anonymous
  • anonymous
What is this TRIG identity called. see attachment.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
anonymous
  • anonymous
@dan815
anonymous
  • anonymous
@LynFran

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anonymous
  • anonymous
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
I'm not sure if there is a specific name for this identity
anonymous
  • anonymous
how would I explain it though
jim_thompson5910
  • jim_thompson5910
I guess you could say "composition of a trig function and another inverse" ? I'm not sure
jim_thompson5910
  • jim_thompson5910
oh you want to know how they got that identity?
anonymous
  • anonymous
yeah
anonymous
  • anonymous
here let me give you an example
jim_thompson5910
  • jim_thompson5910
tan(arccos(x)) let theta = arccos(x) so cos(theta) = x = x/1 agreed?
anonymous
  • anonymous
\[\tan(\cos^-1(1/2)\]
anonymous
  • anonymous
yes agreed
jim_thompson5910
  • jim_thompson5910
so we can create this right triangle |dw:1438215201249:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1438215224912:dw|
jim_thompson5910
  • jim_thompson5910
cos(theta) = x/1 cosine deals with adjacent over hypotenuse
jim_thompson5910
  • jim_thompson5910
so we can add these labels |dw:1438215259342:dw|
jim_thompson5910
  • jim_thompson5910
what is the missing side equal to (in terms of x) ?
anonymous
  • anonymous
1^2-x^2
jim_thompson5910
  • jim_thompson5910
it will be the square root of that
jim_thompson5910
  • jim_thompson5910
|dw:1438215360088:dw|
anonymous
  • anonymous
@jim_thompson5910 how do we gate the base of absolute value of x
jim_thompson5910
  • jim_thompson5910
tan(arccos(x)) turns into tan(theta) this is because I let theta = arccos(x) calculating tan(theta) gives \[\Large \frac{\text{opposite}}{\text{adjacent}} = \frac{\sqrt{1-x^2}}{x}\]
jim_thompson5910
  • jim_thompson5910
for some reason, and I don't know why, they made the denominator be |x| instead of just x. My guess is that they wanted to force tan(arccos(x)) to be positive. However, tan(arccos(x)) is only positive when x is positive
jim_thompson5910
  • jim_thompson5910
So it should be \[\Large \tan(\arccos(x)) = \frac{\sqrt{1-x^2}}{x}\]

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