Inverse trig function inside another trig function? .:Resolved:.

Hello everyone!
i'm so lost! I was doing fine until I got to trig in this calc book, and is not even calc, its a review for calc! I hated pre-calc!
But anyways, i'm having alot of troubles trying to figure this out..
OKay here is an example i don't understand how you solve it without using a calculator! because your not allowed to use a calculator!
Evaluate the expression, draw a right traingle.

Code:

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sin(arcsin(1/2))
and
sin(arctan(3/4))

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How do you find the arcsin of 1/2? or the arctan of 3/4?
what kind of picture do I draw?
also i don't understand:
Solve the equation for x:

Code:

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arcsin(3x-pi) = 1/2

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Any help would be great!!
Thanks for listening! :D

Code:

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  |\
  |o\
2  |  \
  |  \
  |____\
      1

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Here is how to do the 1st one, after this the rest are basically the same.

Ok, draw your 1-2-Sqrt(5) triangle as above.

You know that tan(o) = 1/2 (from the triangle)
=> o = arctan(1/2) by definition

You also know that sin(o) = opp/hyp= 1/sqrt(5), also from the triangle

However, you know that arctan(1/2) actually = o +2*Pi*n
=> arctan(1/2) = o +2*Pi*n

=> sin(arctan(1/2))= +/- 1/sqrt(5)

Thanks for the relpy, but i'm still kind of lost...
sin(arctan(3/4)) = sqr(3)/2
and
sin(arcsin(1/2)) = 1/2 from the inverse function properties,
This is what the answers in the back of the book say, i understand sin(arcsin(1/2))
but i still don't understand the
sin(arctan(3/4)) = sqr(3)/2
how do you know where u draw the triangle at and how do you know what values you put in for the triangle sides?

You know that you want arctan(1/2)

=> you know that if you make a 1-2-sqrt(5) triangle, that the angle marked will be arctan(1/2)

So, with arctan(3/4) do this:

Code:

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  |\
  |o\
4  |  \  5
  |  \
  |____\
      3

---

Quote:

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You know you want the arctan(1/2)

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How do I know that i want the arctan(1/2)? I thought I wanted the arctan(3/4), because the problem is:
sin(arctan(3/4));

I understand though if you draw a 1-2-sqrt(5) triangle you will get tan(o) = 1/2, so if u drew another trangle and replace the 1 with a 3 and the 2 with a 4, you can get tan(o) = 3/4, but with that informatino how do you find the arctan(3/4)? do I use the law of sins or somthing in that nature to find the angle?

Sorry, i'l abbreviate a bit less:

1) With the first Q, you kow you wanted arctan(1/2), so you drew a 1-2-Sqrt(5) triangle, and you knew that the angle opposite the side of length 1 would be arctan(1/2), beacuse it is in a right-angled triangle, the side opposite is 1 and the side adjacent is 2.

2) Now, with the other Q, you want arctan(3/4), so you draw a 3-4-5 triangle, and know that the angle (marked 'o' in the last diagram), which is opposite 3 and adjacent to 4, MUST have tan of 3/4 (i.e. tan(o)=3/4)
=> arctan(3/4) = o ----[actually o +/- n*Pi, for all solutions]

=> you can now easily find sin(arctan(3/4))

N.B. by 'o' i mean the marked angle of last post, not 0 (zero)

Oh, and i just noticed you said:
"you can get tan(o) = 3/4, but with that informatino how do you find the arctan(3/4)?"

if tan(x) = 3/4, x is a solution arctan(3/4)
[for tan, full solution is x +/- n*Pi]

Holly crap on a cracker. I finally understand it! after many long hours of studying ur example and refreshing my mind with trig I finally get what the hell your talking about. yay! Thanks a ton for helping!:D