Solve for X: ARCcos(x)-lnx+ARCsin(x)=pi/2

[wajed]

Homework Statement

ARCcos(x) - ln{xe^[ARCsin(x)]} =pi/2

The Attempt at a Solution

ARCcos(x) - ln{xe^[ARCsin(x)]} = pi/2


ARCcos(x) - lnx + lne^[ARCsin(x)] = pi/2

ARCcos(x) - lnx + ARCsin(x) = pi/2

I can`t go any further, can you help me please?

 

[sylas]

Homework Statement

ARCcos(x) - ln{xe^[ARCsin(x)]} =pi/2

The Attempt at a Solution

ARCcos(x) - ln{xe^[ARCsin(x)]} = pi/2ARCcos(x) - lnx + lne^[ARCsin(x)] = pi/2

ARCcos(x) - lnx + ARCsin(x) = pi/2

I can`t go any further, can you help me please?

Draw yourself a right angle triangle. Label the hypotenuse as "1" (length 1). Label one side (not the hypoenteuse) as "x". Now. What is arccos(x) on this diagram? What is arcsin(x)?

 

[Feezik]

I KNOW KNOW to solve this problem
You can know that arcsin+accos=const

 

[qntty]

you made a sign error in step two which will affect the geometric argument. arccos(x) + arcsin(x) = pi/2 but all we can say about arcsin(x)-arccos(x) is that 0<arcsin(x)-arccos(x)<pi/2

ARCcos(x) - ln{xe^[ARCsin(x)]} = pi/2ARCcos(x) - lnx + lne^[ARCsin(x)] = pi/2

ARCcos(x) - lnx + ARCsin(x) = pi/2

Should be

ARCcos(x) - ln{xe^[ARCsin(x)]} = pi/2ARCcos(x) - lnx - lne^[ARCsin(x)] = pi/2

ARCcos(x) - lnx - ARCsin(x) = pi/2

 

[wajed]

I`m sorry to say this, but ARCcos(x) should be ARCcos(-x)

Gonna read again and try to understand.. then ask

 

[qntty]

no worries, because arccos(-x) = pi - arccos(x). In fact, this solves your problem with the sign

 

[wajed]

Its just a question in a sample test.. and no inverse trig identities are mentioned in my book.. I`m not that smart to figure out that "arccos(x) + arcsin(x) = pi/2"
how could I know that "arccos(x) + arcsin(x) = pi/2"? (if its a straightforward issue, if not then I`ll just take it as it is)

 

[wajed]

"You can know that arcsin+accos=const"
how can I know?

 

[qntty]

Just keep in mind what arccos and arcsin mean. Its just an angle in a triangle and If you remember that cos(x)=sin(90-x) (or the sine of an angle in a triangle is the cosine of the other non-right angle which is easy to see on a unit circle) then you can arrive at the identity. It helps to do more problems with trig identities to get good at spotting things like that

 

[sylas]

"You can know that arcsin+accos=const"
how can I know?

Draw the triangle I suggested earlier. Draw a right angle triangle, and label the hypotenuse as length 1. Label one of the other sides as "x".

Now... where is ARCcos(x) on this diagram? Where is ARCsin(x)? What can you know about their sum?

 

[wajed]

Draw the triangle I suggested earlier. Draw a right angle triangle, and label the hypotenuse as length 1. Label one of the other sides as "x".

Now... where is ARCcos(x) on this diagram? Where is ARCsin(x)? What can you know about their sum?
===================================

oh, lol, I did so before.. but I couldn`t figure out.. now I got it, thank you.
right triangle is 90... if ARCcos(x) is theta1.. then ARCsin(x) is theta2 which is 180-90-theta1=90-theta1

 

[sylas]

Draw the triangle I suggested earlier. Draw a right angle triangle, and label the hypotenuse as length 1. Label one of the other sides as "x".

Now... where is ARCcos(x) on this diagram? Where is ARCsin(x)? What can you know about their sum?
===================================

oh, lol, I did so before.. but I couldn`t figure out.. now I got it, thank you.
right triangle is 90... if ARCcos(x) is theta1.. then ARCsin(x) is theta2 which is 180-90-theta1=90-theta1

Bingo! You got it.