anonymous
  • anonymous
the value of tan(2tan^-1 (1/5)-pi/4) is?
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
Let\[\theta=2\tan^{-1} (1/5)-(\pi/4)\]Form a triangle with tan theta
anonymous
  • anonymous
how?
anonymous
  • anonymous
How what?!

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anonymous
  • anonymous
wrong!
anonymous
  • anonymous
let x= \[\tan^{-1} \frac{1}{5}\]
anonymous
  • anonymous
so our expression becomes \[\tan(2x-\frac{\pi}{4})\]
anonymous
  • anonymous
now apply difference of angle formula
anonymous
  • anonymous
\[\tan(A-B) = \frac{\tan(A)-\tan(B) } {1+\tan(A)\tan(B) } \]
anonymous
  • anonymous
so we get
anonymous
  • anonymous
\[\frac{\tan(2x) -1}{1+\tan(2x)} \]
anonymous
  • anonymous
but ,\[\tan(2x) = \frac{\sin(2x)}{\cos(2x)} = \frac{2\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)} \]
anonymous
  • anonymous
now, we go back to our substitution x= tan^-1(1/5) take tan of both sides , and we get tan(x) = 1/5 now we can draw up a general right angle triangle and mark an angle x , and fill in the lengths 1 and 5 for thre opposite and adjacent sides respectively remember tan = opposite/adjacent
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anonymous
  • anonymous
note the hypotenuse is sqrt(26)
anonymous
  • anonymous
now from that triangle we can find the value of sin(x) and cos(x) , the then you just plug them into the expression we had above
anonymous
  • anonymous
so \[\sin(x) = \frac{1}{\sqrt{26}}\] \[\cos(x) = \frac{5}{\sqrt{26}}\] just by using the definitions
anonymous
  • anonymous
so , our whole expression was\[\tan(2x- \frac{\pi}{4}) = \frac{ \frac{ 2\sin(x)\cos(x) }{\cos^2(x)-\sin^2(x) } -1}{ 1+ \frac{2\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}}\]
anonymous
  • anonymous
then sub in the values for sin(x) and cos(x) above, and simplify
anonymous
  • anonymous
you can do that
anonymous
  • anonymous
good. it worked out well. thanx man!

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