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sin\left(cos^{-1}\left(\frac{x^2}{4-x^2}\right)\right)

Guest Mar 9, 2017
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 #1
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Am I reading it right?

sin(cos^(-1)(x^2/(4 - x^2)))

 

Solve for x:
sqrt(1 - x^4/(4 - x^2)^2) = 0

Square both sides:
1 - x^4/(4 - x^2)^2 = 0

Bring 1 - x^4/(4 - x^2)^2 together using the common denominator (x^2 - 4)^2:
-(8 (x^2 - 2))/(x^2 - 4)^2 = 0

Divide both sides by -8:
(x^2 - 2)/(x^2 - 4)^2 = 0

Multiply both sides by (x^2 - 4)^2:
x^2 - 2 = 0

Add 2 to both sides:
x^2 = 2

Take the square root of both sides:
Answer: |x = sqrt(2)         or          x = -sqrt(2)

Guest Mar 9, 2017
 #2
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\(sin\left(cos^{-1}\left(\frac{x^2}{4-x^2}\right)\right)\)

 

We are looking for the sin of an angle (theta) with a cosine of   x^2 / ( 4 - x^2).....so we have...

 

sin (theta)   =  √ [ 1 - cos^2 (theta) ] =  [ 1 - [ (x^2)/(4-x^2)]^2 ] =  √ [x^4 - 8x^2 + 16 - x^4] / (4 - x^2)   =

 

√ {16 - 8x^2) / (4- x^2)    =  √ [ 4 (4 - 2x^2) ] / (4 - x^2)   =  2√(4 - 2x^2) / (4 - x^2)

 

 

cool cool cool

CPhill  Mar 9, 2017

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