)-asin(1-(-0.5))
ans =  0.52360 - 0.96242i
octave:200> arg(ans)
ans = -1.0725
octave:201> pi/2
ans =  1.5708

I HUNDRED PERCENT AGREE TO YOU!
I tried to mathematically prove that the answer is not possible (hence no solution)
Please check it out.

$$arccos(x) - arcsin(1-x) = 90°$$

$$arccos(x) = 90° + arcsin(1-x)$$

taking cosine of both sides:
$$x = cos90° + cos(arcsin(1-x)$$
$$x = cos[arcsin(1-x)]$$

let arcsin(1-x) = θ ⇔ sinθ = 1-x

y = 1-x x= ? r = 1

$x = \sqrt{(1)^2 - (1-x)^2}$
$x =\sqrt{(1 - (1-2x+x^2)}$
$x = \sqrt{(2x-x^2)}$

taking squares of both sides:
$x^2 = 2x+x^2$

$$x^2$$ cancels out
$$0 = 2x$$ => makes total nonsense! means there is no solution.

 

[acen_gr]

I have a feeling that either (1) the problem is misstated in post #1 OR (2) the teacher has made a mistake.

(1) The post is not misstated.

(2) I feel the same. I can't confront him now because I am not scheduled today for his class. Everyone got wrong with this. So by next time we happen to see each other I'm going to slap a paper (with this solution written) on his face!

EDIT: I probably should not do (2). I definitely don't want to see him again next semester or take this brain-cracking course again T_T

UPDATE: Anyway sir, how do you find out the solution of trigonometric equations through graphs? Like the one you posted in post #27.

 

[Mentallic]

$$arccos(x) = 90° + arcsin(1-x)$$

taking cosine of both sides:
$$x = cos90° + cos(arcsin(1-x)$$

Did you just do what I think you did?

To quote myself from earlier:

$$\cos(A+B)\neq \cos(A)+\cos(B)$$

Remember it's

$$\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)$$

EDIT: By the way, even though your proof isn't correct, at the very end you had

$$0 = 2x$$ => makes total nonsense! means there is no solution.

No way! 2x=0 implies that x=0. Of course this makes sense!

Such things that would be nonsense would be if we divided by 0, or if we tried to take the square root of a negative, or the arcsin of a value outside the range -pi/2 to pi/2, or something like 1=0.