Find the exact value of this inverse function

[1irishman]

Homework Statement

Find the exact value of cos(2arcsin(-1/8))

Homework Equations

make use of the double angle formula

The Attempt at a Solution

let arcsin(-1/8)=theta
then sin theta= -1/8

a=sqrt63=3sqrt7
and that is as far as i could go...please help? Thank you.

 

[Mark44]

Homework Statement

Find the exact value of cos(2arcsin(-1/8))

Homework Equations

make use of the double angle formula

The Attempt at a Solution

let arcsin(-1/8)=theta
then sin theta= -1/8

So now evaluate cos(2theta).

a=sqrt63=3sqrt7
and that is as far as i could go...please help? Thank you.

Where does a come in?

 

[1irishman]

'a' was a side i named

 

[Mark44]

OK. Still, you need to evaluate cos(2theta). One identity is cos(2theta) = cos^2(theta) - sin^2(theta). This identity appears in two other forms, one of which will be useful for this problem.

 

[1irishman]

i'm still lost...

 

[LCKurtz]

i'm still lost...

Draw a picture of the angle θ and label the sides of a little triangle appropriately to have a sine of -1/8. You can read all six trig functions off that triangle and should be able to calculate cos(2θ) from there.

 

[HallsofIvy]

Since you were told "make use of the double angle formula", $cos(2\theta)= cos^2(\theta)- sin^2(\theta)= 1- 2sin^2(\theta)$.

You know that $sin(\theta)= \frac{1}{8}$.

 

[1irishman]

what does cos^2 equal?

 

[LCKurtz]

what does cos^2 equal?

cos²(θ) is just an abbreviation for (cos(θ))².

 

[Mentallic]

what does cos^2 equal?

Once you draw your right-triangle and label an angle $\theta$ then since sin$\theta$ = -1/8, you know what your opposite and hypotenuse sides are, then you can use pythagoras' theorem to find the 3rd side. This will give you cos$\theta$.

 

[HallsofIvy]

cos(2x)$= cos^2(x)- sin^2(x)$$= (1- sin^2(x))- sin^2(x)= 1- 2sin^2(x)$

If x= arcsin(-1/8) what is sin(x)?

I just notice that I had said this same thing several days ago!

 

[1irishman]

sin x is -1/8

 

[Mentallic]

Right, so what is cos(2x) given that cos(2x)=cos^2(x)-sin^2(x) ?

 

[HallsofIvy]

Right, so what is cos(2x) given that cos(2x)=cos^2(x)-sin^2(x) ?

Which, as I pointed out in the post that Irishman's post was in response to (and, I just noticed, several days ago!) , is equal to $1- 2sin^2(x)$.

Irishman, if sin(x)= -1/8, what is $sin^2(x)$? What is $1- 2sin^2(x)$?

 

[Mentallic]

Which, as I pointed out in the post that Irishman's post was in response to (and, I just noticed, several days ago!)

Yes, I know, this identity has been mentioned at least 3 times in this thread now. It just seems like 1irishman is only considering and responding to the most recent reply in this thread each time he logs back on, so I felt the need to hold his hand as we guide him through the process...

If x= arcsin(-1/8) what is sin(x)?

sin x is -1/8