Inverse Trig Question (1 Viewer)

[CubanitaChilena]

Hey this is probably a really easy question, but I haven't found any examples in any of my textbooks on how to do it and we've only brushed really briefly on the inverse functions topic in class. The question is:

Find the exact value of:

sin(sin-1(4/5))

I put the question into my calculator and got 0.8... which could easily be turned into an exact answer, but I'm wondering if there is anymore to the question? Maybe some sort of trig formula is involved?? (we haven't done trig yet... we're heaps behind). I'm under the impression that:

sin-1(4/5) + cos-1(4/5)=π/2 is involved somehow... then again I have no clue. *shrug*

Any help would be appreciated.

Thanks.

 

[lyounamu]

Hey this is probably a really easy question, but I haven't found any examples in any of my textbooks on how to do it and we've only brushed really briefly on the inverse functions topic in class. The question is:

Find the exact value of:

sin(sin-1(4/5))

I put the question into my calculator and got 0.8... which could easily be turned into an exact answer, but I'm wondering if there is anymore to the question? Maybe some sort of trig formula is involved?? (we haven't done trig yet... we're heaps behind). I'm under the impression that:

sin-1(4/5) + cos-1(4/5)=π/2 is involved somehow... then again I have no clue. *shrug*

Any help would be appreciated.

Thanks.

Well, if that's the case, know your range and domain.
Inside the inverse sin, the value must be -1 <_ x <_ 1 (where x is any number)
And the inside the sin, the value must be -pi/2 <_ y <_ pi/2 (where y is any number)

And there is nothing to it apart from that as long as the questions satisfies the following requirement

You will use trig formula when you get question like sin-1(4/5) + cos-1(4/5)

 

[CubanitaChilena]

Wait, I just realised 0.8=4/5 *slaps forehead in idiocy*

So 4/5 is the answer then?

So then sin(sin-1(a)) = a ?

How would that extend to:

cos(sin-1(4/5))

tan(cos-1(12-13))

?

 

[Aerath]

Find the exact value of:

sin(sin-1(4/5))

I put the question into my calculator and got 0.8... which could easily be turned into an exact answer, but I'm wondering if there is anymore to the question? Maybe some sort of trig formula is involved?? (we haven't done trig yet... we're heaps behind).

Since you're going forward, and then backwards, you don't need to be careful.
F[F^-1(x)] = all values of x

You only have to be careful if you go the other way around:
F^-1[F(x)] = x, only if x is within the restriction.

 

[lyounamu]

Wait, I just realised 0.8=4/5 *slaps forehead in idiocy*

So 4/5 is the answer then?

So then sin(sin-1(a)) = a ?

How would that extend to:

cos(sin-1(4/5))

tan(cos-1(12-13))

?

Yep.

Well, it depends. (referring to the next question)

As I said, a must be between certain limits. Look my older post.

cos(sin-1(4/5)) = cos(cos-1(3/5) = 3/5
tan(cos-1(12/13)) = tan(tan-1(5/12) = 5/12

 

[CubanitaChilena]

Okay, that makes sense...

Only, how did you convert sin-1(4/5) to cos-1(3/5)? (the same with cos to tan)

Edit* Nevermind... figured it out (duh *slaps self again*)

Thanks for the help =D

 

[midifile]

Okay, that makes sense...

Only, how did you convert sin-1(4/5) to cos-1(3/5)? (the same with cos to tan)

let x = sin-1(4/5)
therefore sinx = 4/5
then draw a triangle of this and the other side will be 3 (using pythagoris)
using the cos ratio cosx=3/5
therefore x= cos-1(3/5)
so sin-1(4/5) = cos-1(3/5)

The same can be done with cos and tan

 

[conics2008]

Hey buddy regarding your question. you can use a right angel triangle to prove

things involving cos sin and tan.. its helpful.

and see the guy/girl above me thats the correct way on doing these questions. yeah you can also just bash them in your calculator and still get the right answer but the poster above me did the proper working out.