Inverse Function Homework: Simplifying Sin^-1(2Sin^-1(0.8))

In summary, we used the formula for double angle sin(2α) and the Pythagorean identity to simplify the expression ##\sin^{-1}(2\sin^{-1}0.8)## to ##2\sin^{-1}(0.8)\cos^{-1}(0.8)##. We also used the Pythagorean identity to find the value of ##\cos(\alpha)##, which was necessary for the simplification.
  • #1
Karol
1,380
22

Homework Statement


Simplify:
$$\sin^{-1}(2\sin^{-1}0.8)$$

Homework Equations


Inverse sine: ##y=\sin^{-1}(x)~\rightarrow~\sin(y)=x##
$$\sin^2(x)+\cos^2(x)=1$$

The Attempt at a Solution


The inner parenthesis: ##\sin y=0.8## . In the drawing it's alpha's sine.
Snap1.jpg
Now i double the α and the question wants the high edge in the drawing. how to find it?
 
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  • #2
Karol said:

Homework Statement


Simplify:
$$\sin^{-1}(2\sin^{-1}0.8)$$

Homework Equations


Inverse sine: ##y=\sin^{-1}(x)~\rightarrow~\sin(y)=x##
$$\sin^2(x)+\cos^2(x)=1$$

The Attempt at a Solution


The inner parenthesis: ##\sin y=0.8## . In the drawing it's alpha's sine.
View attachment 109084 Now i double the α and the question wants the high edge in the drawing. how to find it?
That problem seems very strange to me.

It would be much more expected to be asked to simplify something like:

## \sin\left(2 \sin^{-1} (0.8)\right) ##
 
  • #3
Karol said:

Homework Statement


Simplify:
$$\sin^{-1}(2\sin^{-1}0.8)$$

Homework Equations


Inverse sine: ##y=\sin^{-1}(x)~\rightarrow~\sin(y)=x##
$$\sin^2(x)+\cos^2(x)=1$$

The Attempt at a Solution


The inner parenthesis: ##\sin y=0.8## . In the drawing it's alpha's sine.
View attachment 109084 Now i double the α and the question wants the high edge in the drawing. how to find it?

Are you sure that's the correct question? It seems undefined to me.
 
  • #4
SammyS said:
That problem seems very strange to me.

It would be much more expected to be asked to simplify something like:

## \sin\left(2 \sin^{-1} (0.8)\right) ##
Looking at the diagram, that is how Karol interpreted it.
@Karol, what formulae do you know for sin(2α) or sin(α+β)?
 
  • #5
$$\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)$$
$$\sin^2(\alpha)+\cos^2(\alpha)=1~\rightarrow~\cos(\alpha)=0.6$$
$$\sin(2\alpha)=2\cdot 0.8 \cdot 0.6$$
 
  • #6
Karol said:
$$\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)$$
$$\sin^2(\alpha)+\cos^2(\alpha)=1~\rightarrow~\cos(\alpha)=0.6$$
$$\sin(2\alpha)=2\cdot 0.8 \cdot 0.6$$
That looks fine, if you're trying to find ##\ \sin\left(2 \sin^{-1} (0.8)\right) \, .##
 
  • #7
Karol said:
$$\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)$$
$$\sin^2(\alpha)+\cos^2(\alpha)=1~\rightarrow~\cos(\alpha)=0.6$$
$$\sin(2\alpha)=2\cdot 0.8 \cdot 0.6$$

Also, if you want to type an implication '##\Rightarrow##', write 'Rightarrow' in Latex instead of 'rightarrow'.
 
  • #8
Thanks everybody, you are great!
 

What does the given inverse function represent?

The given inverse function represents the angle whose sine is equal to 0.8, multiplied by 2, and then the inverse sine of that value.

What is the purpose of simplifying this inverse function?

Simplifying the inverse function allows for a clearer understanding of the original function and its inverse relationship. It can also make it easier to solve for a specific value or perform other mathematical operations.

How do you simplify Sin^-1(2Sin^-1(0.8))?

To simplify this inverse function, first use the double angle identity for sine: Sin(2x) = 2Sin(x)Cos(x). Then, substitute x with Sin^-1(0.8), giving Sin^-1(2(0.8)Cos(Sin^-1(0.8))). Finally, use the inverse sine identity: Sin(Sin^-1(x)) = x, to simplify to 2Sin^-1(0.8).

What is the significance of the value 0.8 in this inverse function?

The value 0.8 represents the sine of the angle being evaluated in the function. It is the input value that results in a sine output of 0.8.

How can this inverse function be applied in real-world situations?

Inverse functions, such as this one, can be used to solve for unknown angles or sides in various problems involving trigonometry or geometry. They are also used in fields such as engineering and physics to model and analyze real-world phenomena.

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