graph y=cos(InverseSin x) (1 Viewer)

[foram]

Graph:

y=cos(InverseSin x)

this question has me totally stumped.

i was thinking inverse sin is pi/2 to -pi/2 and x is 1 to -1. but is that really useful? maybe i can change inverse sin to something else, but i dont know how.

 

[Slidey]

@=arcsin(x)
sin^2(@)+cos^2(@)=1
cos^2(@)=1-sin^2(@)
cos(@)=sqrt(1-x^2)
thus y=sqrt(1-x^2)

So graph a positive semicircle.

 

[pLuvia]

This maybe a bit tedious, but you can find the stat points of the function then the coordinates where it crosses the x and y axis.

Or if else fails....

Trial and error , sub points into the function and then do a rough sketch.

But the latter is for checking to see if the graph has that kind of shape

Or use a trig triangle

 

[namburger]

y=cos(InverseSin x)
Domain: -1< x <1
Range: 0 < y <1 (Y can't be -1 < y < 1 because from our domain, the min value y can be is 0 and max is 1)

y' = -sin(InverseSin x) . 1/root(1-x^2)
= - x/root(1-x^2)
Stationary pt occurs at x=0, sub x = 0 into y=cos(InverseSin x). Stationary pt (0,1)

From y', we can see that at x = +/- 1, the gradient is undefined, therefore a vertical tangent occurs at those pts.
At x = 1, y =0
At x =-1, y =0
At x = 0, y = 1
These coordinates all satisfy the domain and range as previously mentioned. Draw in stationary pt and veritcal tangents and draw the curve, im preety sure it should look like a semicircle

PRO METHOD

@=arcsin(x)
sin^2(@)+cos^2(@)=1
cos^2(@)=1-sin^2(@)
cos(@)=sqrt(1-x^2)
thus y=sqrt(1-x^2)
So graph a positive semicircle.

 

[tommykins]

Sorry if this is completely off topic - but I just found the irony in your sig hilarious.

Haha you're an awesome dude

 

[Gay Captain]

Actual pro method

 

[Slidey]

Or use a trig triangle

Indeed. Draw up a right-angle triangle.

@=arcsin(x)

Put @ as one of the two unknown angles. Now sin(@)=x/1. That is: the length of the hypotenuse is 1 and the length of the opposite side is x. let z be the adjacent side length. cos(@)=z/1. Now, use Pythagoras's theorem:

x^2+z^2=1
so
z=sqrt(1-x^2)
Thus cos(@)=sqrt(1-x^2)/1
So y=cos(cos(sqrt(1-x^2)))
y=sqrt(1-x^2)

Basically the same as I did before, due to the way the sin^2(a) + cos^2(a)=1 identitity is derived (based on the unit circle). However, this way is useful when you have things like:

Simplify:
f(x)=tan(arccos(x+1))

You'd draw a triangle as above:
Adjacent side is length x+1
Hypotenuse is length 1
Opposite is thus sqrt(-2x-x^2)
So arcos(x+1)=arctan(sqrt(-2x-x^2)/(x+1))
Thus f(x)=sqrt(-2x-x^2)/(x+1)