Trigonometric identity for inverse tan

[Kat007]

Hello,

Could you please clarify if this is correct:

If tan^(-1)(x) = Pi/2 - tan^(-1)(1/x)

Then if we have (ax) as the angle where a is a constant, do we get:
tan^(-1)(ax) = Pi/2 - tan^(-1)(a/x)
or does the constant go on the bottom with the x? i.e. or:
tan^(-1)(ax) = Pi/2 - tan^(-1)(1/(ax))

Thank you very much,
Kat

 

[berkeman]

Hello,

Could you please clarify if this is correct:

If tan^(-1)(x) = Pi/2 - tan^(-1)(1/x)

Then if we have (ax) as the angle where a is a constant, do we get:
tan^(-1)(ax) = Pi/2 - tan^(-1)(a/x)
or does the constant go on the bottom with the x? i.e. or:
tan^(-1)(ax) = Pi/2 - tan^(-1)(1/(ax))

Thank you very much,
Kat

I believe that you can answer this yourself by drawing a right triangle (use a 30-60-30 triangle), and writing the tangent in terms of the ratio of the opposite and adjacent sides. Then double the height of the opposite side...

 

[Gigasoft]

Do you think the identity should change if you replace the letter x with the letter y? Is the identity invalid for some value of y? What if y = ax? If you agree that it is still valid, what do you get when you put ax in the place of y?

 

[Kat007]

OK so it seems to work out as the
tan^(-1)(ax) = Pi/2 - tan^(-1)(1/(ax))
Could you please confirm that this is correct?

 

[berkeman]

OK so it seems to work out as the
tan^(-1)(ax) = Pi/2 - tan^(-1)(1/(ax))
Could you please confirm that this is correct?

Which method did you use to conclude that? Giga and I have a bet going...

 

[Kat007]

lol how kind of you.. Jokers heheee
The triangle one.
They both makes sence. Thanks,
Kat

 

[berkeman]

lol how kind of you.. Jokers heheee
The triangle one.
They both makes sence. Thanks,
Kat

Oh fooey. If they both make sense, then neither of us wins the bet. Oh well

 

[Mentallic]

Nah, Giga's make more sense AND no pen and paper required :tongue: