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
Is that correct, and if not, can someone please help me out?
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
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
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