Inverse Trig Functions (1 Viewer)

[Alexluby]

How do you work out questions like:

Inv. tan 1/2 + inv. tan 1/3

i.e. tan^-1 (1/2) + tan^-1 (1/3)

Leaving answer in radian form...

Thx

 

[physician]

How do you work out questions like:

Inv. tan 1/2 + inv. tan 1/3

i.e. tan^-1 (1/2) + tan^-1 (1/3)

Leaving answer in radian form...

Thx

I remember asking my teacher the exact same question.... then i realised the question said calculators may be used... because u had to show that the approximate value = Pi/4...

i can tell u the answer is a decimal... but since u have to leave it in radian form... ummm.. hmm not sure...

maybe ur answer would be something like ~ pi/4 ....

if they were exact triginometric values it wouldn't be so difficult.. sorry dude.. i guess u'll have to wait for someone else to help out...

 

[Alexluby]

yea the answer is in pi/4 ... but is there a methodological way of doing it? Like step and step coz i don't think i will be able to use calculators for it in my coming exam...

 

[shafqat]

let a = tan^-1 (1/2) + tan^-1 (1/3)
then take tan of both sides, and use tan(a-b) formula on RHS
then you get tan a = some number
solve for a

 

[Slidey]

As shaf said:
tan(a+b)=(tana+tanb)/(1-tanatanb)
Let a=atan1/2, b=atan1/3
tan(atan1/2+atan1/3)=(1/2+1/3)/(1-1/6)
atan1/2+atan1/3=atan(5/6/5/6)
atan(1)=pi/4

 

[physician]

let a = tan^-1 (1/2) + tan^-1 (1/3)
then take tan of both sides, and use tan(a-b) formula on RHS
then you get tan a = some number
solve for a

oh yeh

I forgot u could do that...

 

[LaCe]

Easiest way: (i dunno if this is the same as is already up here but here goes):

let a= tan^-1 (1/2)
b= tan^-1 (1/3)

tan(a+b) = (tana + tanb) /(1-tana.tanb)
=1
therefore a+b = pi/4

tan^-1 (1/2) + tan^-1 (1/3) = pi/4
therefore

 

[velox]

For every question like this, just remember your triangles, and taking the trig function of both sides. Pretty easy stuff.

 

[b0b444]

this is not a question you would be asked for an exact value because they can't be read off the exact value triangles or graphs. you would be more likely to get inverse Tan of : 1, -1, sqrt3, -sqrt3, 1/sqrt3 or -1/sqrt3.

 

[acmilan]

this is not a question you would be asked for an exact value because they can't be read off the exact value triangles or graphs. you would be more likely to get inverse Tan of : 1, -1, sqrt3, -sqrt3, 1/sqrt3 or -1/sqrt3.

You can definitely make triangles for it. For example what if i took sin(tan^-1 (1/2) + tan^-1 (1/3) ) instead of tan(tan^-1(1/2) + tan^-1(1/3) )

Construct two triangles.

First one relates to the angle tan^-1(1/2): Opposite side = 1, Adjacent side = 2, Hypotenuse = root 5 units

Second one relates to angle tan^-1(1/3): Opposite: 1, Adjacent = 3, Hypotenuse = root 10

Let
a = (tan^-1(1/2) + tan^-1(1/3))
sin a = sin(tan^-1(1/2) + tan^-1(1/3))
= sin(tan^-1(1/2))cos(tan^-1(1/3)) + cos(tan^-1(1/2)sin(tan^-1(1/3))
= (1/root5)*(3/root10) + (2/root5)*(1/root10)
= 3/root50 + 2/root50
= 5/root50
= 5/5root2
sin a = 1/root2
a = pi/4

Hence (tan^-1(1/2) + tan^-1(1/3)) = pi/4

Youd more likely get a question like this then a simple one such as tan^-1(1/root3) which can be easily done on a calculator

 

[FinalFantasy]

hey b0b444, that cat like figure shown on ur avatar looks quite evil!