I had trig last semester, but I seriously need to review what I "learned".
Anyway, I need help simplifying this experssion:
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:
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?
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
in this question, we could also find the value of tany directly from the triangle we constructed, it would just be opposite/adjacent
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
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).