Simplify the expression

killasnake

Junior Member
Joined
Sep 11, 2005
Messages
55
How do I simplify this expression?

\(\displaystyle tan(2cos^-^1(x/3))\)

this is what I came up with but it's wrong

tan(2arccos((1/3)x))\displaystyle tan(2*arccos((1/3)*x))
 
tan(2θ)=sin(2θ)cos(2θ)=2sinθcosθ2cos2θ1\displaystyle tan(2\theta) = \frac{sin(2\theta)}{cos(2\theta)} = \frac{2sin\theta cos\theta}{2cos^{2}\theta - 1}

Where :

θ=cos1(x3)\displaystyle \theta = cos^{-1}\left(\frac{x}{3}\right)

There may be a better solution to this but I don't see any at the moment ...
 
I would suggest that you draw a sketch. Let theta be the angle whose cosine is x/3. In QI theta has an abscissa of x and a hypotenuse of 3. Use the Pythagorean Th. to determine the measure of the ordinate. I got 9x2\displaystyle \sqrt{9-x^2}. Therefore you can determine that tanθ=9x2x\displaystyle \tan \theta = \frac{\sqrt{9-x^2}}{x}. Combine that with fact that tan2θ=2tanθ1tan2θ\displaystyle \tan 2 \theta = \frac{2\tan\theta}{1-\tan^2\theta} and go from there.
 
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