Inverse trig question: (1 Viewer)

[The Bograt]

Solve:

sin-1 x = cos-1 x (where sin-1 is inverse sine etc)

I can get an answer (1/root2) but I don't think the way I do it is allowed, especially if they ask you to prove it.

Can you just say :. sin x = cos x ?
tan x = 1
x = pi/4
and then take the inverse sin of pi/4 ?

Or is it more complicated than that?

 

[withoutaface]

what I'd do is as follows:

arcsinx=arccosx
cos(arcsinx)=x
root(1-sin²(arcsinx))=x
root(1-x²)=x
1-x²=x²
1=2x²
+/-root(1/2)=x
+/-1/root2=x

 

[BillyMak]

Do it like this, at least it's a proof:

sin^-1x = cos^-1x

let y = cos^-1x
x = cos y

Draw up a triangle, trying my best here....

fffffffffffffffffffffffffff/|
ffffffffffffffffffffffffff/ |
ffffffffffffffffffffff1 /ff|....sqrt(1-x²)
fffffffffffffffffffffff /fff|
ffffffffffffffffffffff/y _|
ffffffffffffffffffffffffffx

Now go back to original eqn.

Take sin of both sides gives x = sin(cos^-1x)
but cos^-1x = y
:. x = sin y

fron triangle, sin y = sqrt(1-x²)

:. x² = 1 - x²

x = +/- 1/2^1/2

 

[BillyMak]

Ahh crap... my triangle didn't work

Edit: Thanks withoutaface

 

[ngai]

x = +/- 1/2^1/2

when x = -sqrt(1/2)
sin^-1x = -pi/4
cos^-1x = 3pi/4

 

[withoutaface]

Shit I was thinking that, works one way but not the other:S

 

[The Bograt]

thanks withoutaface, but I don't understand this step:

root(1-sin2(arcsinx))=x
root(1-x2)=x

 

[withoutaface]

sin(arcsinx)=x
ie (sin of the inverse sin of x)=x

 

[The Bograt]

Thanks for the help man, appreciate it

 

[Li0n]

Solve:

sin-1 x = cos-1 x (where sin-1 is inverse sine etc)

I can get an answer (1/root2) but I don't think the way I do it is allowed, especially if they ask you to prove it.

Can you just say :. sin x = cos x ?
tan x = 1
x = pi/4
and then take the inverse sin of pi/4 ?

Or is it more complicated than that?

thats perfectly correct as well
except you must remember thatn tan^-1 1 = pi/4 and 5pi/4

 

[CrashOveride]

when x = -sqrt(1/2)
sin^-1x = -pi/4
cos^-1x = 3pi/4

You just needed to check your solutions before (because you did some squaring).

Solve:

sin-1 x = cos-1 x (where sin-1 is inverse sine etc)

I can get an answer (1/root2) but I don't think the way I do it is allowed, especially if they ask you to prove it.

Can you just say :. sin x = cos x ?
tan x = 1
x = pi/4
and then take the inverse sin of pi/4 ?

Or is it more complicated than that?

Yes

 

[BillyMak]

You just needed to check your solutions before (because you did some squaring).

Yeah, I would usually check anyways, couldn't be bothered but....

 

[BillyMak]

Great time to be demotivated hey?

 

[Archman]

Solve:

sin-1 x = cos-1 x (where sin-1 is inverse sine etc)

I can get an answer (1/root2) but I don't think the way I do it is allowed, especially if they ask you to prove it.

Can you just say :. sin x = cos x ?
tan x = 1
x = pi/4
and then take the inverse sin of pi/4 ?

Or is it more complicated than that?

well if u du it properly it can work

let y = sin-1 x = cos-1 x
so x = siny = cosy
etc etc
work out y then work out x

 

[CrashOveride]

Bograt i think it should be pointed out that arcsinx=arccosx doesnt mean sinx=cosx

 

[Paroissien]

What do you mean by arc?

 

[Constip8edSkunk]

inverse (sin/cos/tan)

 

[redslert]

What do you mean by arc?

sin^-1x
cos^-1x
tan^-1x

 

[smallcattle]

i didnt know arc was inverse until now...

 

[CrashOveride]

You learnt just in time