# 2 Inverse trig Qs (1 Viewer)

#### MarsBarz

##### Member
1. Sketch y=tan-1(tanx)

2. Prove tan-1(3/4)+sin-1(5/13)=cos-1(33/65)

#### jemsta

##### I sit here alone
i think for question 1 its jus y=x since the inverse and the function just cancel so your left with y=x
for 2nd question
change inverse tan 3/4 into inverse cos(4/5) and change inverse sin (5/13) into inverse cos (12/13) then use the double angle formula for cos
ie cos(x+y) where x= inverse cos (4/5) and and y= inverse cos (12/13)

#### MarsBarz

##### Member
1. What about domain and range restrictions?
2.?

#### MarsBarz

##### Member
For question 2 I lost you at the double angle formula. Can you clarify?

#### gman03

##### Active Member
1: since the image of arctan (inverse tan) is from -pi/2 to pi/2 exclusive, you have a straight interval of y=x between and exclude the two points (-pi/2, -pi/2) and (pi/2, pi/2)

edit: this response is incomplete, please refer to post #12 for further explaination

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#### gman03

##### Active Member
MarsBarz said:
For question 2 I lost you at the double angle formula. Can you clarify?
He tries to show that

cos (x + y) = 33/65

with

cos (x + y) = cos(x)cos(y) - sin(x)sin(y)

where

x = tan-1(3/4)
= sin -1(3/5) = cos -1(4/5)

y = sin-1(5/13) = cos -1 (12/13)

so that

cos (x + y) = 4/5 * 12 /13 - 5 / 13 * 3 / 5 = (48 - 15) / 65

Afterward you take cos -1 of cos (x + y).

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#### MarsBarz

##### Member
gman03 said:
1: since the image of arctan (inverse tan) is from -pi/2 to pi/2 exclusive, you have a straight interval of y=x between and exclude the two points (-pi/2, -pi/2) and (pi/2, pi/2)
So does that mean that you would have two vertical asymptotes at -pi/2 and pi/2 ?
Would the line y=x be continuous after those two points?

Thanks for the second question.

Graph...........

#### MarsBarz

##### Member
Cheers. I think that I finally figured it out. Can you just confirm that the following is the correct working for such a question:

y=tan-1(tanx)
tany=tanx
y=x

Taking into account that the range of y=tan-1A is -PI/2 < y < PI/2
Since we have no restriction on the domain of tanx, we sub in the minima and maxima values of the range into y=x, which gives us the domain -PI/2 < x < PI/2.
Which gives us our discontinuous y=x graph.

See I'm a little fuzzy on getting the domain for this curve. Is what I have done even correct?

Seeking final confirmation. (Sorry for being paranoid but that question has bugged me for a while and I have been given conflicting answers from teachers and tutor).

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#### Trev

##### stix
It's a better explanation than I would ever bother giving.

#### MarsBarz

##### Member
I'm still not confident with those graphs and I am still unsure that the solution which you provided is correct.

I did it again and this is what I found:

y = tan-1(tanx)

Consider y = tan-1A
The range is always -PI/2 < y < PI/2.
So we have determined that the range of the original curve is -PI/2 < y < PI/2.

Now we must find the domain of this curve.
Consider y= tan-1A
The domain is all A.
So all tanx for the original curve.
Does that correspond to all x? What happens at asymptotes of tanx, and does that mean that the graph would not just stop at the x values of -PI/2 and PI/2 like in your diagram?

Thinking about this I think that we still get vertical asymptotes at x values of PI/2, 3PI/2 etc...

Somebody help lol!! I'm getting more confused by the minute.

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#### gman03

##### Active Member
MarsBarz said:
I'm still not confident with those graphs and I am still unsure that the solution which you provided is correct.

I did it again and this is what I found:

y = tan-1(tanx)

Consider y = tan-1A
The range is always -PI/2 < y < PI/2.
So we have determined that the range of the original curve is -PI/2 < y < PI/2.

Now we must find the domain of this curve.
Consider y= tan-1A
The domain is all A.
So all tanx for the original curve.
Does that correspond to all x? What happens at asymptotes of tanx, and does that mean that the graph would not just stop at the x values of -PI/2 and PI/2 like in your diagram?

Thinking about this I think that we still get vertical asymptotes at x values of PI/2, 3PI/2 etc...

Somebody help lol!! I'm getting more confused by the minute.

I apologise for giving you incomplete answer. You are RIGHT about the domain (x values). The function is actually a PERIODIC function since tan x repeats itself every pi, so the graph actually corresponds to parallel slopes. It does not have any asymptotes.

To find the graph of function, you should start from looking from the values of x.
1. x in real

But the domain of tan x excludes x = n*pi + pi/2 where n is an integer.

Now let y(x) = tan-1 ( tan x )
then y(x+pi) = tan-1 ( tan (x+pi) ) = tan-1 ( tan x ) = y(x)
so y is pi-periodic.

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#### MarsBarz

##### Member
I don't quite understand the logic behind that second last line. Other than that you're graph looks perfect. Finnaly someone with a solution that looks correct hehehe...

Thanks for the help by the way, appreciate it.