anonymous
  • anonymous
cos(inv) {(a+bcosx)/(b+acosx)} Differentiate...
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
.Sam.
  • .Sam.
cos^(-1)?
anonymous
  • anonymous
\[\cos^{-1} ((a+b \cos x)/(b+a \cos x))\]
.Sam.
  • .Sam.
Quotient rule

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.Sam.
  • .Sam.
\[\huge -\frac{\frac{a+b \cos (x)}{\sqrt{1-x^2}}-b \sin (x) \left(b+\cos ^{-1}(x)\right)}{\left(b+\cos ^{-1}(x)\right)^2 \sqrt{1-\frac{(a+b \cos (x))^2}{\left(b+\cos ^{-1}(x)\right)^2}}}\] \[\text{you can simplify further by multiplying \top and bottom with} \sqrt{1-x^2}\]
anonymous
  • anonymous
isnt there an easier method???
anonymous
  • anonymous
That's what I am pondering about
anonymous
  • anonymous
using .sam's method,would'nt it be too long??
anonymous
  • anonymous
\[x = \cos^{-1}\frac{t}{ab}\]\[\frac{a+b\cos x}{b + a\cos x} = \frac{a + b\cos \cos^{-1}\frac{t}{ab}}{b + a\cos^{-1}\cos \frac{t}{ab}}\]Can we do this? lol
anonymous
  • anonymous
wat's t ???
anonymous
  • anonymous
A Variable
anonymous
  • anonymous
can you make one thing clear ...Which is correct??? \[\cos^{-1} (\cos x) =x\]\ or \[\cos (\cos^{-1} x) =x\]
anonymous
  • anonymous
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
anonymous
  • anonymous
It could have been easier if one of the cos was sin.
anonymous
  • anonymous
please answer my question which is true??? cos−1(cosx)=x or cos(cos−1x)=x
.Sam.
  • .Sam.
cos−1(cosx)=x
Diyadiya
  • Diyadiya
Both are true @Sarkar
anonymous
  • anonymous
i only knew the first one was...how come the second one is true???
Diyadiya
  • Diyadiya
http://www.wolframalpha.com/input/?i=cos%28cos%5E%28-1%29x%29%3D
anonymous
  • anonymous
You wouldn't know a lot, that doesn't mean they are all false.
anonymous
  • anonymous
in my text book,the first is only given..
anonymous
  • anonymous
Use your head to get the second one.
Mani_Jha
  • Mani_Jha
\[\cos^{-1} x=a \] \[\cos a=x\] \[\cos(\cos^{-1} x)=x\] Get it?
anonymous
  • anonymous
yeah i figured it out..
Mani_Jha
  • Mani_Jha
I just substituted a = \[\cos^{-1} x\]
anonymous
  • anonymous
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!

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