Exact Value of Trigonometric Equations

[Sadriam]

Homework Statement

Find the exact value of the given expressions.

sin (2 sin[SUP]-1[/SUP] (4/5))
or 
sin (2 arcsin (4/5))

Just two different ways of writing it

Homework Equations

maybe

The Attempt at a Solution

I've only gotten this far:

sin (2 arcsin (4/5)) = 2 sin (arcsin (4/5))

I think i might need to use: if -1 <= x <= 1, then sin(arcsin (x)) = x and cos(arccos (x)) = x. But I am not entirely sure. Any help explaining this to me?

 

[micromass]

You seem to think that \sin(2\alpha)=2\sin(\alpha). This is not true. It is one of those mistakes that newbies like to make, but you need to remember that it is not true.

That said, do you know the actual formula for \sin(2\alpha)?

 

[jedishrfu]

You need to use the identity: sin(2x) = 2 (sin x) (cos x) where x = arcsin(4/5)

 

[Sadriam]

Okay thank you. Yes i am familiar with sin 2x = 2 sin(x)cos(x). So would this be correct:

sin (2 arcsin (4/5)) = 2 sin (arcsin (4/5)) * cos (arcsin (4/5))

 

[jedishrfu]

Okay thank you. Yes i am familiar with sin 2x = 2 sin(x)cos(x). So would this be correct:

sin (2 arcsin (4/5)) = 2 sin (arcsin (4/5)) * cos (arcsin (4/5))

Yes, now you can continue to simply the factors.

 

[Sadriam]

Would my next step be this:

[U]sin (2 arcsin (4/5))[/U]    = cos (arcsin (4/5))
2 sin (arcsin (4/5))

1 = cos (arcsin (4/5)) ?

 

[Michael Redei]

Would my next step be this:

[U]sin (2 arcsin (4/5))[/U] = cos (arcsin (4/5))
2 sin (arcsin (4/5))

If you want the value of sin(2 arcsin(4/5)), you'd be better off calculating 2 sin(arcsin(4/5)) and cos(arcsin(4/5)) and multiplying those results. The first should be easy, and for the second I'd try using some equation that connects sin(x) and cos(x) and use that to get an expression for cos(x)=...

 

[Sadriam]

Could the way i did be used? If so, would it become:

1 = cos (arcsin (4/5))

And I'm a little confused about arcsin. Does arcsin mean "some side" over "some side" just like cos means adj / hyp ?

 

[Michael Redei]

Could the way i did be used?

You shouldn't lose sight of what you're aiming for, i.e. the value of sin(2 arcsin(4/5)), so putting that in a fraction will lead you away from your goal. Ans you still seem to think that 2 sin(...) is the same as sin(2 ...) which it isn't.

And I'm a little confused about arcsin. Does arcsin mean "some side" over "some side" just like cos means adj / hyp ?

No. arcsin is the inverse of sin, so when y=sin(x), you have arcsin(y)=x. It's like y=x² meaning the same as √y=x, and like that, it's true only if you keep to certain values for x and y. (IN your case, you only look at one value, 4/5, for the arcsin function, so you shouldn't have any problems.)

 

[Mark44]

No. arcsin is the inverse of sin, so when y=sin(x), you have arcsin(y)=x. It's like y=x² meaning the same as √y=x

No, not quite. √(x²) is not the same as x.

, and like that, it's true only if you keep to certain values for x and y. (IN your case, you only look at one value, 4/5, for the arcsin function, so you shouldn't have any problems.)

 

[LCKurtz]

Okay thank you. Yes i am familiar with sin 2x = 2 sin(x)cos(x). So would this be correct:

sin (2 arcsin (4/5)) = 2 sin (arcsin (4/5)) * cos (arcsin (4/5))

Remember that arcsin(x) is the [principle value] of the angle whose sine is x. So when you want to calculate sin (arcsin (4/5)) you are asking for the sine of the angle whose sine is x. Much like asking for the color of a horse whose color is brown.

 

[pierce15]

So you're this far: sin(2u)=2sin(u)cos(u) and here we are working with u=arcsin of 4/5. So far we have:
2sin(arcsin(4/5))cos(arcsin(4/5))=8cos(arcsin(4/5))/5

So now the problem is determining cos(arcsin(4/5)). The first thing that you need to do is think about what arcsin(4/5) really is. This value is the measure of the angle formed in a right triangle whose opposite side is 4 and the hypotenuse is 5, because sin(x)=\frac{opposite}{hypotenuse}

Think about this triangle. We know that it is a right triangle, and we know two of the sides. Thus, you can calculate the third side using the pythagorean theorem. After that, you can use the definition of cosine. cos(arcsin(4/5))=\frac{adjacent}{hypotenuse}
We know that the hypotenuse is 5 and the adjacent side is the one that we have not yet found. Can you finish the problem from here?