Famous Inverse Trig Questions (1 Viewer)

[velox]

Anyone know of any famous inverse trig questions that come up time and time again? Post them in here

 

[Idyll]

If you have the Cambridge 3u textbooks have a look in those. They'll have about all the inverse trig questions you could imagine.

 

[velox]

Yeh i just got it today. Ill have the fun task of doing them tomorrow.

 

[velox]

How would you do this?

 

[FinalFantasy]

sec( sin^(-1) (-1\3) )
let sin ^(-1) (-1\3) =@
sin @=1\3
cos @=root 8\3
sec@=3\root8
sec( sin^(-1) (-1\3) )=sec @=3\root 8

forgot put da 3 there

 

[velox]

yeh thats what i got, fitz got something equivalent but in a different form (3root2/4), just wondering how he did it. thanks anyway

 

[shafqat]

find the six solutions of the equation sin (2cos ^ -1 ( cot (2tan ^ -1 x) ) ) = 0.

 

[FinalFantasy]

yeh thats what i got, fitz got something equivalent but in a different form (3root2/4), just wondering how he did it. thanks anyway

3\root8=3\2root2 * root 2\root2=3root2\4
i just didn't bother simplifying

 

[FinalFantasy]

find the six solutions of the equation sin (2cos ^ -1 ( cot (2tan ^ -1 x) ) ) = 0.

sin (2cos ^ -1 ( cot (2tan ^ -1 x) ) ) = 0
let A=2tan^-1 x
tan^-1 x=A\2
tan (A\2)=x
cot A=1\tanA=(1-tan² (A\2))\2tan(A\2)=(1-x²)\2x
now sin(2cos^-1 (1-x²)\2x )=0
let 2cos^-1 (1-x²)\2x)=B
cos^-1 (1-x²)\2x=B\2
cos (B\2)=(1-x²)\2x

now sin B=0
2sin (B\2)cos (B\2)=0
[2sqrt (4x²-(1-x²)²)]\2x * (1-x²)\2x=0
(1-x²)sqrt(4x²-(1-x²)²) \ 2x²=0
(1-x²)sqrt(4x²-(1-2x²+x^4))\2x²=0
(1-x²)sqrt(6x²-x^4-1)\2x²=0
(1-x²)(6x²-x^4-1)=0
6x²-x^4-1-6x^4+x^6+x²=0
x^6-7x^4+7x²-1=0
den solve for x..

is dat right shafqat?

 

[velox]

ah crap i should sleep, didnt realise he just rationalised.

 

[velox]

What about this one?

 

[velox]

nup........

 

[k3tan]

2 / rt 9 - 4x<sup>2

 

[velox]

yep. How did u get the 2 in there?

is it just there cos its the differential of whats in the brackets?

 

[FinalFantasy]

d\dx (sin^-1 (2x\3))=2\3* 1\sqrt(1-(2x\3)²)
do some stuff den u get there

 

[velox]

okies thanks

 

[velox]

I have another one, im not too sure how they got the answer in cambridge
Find inverse function of y = (x-2)/(x+2) and show that f^-1(f(x)) = x and f(f^-1(x)) = x. I dont get how to do the last part. ( f^-1(f(x)) = x and f(f^-1(x)) = x)

 

[acmilan]

f(x) = (x - 2)/(x + 2) and f^-1(x) = -2(x+1)/(x-1)

to find f^-1(f(x)), for every x in f^-1(x), you sub in f(x). ie.

f^-1(f(x))
= -2(f(x) + 1)/(f(x) - 1)
= -2[(x - 2)/(x + 2) + 1]/[(x - 2)/(x + 2) - 1]
= -2[(x - 2 + x + 2)/(x+2)]/[(x - 2 - x - 2)/(x + 2)]
= -2[2x/(x+2)]/[-4/(x+2)]
= -2[2x/(x+2)]*[(x+2)/-4]
= 2x/2
= x

Same thing for f(f^-1(x)), sub in f^-1(x) wherever x is in f(x)

 

[velox]

= -2(f(x) + 1)/(f(x) - 1)

Thanks acmilan, i needed that line ^ .

 

[velox]

x² + 2x + 4
x>=-1
How do u find the inverse function of that?