State the value of the root of equation cos^-1 x = sin^-1 x

[Monkeyseat]

Hi,

Question

Show that the equation cos^-1 x = sin^-1 x has only one real root and state its value to three significant figures.

Working

I've sketched a graph:

Therefore, there is only one root.

However, I have no idea how to solve "cos^-1 x = sin^-1 x" to find the value of that root!

I thought maybe I could do this:

cos(cos^-1 x) = cos(sin^-1 x)
x = cos(sin^-1 x)

But then I'm stumped. Sorry for the lack of working, I just don't know where to go with this. The book says the answer is 0.707, if that helps.

I realised that cos(pi/4) is 0.707 and sin(pi/4) is also 0.707, but I don't know how that explains anything or ties in with the question... The equation is cos^-1 x = sin^-1 x not cosx = sinx. I just found that answer by putting numbers in on the calculator, no real logic used (although the interception point seems to be where y = pi/4, but my sketch is not accurate so I couldn't say for sure and I don't know what it would mean if y = pi/4 anyway).

Thanks for any help.

 

[mmm4444bot]

Hello Monkey Seat:

Use the definitions of the inverse trigonometric functions to help understand.

You're given the following.

arcsin(x) = y

arccos(x) = y

This means the following.

sin(y) = x

cos(y) = x

From this, you should be thinking about angles y for which the values of sine and cosine are the same.

Next, from your graph, you should realize that 0 <= y <= Pi/2.

In other words, you are looking for the only angle between 0 and Pi/2 for which the sine and cosine values are the same.

The rest you can do in your head.

Cheers,

~ Mark

 

[Monkeyseat]

Thanks for the reply. Sorry that it has taken so long for me to get back to you.

Okay so from 'guessing' cos(pi/4) = sin(pi/4). (I don't know how you would actually 'find' this, I just figured it out by trying numbers! How would you?)

But what we are looking for is cos^-1 (x) = sin^-1 (x), not cos(x) = sin(x). So how would I go about solving cos^-1 (x) = sin^-1 (x)?

Sorry to ask so many questions. I just find this question a bit confusing. I do not recall coming across arcsin(x) or arccos(x) yet.

Thanks.

 

[Deleted member 4993]

Thanks for the reply. Sorry that it has taken so long for me to get back to you.

Okay so from 'guessing' cos(pi/4) = sin(pi/4). (I don't know how you would actually 'find' this, I just figured it out by trying numbers! How would you?)

But what we are looking for is cos^-1 (x) = sin^-1 (x), not cos(x) = sin(x). So how would I go about solving cos^-1 (x) = sin^-1 (x)?

Sorry to ask so many questions. I just find this question a bit confusing. I do not recall coming across arcsin(x) or arccos(x) yet.

Thanks.

Did you use paper and pencil and work through the solution? Or did you just stare at the screen?

Mark said

cos^-1 (x) = sin^-1(x) = y

and he suggested a method to find y for you

I don't know how you would actually 'find' this, I just figured it out by trying numbers! How would you?

By simple mathematics (and pencil & paper) of course....

sin(y) = cos(y)

sin(y) = sin (?/2 - y)

y = ?/2 - y

y = ?/4

So now what is the value of 'x'?

 

[mmm4444bot]

… I do not recall coming across arcsin(x) or arccos(x) yet …

Really? :?

How did you sketch your graph?

 

[Deleted member 4993]

I suppose s/he is referring to the term "arc" as opposed to "^-1" - as not coming across yet.

 

[nikkor180]

Greetings:

Algebraically, given cos^-1(x) = sin^-1(x), we can begin, as did you, by taking the cosine of both sides for x = cos[sin^-1(x)]. Now, by definition, sin-1(x) is the angle whose sine is x. If the sine is x, then from the identity sin^2(u) + cos^2(u) = 1, it follows that the cosine is sqrt[1 - x^2] (discussion as to why not +/- sqrt to follow). Therefore, cos[sin^-1(x)] = sqrt[1 - x^2].

Applied to the problem at hand gives us, x = sqrt[1 - x^2] ==> x^2 = 1 - x^2 ==> 2x^2 = 1 ==> x = [sqrt(2)] / 2.

As to uniqueness, the sin^-1 and cos^-1 functions have domains [-1,1] and ranges [-pi/2, pi/2] and (0, pi) respectively. Therefore the only angles of interest (with regard to the given equation) are in the intersection of [-pi/2, pi/2] and [0, pi] which, of course, is [0, pi/2] - i.e., first quadrant.

Finally, an easy means by which to determine values of such expressions as cos[sin^-1(x)] is to make a quick sketch of a right triangle and label one of the two non-right interior angles "sin^-1(x)". That done, it follows that you would label the side opposite, x, and hypotenuse, 1. By Pythagoras, the adjacent side must have length sqrt[1-x^2]. From the diagram, you can then identify any and all values of the corresponding six trigonometric functions.

I hope this helps.

Best regards,

Rich B.

 

[Monkeyseat]

Thanks for the reply. Sorry that it has taken so long for me to get back to you.

Okay so from 'guessing' cos(pi/4) = sin(pi/4). (I don't know how you would actually 'find' this, I just figured it out by trying numbers! How would you?)

But what we are looking for is cos^-1 (x) = sin^-1 (x), not cos(x) = sin(x). So how would I go about solving cos^-1 (x) = sin^-1 (x)?

Sorry to ask so many questions. I just find this question a bit confusing. I do not recall coming across arcsin(x) or arccos(x) yet.

Thanks.

Did you use paper and pencil and work through the solution? Or did you just stare at the screen?

Yes I had tried to work through this problem before but I wasn't entirely sure what mmm4444bot meant at the time.

Mark said

cos^-1 (x) = sin^-1(x) = y

and he suggested a method to find y for you

I don't know how you would actually 'find' this, I just figured it out by trying numbers! How would you?

By simple mathematics (and pencil & paper) of course....

sin(y) = cos(y)

sin(y) = sin (?/2 - y)

y = ?/2 - y

y = ?/4

So now what is the value of 'x'?

I understand now. Thanks.

I suppose s/he is referring to the term "arc" as opposed to "^-1" - as not coming across yet.

Correct, although I know now. I just knew them as inverse sin, inverse cos and inverse tan.

Thanks mmm4444bot, Subhotosh Khan and nikkor180 for helping, much appreciated. I had seen a few things before like what you discussed nikkor180, so thanks for going through it for me.

Thanks all.