# How to solve these questions about inverse trig functions?

#### hndalama

##### Junior Member
a.) Given that y=arctan(x), express sin(y) + cos(y) in terms of x

b.) Given that x =tany, find arccotx in terms of y

The book gives the solutions but can someone explain how to solve these questions?

#### Subhotosh Khan

##### Super Moderator
Staff member
a.) Given that y=arctan(x), express sin(y) + cos(y) in terms of x

y = tan-1(x) → x = tan(y) → sin(y) = x/√(1+x2) & cos(y) = 1/√(1+x2)

b.) Given that x =tany, find arccotx in terms of y

The book gives the solutions but can someone explain how to solve these questions?
Continue....

#### hndalama

##### Junior Member
I don't understand how you're getting from x = tan(y) to sin(y) = x/√(1+x2) & cos(y) = 1/√(1+x2)

#### stapel

##### Super Moderator
Staff member
I don't understand how you're getting from x = tan(y) to sin(y) = x/√(1+x2) & cos(y) = 1/√(1+x2)

What did you get when you drew the right triangle, labelled the one angle as "y", noted that x = x/1, labelled the two appropriate sides of the triangle, applied the Pythagorean Theorem, etc, etc? What have you done? What has this led? Where are you stuck?

Please be complete. Thank you! #### hndalama

##### Junior Member
I didn't realise I could draw a right triangle from the information given. Knowing that now, part a was simpler than I thought. But how can part b be solved using the triangle?

#### stapel

##### Super Moderator
Staff member
I didn't realise I could draw a right triangle from the information given. Knowing that now, part a was simpler than I thought. But how can part b be solved using the triangle?
The second one works in exactly the same way as the first one. Did you even try to do it? If so:

What did you get when you drew the triangle, you labelled one angle with the indicated variable, you noted which sides of the triangle had to have been used to find the tangent value, you converted the tangent value into a fraction, you labelled the two sides with the appropriate variable and value, you applied the same Theorem given to you last time, and you found the expression for the third side? Where are you stuck in your work?

Please be complete. Thank you! #### hndalama

##### Junior Member
after applying the Pythagoras theorem to get the value of the hypotenuse in terms of x, i don't know what to do to derive arccotx from this information.
In part a, it was simply a matter of applying soh cah toa to find siny and cosy in terms of x but this doesn't apply to arccotx. the question is also asking us to express arccotx in terms of y, which is different to what question a is asking.

I tried using the theorem tany = 1/coty but using this I could only derive y=arccot(1/x). This is where I am stuck, How can I use the triangle to find 'arccotx in terms of y?'

#### Eunice StJariel7

##### New member
Hey,
So I understand why it’s so difficult. You are right. Using the CAST diagram with the triangle and Pythagoras’ might not work this time.

I found out that by definition, arccotx is (pi/2) minus arctanx. So you must show that somehow arctanx is y. So then, if x=tany, then multiplying both sides by tan^-1 you get tan^-1(x)=y. So tan inverse equals y. Here we see that tan^-1(x) is the same as arctanx. Since arccotx=(pi/2) minus arctanx, arccotx=(pi/2) minus y.

I know it’s 2 years late but I am studying the subject and was just struggling with this question as well. Hopefully you or someone else finds this table helpful;

I found it at https://brownmath.com/twt/inverse.htm

nverse trig functions:
functionderived fromdomainrange
Arcsininverse of sine function[−1, +1]Q IV, I: [−π/2, +π/2]
ArccosArccos x = π/2 − Arcsin x[−1, +1]Q I, II: [0, π]
Arctaninverse of tangent functionall realsQ IV, I: (−π/2, +π/2)
ArccotArccot x = π/2 − Arctan xall realsQ I, II: (0, π)
ArcsecArcsec x = Arccos(1/x)(−∞, −1] and [1, ∞)Q I, II: [0, π]
ArccscArccsc x = Arcsin(1/x)(−∞, −1] and [1, ∞)Q IV, I: [−π/2, +π/2]