Inverse trig functions (1 Viewer)

[sasquatch]

Hello, i can do this question but im am just curious to if there is another simpiler method.

Prove tan^-1(1/4) + tan^-1(3/5) = pi/4

This is my method,


LHS = tan^-1[tan[tan^-1(1/4) + tan^-1(3/5)]
= tan^-1[tan[{tan^-1(1/4)] + tan[tan^-1(3/5)}/{1 - tan[tan^-1(1/4)tan[tan^-1(3/5)}]
=tan^-1[ {(1/4) + (3/5)} / {1 - (1/4)(3/5)}]
= tan^-1[(17/20)/(17/20)]
= tan^-1(1)
= pi/4
= RHS


Thanks...

 

[Trev]

That method is pretty simple anyway.

 

[sasquatch]

Oh well ok...

 

[STx]

hmm, well kinda:

Let tan^-1(1/4)=α and tan^-1(3/4)=β
i.e. α+β=pi/4
take tan of both sides,
tan(α+β)=tanα+tanβ/1-tanα.tanβ = tan(pi/4)=1
LHS = [1/4+3/5]/[1-3/20]
= 1
=pi/4
=RHS
Does that make sense?

 

[Trev]

That's exactly the same method just written slightly different.
If you wrote that I doubt you would get full marks because you have said "=1; =pi/4; =RHS" which has been written incorrectly. You've also changed the RHS, most markers prefer only one side of the equation to be manipulated.

 

[STx]

Ahk, woops:

Let tan^-1(1/4)=α and tan^-1(3/4)=β
i.e. α+β=pi/4
take tan of both sides,
tan(α+β)=tanα+tanβ/1-tanα.tanβ = tan(pi/4)
=1
LHS = tanα+tanβ/1-tanα.tanβ
= [1/4+3/5]/[1-3/20]
= 1
therefore α+β = tan^-1(1/4) + tan^-1(3/4)
= tan^-1(1)
= pi/4
#

 

[Riviet]

Any of the above methods are just as valid. Neither seem to be faster or save more time, so it's really just a matter of personal choice.