- #26

Simon Bridge

Science Advisor

Homework Helper

- 17,874

- 1,655

Yep - that identity is true for all x, so if you subtract the relation you are investigating you get ... arcsin(x)+arcsin(1-x)=0 ... but I'm working on something involving the 1-2-root-3 triangle.

[edit]

BUT ... plug it in: I shouldn't do this at 3am :(

arccos(0.5)=pi/3

arcsin(0.5)=pi/6

pi/3 + pi/6 = pi/2 ... but that is not the relation: pi/3-pi/6=pi/6

then x and 1-x correspond to the sides of a right-angle triangle.

set θ=arccos(x) and 90-θ = arcsin(1-x) are the complimentary angles right.

so x=cosθ and 1-x=sin(90-θ)=cosθ

so 1-x = x => x=1/2

The reason this got convoluted is because of the policy of helping the OP with what they are doing rather than doing the problem for them.

That was the first thing I thought of.

[edit] don't get excited - I said

It helps understand the work so far ... see next:

It has to be - see bottom of post #24 ;) your post crossed by edit.UPDATE: I think x could be equal to 1/2. What do you think?

[edit]

BUT ... plug it in: I shouldn't do this at 3am :(

arccos(0.5)=pi/3

arcsin(0.5)=pi/6

pi/3 + pi/6 = pi/2 ... but that is not the relation: pi/3-pi/6=pi/6

**if**arcos(x) and arcsin(1-x) are complimentary angles.then x and 1-x correspond to the sides of a right-angle triangle.

set θ=arccos(x) and 90-θ = arcsin(1-x) are the complimentary angles right.

so x=cosθ and 1-x=sin(90-θ)=cosθ

so 1-x = x => x=1/2

The reason this got convoluted is because of the policy of helping the OP with what they are doing rather than doing the problem for them.

That was the first thing I thought of.

[edit] don't get excited - I said

**if**... to use the above for your problem, you need a slight modification.It helps understand the work so far ... see next:

Last edited: