Inverse (1 Viewer)

[currysauce]

Find the exact values of x and y which satisfy the simultaneous equations

INVsinx - INVcosy = pi/12

INVcosx + INV siny = 5pi/12

AND

Show that the two curves y=INVcosx and y= 2INVtan(1-x) cut the y-axis at the same point and have a common tangent at this point

 

[香港!]

Find the exact values of x and y which satisfy the simultaneous equati

INVsinx - INVcosy = pi/12

INVcosx + INV siny = 5pi/12

sin^-1 (y)+cos^-1 (y)+cos^-1 (x)-sin^-1 (x)=5pi\12-pi\12=4pi\12=pi\3
(pi\2)+cos^-1 (x)-sin^-1 (x)=pi\3
cos^-1 (x)+sin^-1 (x)-sin^-1 (x)=-pi\6+sin^-1 (x)
pi\2=-pi\6+2sin^-1 (x)
2pi\3=2sin^-1 (x)
sin^-1 (x)=pi\3
x=sin (pi\3)=sqrt (3) \ 2

sin^-1 (x)-cos^-1 (y)=pi\12
pi\3-cos^-1 (y)=pi\12
cos ^-1 (y)=pi\4
.: y=cos (pi\4)=1\sqrt 2

 

[香港!]

"Show that the two curves y=INVcosx and y= 2INVtan(1-x) cut the y-axis at the same point and have a common tangent at this point"
y=cos^-1 (x)
y=2tan^-1 (1-x)

At y-axis, x=0

y=cos^-1 (0)=pi\2
y=2tan^-1 (1-0)=2pi\4=pi\2
so they cut the y-axis at same point

y=cos^-1 (x)
y'=-1\sqrt(1-x²)
at (0,pi\2)
y'=-1

y=2tan^-1 (1-x)
y'=-2\[1+(1-x)²]
at (0,pi\2)
y'=-2\2=-1

.: they have common tangent

 

[currysauce]

thanks alot

new question

evaluate INT from 1.5 to 0 dx / root(9-2x²) in exact form


i get this

1/root(2) sin^-1 (root(2)x/3) then when i substitute, don't get the answer

 

[word.]

Subbing it does give the answer ...

pi / 4sqrt[2]

 

[currysauce]

more questions!!

Find the inverse of f(x) = (x-4)/(x-1) ((x not = 2))


AND

exact value of sin(2tan^(-1)1/2)

 

[香港!]

more questions!!

Find the inverse of f(x) = (x-4)/(x-1) ((x not = 2))


AND

exact value of sin(2tan^(-1)1/2)

first one...

let y=f(x) = (x-4)/(x-1)
inverse:
x=(y-4)\(y-1)
xy-x=y-4
y(x-1)=x-4
y=...


"exact value of sin(2tan^(-1)1/2)"
let tan^-1 (1\2)=A
tanA=1\2
so u get sin(2A)=2sinAcosA
use pythagorus... den da stuffs in