Proving inverse trig functions

[Vikash]

How do I prove this : where 0 ≤ x ≤ 1 arccosx = arcsinx √1-x².

is it that we can conclude since they from similar triangles in the same quadrants so they are equal? Is it one way we can prove this result? Thanks help is really appreciated.

 

[Steven G]

You say so they are equal. Do you mean that arccosx= arcsin sqrt( 1- x^2)? Yes, of course they are equal since you are told that. If you mean that something else are equal then please state what you mean.

Draw a right triangle that lives up to arccosx = [math]\theta_1[/math], label the triangle and see what you can do from there.

 

[Deleted member 4993]

How do I prove this : where 0 ≤ x ≤ 1 arccosx = arcsinx √1-x².

is it that we can conclude since they from similar triangles in the same quadrants so they are equal? Is it one way we can prove this result? Thanks help is really appreciated.

Did you mean

Prove:

arccos(x) = arcsin(√(1-x²)) ......................for 0 ≤ x ≤ 1

In your OP you have an extra 'x' stuck in there. And you need those parentheses to indicate correct grouping.

Please share your work.

 

[Dr.Peterson]

How do I prove this : where 0 ≤ x ≤ 1 arccosx = arcsinx √1-x².

is it that we can conclude since they from similar triangles in the same quadrants so they are equal? Is it one way we can prove this result? Thanks help is really appreciated.

It would help if you would show what similar triangles you are talking about; we might be able to just say "yes", but as it is, we can't yet.

You presumably want to prove that [MATH]\arccos(x) = \arcsin\left(\sqrt{1- x^2}\right)[/MATH]. I would start by letting [MATH]\theta = \arccos(x)[/MATH], which is in the first quadrant, so that [MATH]\cos(\theta) = x[/MATH]. Then you need to show that [MATH]\sin(\theta) = \sqrt{1- x^2}[/MATH].

 

[Romsek]

Consider a right triangle with hypotenuse length [MATH]1[/MATH] and adjacent side length [MATH]x[/MATH].
The opposite side is given by [MATH]\sqrt{1-x^2}[/MATH]
See if that causes an aha moment.

 

[Vikash]

Consider a right triangle with hypotenuse length [MATH]1[/MATH] and adjacent side length [MATH]x[/MATH].
The opposite side is given by [MATH]\sqrt{1-x^2}[/MATH]
See if that causes an aha moment.

Yes.....same TRIANGLES IN SAME QUADRANT SO EQUAL!!!!??

 

[Vikash]

Thanks everyone?