inverse trig == (1 Viewer)

[pharmaceutical]

show that each of the following approximate to the value of pi/4

1. 2 tan-1 (1/3) + tan-1 (1/7)
2. 4 tan-1 (1/5) + tan-1 (1/229)

if you show me how to do the first one i might be able to do the second.

 

[zingerburger]

I'd probably just do it with a calculator. Just type the two inverse trig functions in each question separately, round them off to your choice of decimal places and add them together, which should equal about 45 degrees, or pi/4. That should be enough, I think.

 

[Shadowulf]

show that each of the following approximate to the value of pi/4

1. 2 tan-1 (1/3) + tan-1 (1/7)
2. 4 tan-1 (1/5) + tan-1 (1/229)

if you show me how to do the first one i might be able to do the second.

is the answer for question 1.... 1??
the way i did it was..

let theta (-) = tan-1(1/3)
therefore, tan(-) = 1/3

Let alpha (a) = tan-1 (1/7)
therefore tan(a) = 1/7

Now, 2(-) + (a)
tan on both sides, tan[2(-) + (a)]
expand it for equation [1]

[tan2(-) + tan(a)] / [1 - tan2(-)tan(a)]

you now work out tan2(-) to be:
[2tan(-)] / [1- tan^2(-)]
sub in what you had before, (i.e. tan(-) = 1/3)
to get [2/3] / [1- 1/9] = 6/8 = 3/4

now you can sub it into the equation [1], along with what you worked out before: (i.e. tan(a) = 1/7)
you now have:

[3/4 + 1/7] / [1 - (3/4)(1/7)] which works out to be [25/28] / [25/28]
which ends up to be 1

and then you can write your conclusion from there

NOTE: Im not sure if i worked it out the right way or not, i dont even know if i got it right oO" but i would also like to know if i did it correctly

 

[ellen.louise]

show that each of the following approximate to the value of pi/4

1. 2 tan-1 (1/3) + tan-1 (1/7)
2. 4 tan-1 (1/5) + tan-1 (1/229)

if you show me how to do the first one i might be able to do the second.

you have to use the results with tan^-1(tanx) etc. i'll do it tomorrow and post it if no one else has.

 

[pharmaceutical]

wow you nailed it shadowulf. No zingerburger == only general use calculators =.=.

though i would like to see how ellen does it or is it the same as shadow oO