Question

cos(inv) {(a+bcosx)/(b+acosx)}
Differentiate...

Answer 1

cos^(-1)?

Answer 2

\[\cos^{-1} ((a+b \cos x)/(b+a \cos x))\]

Answer 3

Quotient rule

Answer 4

\[\huge -\frac{\frac{a+b \cos (x)}{\sqrt{1-x^2}}-b \sin (x) \left(b+\cos ^{-1}(x)\right)}{\left(b+\cos ^{-1}(x)\right)^2 \sqrt{1-\frac{(a+b \cos (x))^2}{\left(b+\cos ^{-1}(x)\right)^2}}}\]

\[\text{you can simplify further by multiplying \top and bottom with} \sqrt{1-x^2}\]

Answer 5

isnt there an easier method???

Answer 6

That's what I am pondering about

Answer 7

using .sam's method,would'nt it be too long??

Answer 8

\[x = \cos^{-1}\frac{t}{ab}\]\[\frac{a+b\cos x}{b + a\cos x} = \frac{a + b\cos \cos^{-1}\frac{t}{ab}}{b + a\cos^{-1}\cos \frac{t}{ab}}\]Can we do this? lol

Answer 9

wat's t ???

Answer 10

A Variable

Answer 11

can you make one thing clear ...Which is correct???
\[\cos^{-1} (\cos x) =x\]\
or
\[\cos (\cos^{-1} x) =x\]

Answer 12

Oh sorry\[\frac{a+b\cos x}{b + a\cos x} = \frac{a + b\cos \cos^{-1}\frac{t}{ab}}{b + a\cos\cos^{-1} \frac{t}{ab}} = \frac{a + \frac t a}{b + \frac t b}\]Maybe

Answer 13

It could have been easier if one of the cos was sin.

Answer 14

please answer my question which is true???
cos−1(cosx)=x
or
cos(cos−1x)=x

Answer 15

cos−1(cosx)=x

Answer 16

Both are true @Sarkar

Answer 17

i only knew the first one was...how come the second one is true???

Answer 18

http://www.wolframalpha.com/input/?i=cos%28cos%5E%28-1%29x%29%3D

Answer 19

You wouldn't know a lot, that doesn't mean they are all false.

Answer 20

in my text book,the first is only given..

Answer 21

Use your head to get the second one.

Answer 22

\[\cos^{-1} x=a \]
\[\cos a=x\]
\[\cos(\cos^{-1} x)=x\]
Get it?

Answer 23

yeah i figured it out..

Answer 24

I just substituted a = \[\cos^{-1} x\]

Answer 25

If you expand out the numerator, some terms cancel, and then we can factor itl:
-{bsin(x)[acos(x) + b] - asin(x)[a + bcos(x)]}
= -bsin(x)[acos(x) + b] + asin(x)[a + bcos(x)]
= -absin(x)cos(x) - b^2*sin(x) + a^2*sin(x) + absin(x)cos(x)
= -b^2*sin(x) + a^2*sin(x)
= (a^2 - b^2)sin(x)

In the radicand, make a common denominator, expand it, and simplify it:
1 - [acos(x) + b]^2/[a + bcos(x)]^2
= [a + bcos(x)]^2/[a + bcos(x)]^2 - [acos(x) + b]^2/[a + bcos(x)]^2
= {[a + bcos(x)]^2 - [acos(x) + b]^2}/[a + bcos(x)]^2
= {[a^2 + 2abcos(x) + b^2*cos^2(x)] - [a^2*cos^2(x) + 2abcos(x) + b^2]}/[a + bcos(x)]^2
= [a^2 + 2abcos(x) + b^2*cos^2(x) - a^2*cos^2(x) - 2abcos(x) - b^2]/[a + bcos(x)]^2
= [a^2 + b^2*cos^2(x) - a^2*cos^2(x) - b^2]/[a + bcos(x)]^2
= [{a^2 - b^2} - (a^2 - b^2)cos^2(x)]/[a + bcos(x)]^2
= (a^2 - b^2)[1 - cos^2(x)]/[a + bcos(x)]^2
= (a^2 - b^2)[sin^2(x)]/[a + bcos(x)]^2

So in the denominator, we would have:
[a + bcos(x)]^2*√{(a^2 - b^2)[sin^2(x)]/[a + bcos(x)]^2}
= [a + bcos(x)]^2*sin(x)√(a^2 - b^2)/[a + bcos(x)]
= [a + bcos(x)]*sin(x)√(a^2 - b^2)

Taking the quotient of these two, we get:
(a^2 - b^2)sin(x)/{[a + bcos(x)]*sin(x)√(a^2 - b^2)}
= (a^2 - b^2)/{[a + bcos(x)]*√(a^2 - b^2)}
= √(a^2 - b^2)/[a + bcos(x)]

I hope this helps!