Famous Inverse Trig Questions (1 Viewer)

[FinalFantasy]

x² + 2x + 4
x>=-1
How do u find the inverse function of that?

y=x²+2x+4
inverse: x=y²+2y+4
y²+2y+(4-x)=0

y=[-2+-sqrt (4-4(4-x) ) ]\2
=[-2+-sqrt(4x-12) ]\2
=-1+-sqrt(x-3)

.: y=-1+sqrt (x-3) since y>=-1

 

[Slidey]

x^2 + 2x + 4=y

Complete the square:
(x+1)^2 + 3=y
(x+1)^2=y-3
x+1=sqrt(y-3)
x=+sqrt(y-3) - 1
Now you had a restricted domain, x>=-1
x^2 + 2x + 4=y
(-1)^2 + 2*(-1) + 4=y=1-2+4=3
so y>=3
Thus the restricted domain for the inverse function:
y=+sqrt(x-3)-1 is x>=3
Which basically equates to the function:
y=sqrt(x-3)-1 (positive square root only)

 

[velox]

y²+2y+(4-x)=0

y=[-2+-sqrt (4-4(4-x) ) ]\2

FF how did you get from the first step to the second?

 

[FinalFantasy]

y²+2y+(4-x)=0

y=[-2+-sqrt (4-4(4-x) ) ]\2

FF how did you get from the first step to the second?

y=[-b+-sqrt(b²-4(a)(c))]\2a --->quadratic formula
where a is coefficient of the y² term, b is coefficient of y term and c is the constant

 

[velox]

thanks didn't think of that.

 

[velox]

What about this one....

g(x) = 2x / (1+x^2)

Is there a general way of finding them cos i have trouble with these, and the cambridge book doesnt explain the method.

 

[Slidey]

General method? Solve for x.

You know how if they give you something like:
ax+by+c=0, and say solve for y, you get a monic y to one side? y=-(ax+c)/b Do that here, but for x instead. Completing the square is usually supremely useful, as the quadratic formula gets very ugly.

y=2x/(1+x^2)
y+yx^2=2x
y+yx^2-2x=0
x^2-2x/y+1=0 (now complete the square)
(x-1/y)^2+1-1/y^2=0
(x-1/y)^2=1/y^2-1=(1-y^2)/y^2
x-1/y=sqrt(1-y^2)/y
x=1/y+sqrt(1-y^2)/y

x=(1+sqrt(1-y^2))/y

Thus, the inverse function you want is:

y=(1+sqrt(1-x^2))/x

I chose the positive square root since it gives the largest domain (to both negative and positive infinity), while the negative square root will give you the restricted bit in the middle.

 

[velox]

slide where did the 1/y^2 come from?

 

[Slidey]

x^2-2x/y+1=0 (now complete the square)
(x-1/y)^2+1-1/y^2=0

There?

If you expand the expression (x-1/y)^2 you get x^2-2x/y+1/y^2. But we never had a 1/y^2, so we need to subtract it.

OR do you mean here?

1/y^2-1=(1-y^2)/y^2

I found a common denominator.

 

[yrtherenonames]

completing the square

 

[nit]

Originally posted by Velox
Is there a general way of finding them cos i have trouble with these, and the cambridge book doesnt explain the method.

well g(x)=y, say;
you seek g^-1(y)=x - ie the inverse function;
so your task is simply to solve for x in terms of y, and since x and y are dummy variables, convert the expression into ones incorporating x as opposed to y.

This applies in general to most invertible functions you'd encounter, if solvable in this manner.

 

[velox]

Thanks nit, but i meant about the ones where you can't just solve for x, i.e. like the one posted above

The first part of this question is easy but how do u do the second part?
http://img284.imageshack.us/img284/9695/inv1b1ps.jpg

 

[yrtherenonames]

Isn't it just pi/4?

Let a=tan^-1 (1/2)
Let b=tan^-1 (1/3)
Then tan(a+b)=1 (from part one)
Therefore a+b=pi/4
So tan^-1(1/2)+tan^-1(1/3)=pi/4

 

[Slidey]

Try these:

Show:
arctan(4)-arctan(3/5)=pi/4
arcsin(3/5)+arctan(7/24)=arccos(3/5)

Find exactly:
sin(2arctan(4/3))
cos(arctan(4/3)-arccos(5/13))
sin(arccos(3/5)+arctan(-3/4))

Prove that:
arcsinx=arccos(sqrt(1-x^2)) for 0<=x<=1

 

[FinalFantasy]

Try these:

Show:
arctan(4)-arctan(3/5)=pi/4
arcsin(3/5)+arctan(7/24)=arccos(3/5)

Find exactly:
sin(2arctan(4/3))
cos(arctan(4/3)-arccos(5/13))
sin(arccos(3/5)+arctan(-3/4))

Prove that:
arcsinx=arccos(sqrt(1-x^2)) for 0<=x<=1

to show
"arctan(4)-arctan(3/5)=pi/4"
let A=tan^-1 4 and B=tan^-1 (3\5)
tan A=4 and tan B=3\5
tan(A-B)=(tanA-tanB)\(1+tanAtanB)
=(4-3\5)\(1+12\5)
=1
.: A-B=pi\4
hence tan^-1 (4)-tan^-1 (3\5)=pi\4

to show: "arcsin(3/5)+arctan(7/24)=arccos(3/5)"
let sin^-1 (3\5)=A and tan^-1 (7\24)=B
sinA=3\5 and tan B=7\24
cos (A+B)=cosAcosB-sinAsinB
=(4\5)(24\25)-(3\5)(7\25)
=3\5
.: cos^-1 (3\5)=A+B
hence sin^-1 (3\5)+tan^-1 (7\24)=cos^-1 (3\5)

Find exactly:
"sin(2arctan(4/3))"
let tan^-1 (4\3)=A
tanA=(4\3)
.: sin(2arctan(4/3))"=sin(2A)=2sinAcosA=2(3\5)(4\5)

"cos(arctan(4/3)-arccos(5/13))"
let tan^-1 (4\3)=A and cos^-1 (5\13)=B
tanA=3\4 and cos B=5\13
.: cos(arctan(4/3)-arccos(5/13))=cos (A-B)=cosAcosB+sinAsinB
=(4\5)(5\13)+(3\5)(12\13)



"sin(arccos(3/5)+arctan(-3/4))"
let A=cos^-1 (3\5) and B=tan^-1 (-3\4)
cosA=3\5 and tanB=-3\4
.: sin(arccos(3/5)+arctan(-3/4))=sin(A+B)
=sinAcosB+cosAsinB
=(4\5)(4\5)+(3\5)(-3\5)
=7\25

is dat rite

 

[velox]

Isn't it just pi/4?

Let a=tan^-1 (1/2)
Let b=tan^-1 (1/3)
Then tan(a+b)=1 (from part one)
Therefore a+b=pi/4
So tan^-1(1/2)+tan^-1(1/3)=pi/4

lol damn, i thought there was some work involved in the second part. Didnt realise it was that obvious ><

 

[FinalFantasy]

Prove that:
arcsinx=arccos(sqrt(1-x^2)) for 0<=x<=1

to prove sin^-1 x=cos^-1 (sqrt(1-x²) )
let A=sin^-1 x and B=cos^-1 ((sqrt(1-x²))
sinA=x and cos B=sqrt(1-x²)
by pythagorus cosA=sqrt(1-x²)=cosB
and by pythagorus sinB=x=sinA
.: A=B
.: sin^-1 x=cos^-1 ((sqrt(1-x²))

is dat rite?

 

[velox]

pretty sure they're right. Slide got them out of either cambridge or fitz, i've done them somewhere b4.

 

[FinalFantasy]

how do u do this one:
Show:
arcsin (1/sqrt5) + arcsin (1/sqrt10) = pi/4

let A=sin^-1 (1\sqrt 5) and B=sin^-1 (1\sqrt 10)
sinA=1\sqrt 5 and sinB=1\sqrt 10
sin(A+B)=sinAcosB+cosAsinB
=(1\sqrt5)(3\sqrt 10)+(2\sqrt 5)(1\sqrt 10)
=3\sqrt 50+2\sqrt 50
=5\sqrt50
=5\5root2
=1\root2
=sin pi\4

.: arcsin (1/sqrt5) + arcsin (1/sqrt10) = pi/4

 

[velox]

thanks, just got it so i deleted the post. =)