# Thread: simplifying inverse trig functions

1. ## simplifying inverse trig functions

I had trig last semester, but I seriously need to review what I "learned".

Anyway, I need help simplifying this experssion:

tan(sin^-1x)

by following the example in my book (because my professor hasn't even show us how to do this!), this is how far I've gotten:

tan(sin^-1x)
y = sin^-1x
sin y = x and -n/2 < y < n/2

1 + cot^2y = csc^2y . . . by the pythagorean identities
csc y = (square root of)1 + x^2 . . . I'm pretty sure I messed up on this step

tan(sin^-1x) = tan y = 1/csc y = 1/ (square root of)1 + X^2

2. Originally Posted by zachb
I had trig last semester, but I seriously need to review what I "learned".

Anyway, I need help simplifying this experssion:

tan(sin^-1x)

by following the example in my book (because my professor hasn't even show us how to do this!), this is how far I've gotten:

tan(sin^-1x)
y = sin^-1x
sin y = x and -n/2 < y < n/2

as sin(y) = x, cos^2(y) = 1 - sin^2(y) = 1 - x^2

so tan( sin^-1(x)) = tan(y) = sin(y)/cos(y)

but cos(y) = +/- sqrt(1-x^2), and as tan( sin^-1(x)) has the same sign as
x:

tan( sin^-1(x)) = tan(y) = sin(y)/cos(y) = x/sqrt(1-x^2)

RonL

3. Can you explain why you have to find cos(y) instead of csc(y)?

In my book, for the expression, sin(tan^-1x), it finds sec(y), so I figured that for this problem I would have to find csc(y).

4. Originally Posted by zachb
I had trig last semester, but I seriously need to review what I "learned".

Anyway, I need help simplifying this experssion:

tan(sin^-1x)

by following the example in my book (because my professor hasn't even show us how to do this!), this is how far I've gotten:

tan(sin^-1x)
y = sin^-1x
sin y = x and -n/2 < y < n/2

1 + cot^2y = csc^2y . . . by the pythagorean identities
csc y = (square root of)1 + x^2 . . . I'm pretty sure I messed up on this step

tan(sin^-1x) = tan y = 1/csc y = 1/ (square root of)1 + X^2

CaptainBlack is completely right, but his method does not require the use of Pythagoras' theorem as your book does. Which is fine, sometimes you can't bother with drawing a triagle and all that stuff. But just in case you are partial to the Pythagoras' way of doing this, and for your general edification (it's always best to know more than one way of doing a problem), I will do it using Pythagoras. (See diagram below).

so tan(sin^-1x)
let y = sin^-1x
then sin(y) = x ...(or more explicitly, x/1, this is how we draw the triagle, we write over 1 since we want a ratio, then we can say sine is opposite/hyptoenuse, so we know the opposite side to the angle y is x, ad the hypotenuse is 1).

then by Pythogoras, cosy = sqrt(1 - x^2) ........(see diagram below)

so now tan(sin^-1x)
= tan(y) .............since we called sin^-1x the name y
= sin(y)/cos(y) ........a trig identity
= x/sqrt(1 - x^2) .....just as the brilliant CaptainBlack said

5. here is the triangle:

6. Originally Posted by zachb
Can you explain why you have to find cos(y) instead of csc(y)?

In my book, for the expression, sin(tan^-1x), it finds sec(y), so I figured that for this problem I would have to find csc(y).
there was no need for cscy, we are finding an exact value here, not another trig identity. your book probably used cscy because tany = secy/cscy so they could just solve for secy. here we need no such association, we dont care what secy is, we just want the value of tany. got it?

in this question, we could also find the value of tany directly from the triangle we constructed, it would just be opposite/adjacent

7. Originally Posted by zachb
Can you explain why you have to find cos(y) instead of csc(y)?
You don't have to find cos(y) rather than csc(y), its a matter of
convienience. Also the trig identities for cos and sin are more familiar than
those for csc and sec, so its prefereable to work with the former rather than
the latter other things being equal.

RonL

8. Originally Posted by Jhevon
CaptainBlack is completely right, but his method does not require the use of Pythagoras' theorem as your book does.
Just to be picky, as I use the identity:

cos^2(y) = 1 - sin^2(y),

which is Pythagoras' theorem I do require it!

RonL

9. Originally Posted by CaptainBlack
Just to be picky, as I use the identity:

cos^2(y) = 1 - sin^2(y),

which is Pythagoras' theorem I do require it!

RonL
well, maybe, i've always thought of sin^2(x) + cos^2(x) = 1 as a corollary of Pythagoras' theorem. it's result of the theorem applied to the unit circle. but hey, what do i know. let it be as you say, Captain

10. Originally Posted by Jhevon
well, maybe, i've always thought of sin^2(x) + cos^2(x) = 1 as a corollary of Pythagoras' theorem. it's result of the theorem applied to the unit circle. but hey, what do i know. let it be as you say, Captain
Take an arbitary right triangle with one angle theta != 90, and hypotenuse h.

Then Pythagoras and the definitions of sin and cos in terms of the sides
of right triangles tells us that:

h^2 [sin^2(theta) + cos^2(theta)] = h^2

so sin^2(theta) + cos^2(theta) = 1.

So Pythagoras implies sin^2(theta) + cos^2(theta) = 1.

Now suppose we have for a right triangle as above that:

sin^2(theta) + cos^2(theta) = 1

then use the definitions that sin(theta)=a/h and cos(theta) = b/h, where a
and b are the lengths of the oposite and adjacent legs of the triangle, then:

a^2/h^2 + b^2/h^2 = 1

or:

a^2 + b^2 = h^2

So the identity sin^2(theta) + cos^2(theta) = 1 implies Pythagoras theorem.

Hence sin^2(theta) + cos^2(theta) = 1 is equivalent to Pythagoras' theorem
(and so is just a restatement of it in terms of trig functions).

RonL