Inverse trigonometry?

Question

If anyone could help with any/all of the following questions it would be greatly appreciated. I am so confused..

**arc_ = the inverse if you didn't know**

1. arctan(2x-3)=pi/4
2. arcsin(3x-pi)=1/2
3. arctan(2x)= -1
4. arcsin(the square root of 2x)=arccos(the square root ofx)

5. find tan [arcsec(root 5)/2]
6. find sin (arc sec(x))
7. find csc [arctan(x/root2)]

also if anyone knows the domain and range of csc(x), sec(x), and cot(x) that would be helpful too.

Thanks!

Answer 1

1. arctan(2x-3)=pi/4
This is the same as tan(pi/4) = 2x - 3
But tan(pi/4) = 1 = 2x - 3 and 2x = 4
so: x = 2

2. arcsin(3x-pi)=1/2
This is the same as sin(1/2) = 3x - pi
sin(1/2) = 0.4794 = 3x - pi
x = (0.4794 + pi)/3 = 1.207

3. arctan(2x)= -1
tan(-1) = 2x = -1.557
x = -0.7787

4. arcsin(the square root of 2x)=arccos(the square root ofx)
Form a triangle since these are the same angle. The right hand side gives an adjacent side of SQRT(x) with a hyptoeneuse of 1 and the left hand side gives an opposite side of SQRT(2x) with a hypoteneuse of 1. So using c^2 = a^2 + b^2:
1 = 2x + x = 3x and x=1/3
But I have made this arbitrary by picking 1 as the hypoteneuse and it can really be anything. Now the angle formed from this is tan(Angle) = opposite/adjacent so:
tan(Angle) = SQRT(2x)/SQRT(x) = SQRT(2) = 1.414
Angle = 54.74 degrees = 0.955 radians

5. find tan [arcsec(root 5)/2]
The sec is 1/cos or (hypoteneuse/adjacent) so SQRT(5) is the hypoteneuse of a triangle and 2 is the side adjacent the angle formed by arc sec(SQRT(5)/2). The other side of the triangle is then SQRT(5 - 4) = 1. So:
tan (arc sec(SQRT(5)/2)) = 1/2

6. find sin (arc sec(x))
The sec is 1/cos so x is the hypoteneuse of a triangle and 1 is the side adjacent the angle formed by arc sec(x). The other side of the triangle is then SQRT(x^2 - 1).
That means the sin (arc sec(x)) is then: SQRT(x^2 - 1)/x

7. find csc [arctan(x/root2)]
As the others. The tan is x/SQRT(2) so form a triangle. Opposite side is x and adjacent is SQRT(2). So the hypoteneuse is SQRT(x^2 + 2).
Since csc is 1/sin = hypoteneuse/opposite
csc[arc tan(x/SQRT(2)) = SQRT(x^2 + 2)/x


The domain and range of csc and sec are the same. these are the inverse of the sin and cos respectively and these have values from -1 to 1. Now 1/w where w goes from 0 to 1 gives you +Inf to1 and when w goes from 0 to -1 you get -Inf to -1. So the domain and range of these are:
(-Inf, -1] and [1,+Inf)

The cot domain and range is the same as those of the tangent. So the domain and range are (-Inf,+Inf)

Answer 2

To you understand better the mean of "arctan", "arcsin", "arcsec", and so on, read them as "arc whose tan", "arc whose sin", ...

2. arcsin(3x-pi)=1/2
You can read like this: The angle whose sin is 1/2 is 3x-pi
First: the angle whose sin is 1/2 is (pi/3)rad or 45° (0 Second: then 3x-pi = pi/3
3x = pi/3 + pi
3x = 4pi/3
x = 4pi/9
Now, try to do the other ones.
I hope it can help you.
I'm Brazilian and I don't know to write in english very well. So, I'm sorry for the mistakes!!!
My email: [e-mail]

Answer 3

Do you know algebra yet?

Youre supposed to reverse procedure/function to both sides in order to isolate variables.

This is barely a trig question. Its more of a question that tests your conceptual understanding of algebra.

arctan(2x - 3) = π/4

Take the tan of both sides.
tan arctan (2x - 3) = tan π/4

cancel out function and inverse function
2x - 3 = tan π/4

All Im doing is rearranging what is there. Basic Algebra! I dont see where your issue is coming from

The only questions here that are even remotely confusing are questions 4 on up.

Answer 4

The domain of sec is 'all real numbers *except* [(2n+1)*pi]/2 for n = -1, 0, 1, .... The range is (-infinity,-1] U [1, infinity).

still answering

edit--got sidetracked at work...that happens from time to time