Inverse Functions (1 Viewer)

[FDownes]

I'm having some trouble with a few inverse trigonometry questions. Hopefully someone here can explain the process to me so I can figure out future questions for myself. Currently I'm stuck on;

Find the exact value of sin[tan^-1(3/7)]

It can't simply be evaluated using a calculator since the answer you get is irrational. Please help!

 

[namburger]

Let A = [tan^-1(3/7)]
Tan A = 3/7

Draw triangle: such that the opposite angle is 3 and adjacent angle is 7. therefore hypotunese is sqrt(58)

sin[tan-1(3/7)] = Sin A

Sin A = opposite/hypotunese
= 3/sqrt(58)

 

[FDownes]

Ah, that clears that up. Thanks very much.

 

[FDownes]

Time for another question;

a) Write sin 2Θ in terms of Θ.

b) Hence find the exact value of sin [2cos^-1(3/5)].

The first part is easy; the answer is of course 2 sin Θ cosΘ. It's the second part that's giving me problems.

 

[foram]

Time for another question;

a) Write sin 2Θ in terms of Θ.

b) Hence find the exact value of sin [2cos^-1(3/5)].

The first part is easy; the answer is of course 2 sin Θ cosΘ. It's the second part that's giving me problems.

y= sin (2cos^-1 (3/5))
let cos^-1 (3/5) = @

y= sin(2@)
y= 2 sin@ cos@
y= 2 sin@ (3/5)

sin^2 @ = 1 - cos^2 @
= 1 - (3/5)^2
= 16/25

sin@ = 4/5

y= 2 sin@ (3/5)
y= 2 x 4/5 x 3/5
y= 24/25

EDIT: how did i end up with 1 - 9/25 = 1/225? I must have pressed the wrong buttons on my calculator. oops.

 

[Mark576]

(b) sin2Θ = 2sinΘ.cosΘ
Let cos^-1(3/5) = A => cos A = 3/5
- Same procedure as before, sketch a triangle where cos A = 3/5, then sin A = 4/5.
Then sin2A = 2sinA.cosA = 2.(4/5).(3/5) = 24/25

 

[FDownes]

Thanks, but as always, I've got another question;

Find the derivative of 1/(tan^-1x).

Should be pretty simple, but I'm not sure how to approach it.

 

[foram]

Thanks, but as always, I've got another question;

Find the derivative of 1/(tan^-1x).

Should be pretty simple, but I'm not sure how to approach it.

quotient rule if you're lazy. or else, take it to the power of -1 and use the chain rule.

1/ tan^-1 (x) = [tan^-1 (x)]^-1
differentiating

=-[tan^-1 (x)]^-2 .(1/1+x^2)

=-1/(1+x^2).[tan^-1 (x)]^2

 

[lyounamu]

Thanks, but as always, I've got another question;

Find the derivative of 1/(tan^-1x).

Should be pretty simple, but I'm not sure how to approach it.

I learnt this today!!!

y = 1 / tan^-1 (x)

dy/dx = (tan^-1 (x) . d/dx (1) - 1 . d/dx (tan^-1 (x)) / ((tan^-1 (x)^2)
= (-1 . 1/(1+x^2) . d/dx (x)) / ((tan^-1 (x)^2)
= -1 / (1 +x^2) ((tan^-1(x)^2)

 

[foram]

I learnt this today!!!

shouldn't you be doing binomial by now?

 

[lyounamu]

shouldn't you be doing binomial by now?

It's different depending on school. We have done everything except Binomial and Inverse Trig and some of Physical Application of Calculus.

 

[lionking1191]

Please click one of the Quick Reply icons in the posts above to activate Quick Reply.

 

[lionking1191]

oops..

3 left? wow, we are even slower than i thought..

 

[FDownes]

Thanks, now it's time for another;

Find the derivative of tan^-1(1/x).

Again, should be simple.

 

[namburger]

Thanks, now it's time for another;

Find the derivative of tan^-1(1/x).

Again, should be simple.

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

pluging into the formula

 

[FDownes]

Ah, now I see my mistake. Thanks.

 

[FDownes]

Okay, got another one;

Find the first derivative of y = cos^-1(sin x) in the domain -pi ≤ x ≤ pi.

The answer, according to the textbook, is -1 for -pi/2 < x < pi/2, 1 for -pi ≤ x < -pi/2 or pi/2 < x ≤ pi.

 

[lyounamu]

Okay, got another one;

Find the first derivative of y = cos^-1(sin x) in the domain -pi ≤ x ≤ pi.

The answer, according to the textbook, is -1 for -pi/2 < x < pi/2, 1 for -pi ≤ x < -pi/2 or pi/2 < x ≤ pi.

y = cos<SUP>-1</SUP>(sin x)

dy/dx = -1/(root of (1-sin^2 x)) . d/dx (sinx)
= -1/(root of cos^2 x) . -cos x
= cos x / absolute value of (cos x)
= 1 when cos x > 0 (and hence pi >_ x > pi/2 or -pi >_ x > -pi/2)
OR -1 when cos x < 0 (and hence -pi/2 <x < pi/2)

 

[namburger]

y = cos^-1(sin x)

dy/dx = 1/(root of (1-sin^2 x)) . d/dx (sinx)
<x>

there should be a minus
</x>

 

[lyounamu]

there should be a minus
</X>

Thanks. I just saw that. I will fix it.