Inverse trig functions (1 Viewer)

[Timothy.Siu]

yeah ok, i'll be careful

 

[Trebla]

omg time warp due to clock changes!!! LOL

 

[Timothy.Siu]

omg time warp due to clock changes!!! LOL

luls

 

[cutemouse]

Hello,

I have a question with this. I'd appreciate it if you could show steps.

Evaluate:

1) cos(2sin^-1(sqrt(3))/2)
2) cos(2cos^-1(sqrt(5))/13)

Thanks

 

[azureus88]

first one should be fairly obvious as sin<SUP>-1</SUP>((sqrt3)/2) =pi/3 so cos(2pi/3)= -0.5

2) let @ = cos<SUP>-1</SUP>((sqrt5)/13)
cos@ = (sqrt5)/13)

cos[2cos<SUP>-1</SUP>(sqrt5)/13)]
= cos 2@
= 2cos^2@ - 1
= 2(5/169) -1
= -159/169

 

[Pupzrollin]

Hello,

I have a question with this. I'd appreciate it if you could show steps.

Evaluate:

1) cos(2sin^-1(sqrt(3))/2)

Well for the first one:
let a=sin-1(sqrt(3))/2)
Now draw a right-angled triangle showing this information as seen in the attachment


cos2a=2cos^2(a)-1
=2(0.5)^2 -1
=-0.5

 

[cutemouse]

Thanks for that. I have a few more questions. Would appreciate if you could assist me.

Show that:
1) tan^-14 - tan^-1(3/5) = pi/4
2) tan^-1(5/12) + cos^-1(5/13) = pi/2

Evaluate the without aid of tables:
3) cos[sin^-1(5/13)+sin^-1(4/5)]
4) sin[2tan^-1(4/3)]

Thanks again!

 

[Trebla]

Thanks for that. I have a few more questions. Would appreciate if you could assist me.

Show that:
1) tan^-14 - tan^-1(3/5) = pi/2
2) tan^-1(5/12) + cos^-1(5/13) = pi/2

Evaluate the without aid of tables:
3) cos[sin^-1(5/13)+sin^-1(4/5)]
4) sin[2tan^-1(4/3)]

Thanks again!

1)
Let x = tan^-14
=> tan x = 4
Note that:
tan (x - π/4) = (tan x - tan π/4) / (1 + tan x.tan π/4)
= (4 - 1) / (1 + 4(1))
= 3/5
x - π/4 = tan^-1(3/5)
x - tan^-1(3/5) = π/4
.: tan^-14 - tan^-1(3/5) = π/4

2)
Let x = cos^-1(5/13)
=> cos x = 5/13
=> tan x = 12/5 (draw a right-angled triangle to see this)
So:
cot x = 5/12
tan (π/2 - x) = 5/12
π/2 - x = tan^-1(5/12)
tan^-1(5/12) + x = π/2
.: tan^-1(5/12) + cos^-1(5/13) = π/2

3)
Let x = sin^-1(5/13) and y = sin^-1(4/5)
=> sin x = 5/13 and sin y = 4/5
From drawing a right-angled triangle:
cos x = 12/13 and cos y = 3/5
Note that:
cos(x + y) = cos x.cos y - sin x.sin y
= (12/13)(3/5) - (5/13)(4/5)
= 16/65
.: cos[sin^-1(5/13) + sin^-1(4/5)] = 16/65

4)
Let x = 2tan^-1(4/3)
=> tan (x/2) = 4/3
Recall the t-formula for sin x
sin x = 2t / (1 + t²)
= 2tan (x/2) / (1 + tan²(x/2))
= 2(4/3) / (1 + 16/9)
= 24/25
.: sin [2tan^-1(4/3)] = 24/25

 

[cutemouse]

I think I know how to do 3 and 4, so it'd be great if you could help me with just 1 and 2.

Thanks

 

[azureus88]

i'll just do question 2 since its a bit different.

2) let @ = tan<SUP>-1</SUP>(5/12)
tan@ = 5/12

let b = cos<SUP>-1</SUP>(5/13)
cos b = 5/13
tan b = 12/5 using the triangle method suggested by Pupzrollin

(note that tan<SUP>-1</SUP> (12/5) = cos<SUP>-1</SUP>(5/13) =b)

tan (@+b) = [(5/12)+(12/5)]/[1-(5/12)(12/5)]

(note that the denominator is 0 and so tan (@+b) = infinite)

hence @+b = pi/2
so tan<SUP>-1</SUP>(5/12) + cos<SUP>-1</SUP>(5/13) = pi/2

3) let @ = sin<SUP>-1</SUP>(5/13) and b = sin<SUP>-1</SUP>(4/5), then use compound angles

you should have a go at the rest, they follow similar procedures.

 

[cutemouse]

Thanks Trebla, I actually updated by post while you were posting yours, so could you please do Q1 for me please?

Thanks alot

 

[Trebla]

Edited

 

[cutemouse]

Thank you Trebla,

Just one quick question though, in Q1 where did you get (x - π/4) from?

 

[Trebla]

Thank you Trebla,

Just one quick question though, in Q1 where did you get (x - π/4) from?

I used a pre-existing result of compound angles out of the blue.
tan (x - π/4) = (tan x - tan π/4) / (1 + tan x.tan π/4)

 

[cutemouse]

I see... Is there a list of these results that would be good to remember? And what relevance does that result have to the question? Please excuse my ignorance

EDIT: I see the relevance, but please let me know if there are a list of results to remember. Thanks

 

[Trebla]

There really isn't a result to remember as such. It's just something you can come up with which helps you forecast the method to the answer. It's something that comes with practice.

So in the specific question there's a π/4, so that has to come from somewhere and it's a nuisance to use two variables so you need the other inverse tan expresion in terms of x (the original variable you allowed) somehow. It turns out the compound angle formula conveniently provides that.

 

[Timothy.Siu]

lol, for those ones are u allowed to just tan both sides? hmm coz thats wat i did when i did them but i'm sure if ur allowed to.

 

[Trebla]

lol, for those ones are u allowed to just tan both sides? hmm coz thats wat i did when i did them but i'm sure if ur allowed to.

Doesn't that mean you've assumed the result you are trying to prove?

 

[Timothy.Siu]

probably,

like for this one

the Q is 1) tan^-14 - tan^-1(3/5) = pi/4

i say, tan[tan^-14 - tan^-1(3/5)]=[working out]=tan pi/4
and , then i say therefore tan^-14 - tan^-1(3/5) = pi/4

 

[Trebla]

probably,

like for this one

the Q is 1) tan^-14 - tan^-1(3/5) = pi/4

i say, tan[tan^-14 - tan^-1(3/5)]=[working out]=tan pi/4
and , then i say therefore tan^-14 - tan^-1(3/5) = pi/4

In that case, as long as you go from LHS --> RHS without assuming the result, it should be fine.