anonymous
  • anonymous
How do you know that tan(1/2) = pi/4 and tan(1/4) = 1.. etc. etc?
Mathematics
schrodinger
  • schrodinger
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lgbasallote
  • lgbasallote
hmm tan (1/2) = pi/4?
anonymous
  • anonymous
my bad, tan(1) = pi/4. are there any tricks to remembering them? and I mean arctan!
lgbasallote
  • lgbasallote
ahh are you familiar with the 45-45-90 triangle?

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lgbasallote
  • lgbasallote
|dw:1346929711704:dw| so as you can see from my triangle... \[\tan (\frac \pi 4 ) = 1\] therefore.. \[\tan^{-1} (1) = \frac \pi 4\] does that help?
anonymous
  • anonymous
yes.. great :D is there a way to remember them for like cos/sin?
lgbasallote
  • lgbasallote
yes. this is how you remember it. using triangles. sin and cosine are also from triangles
lgbasallote
  • lgbasallote
just remember these two triangles |dw:1346929906503:dw| and |dw:1346929946985:dw|
anonymous
  • anonymous
A 0 pi/6 pi/4 pi/3 pi/2 SinA
anonymous
  • anonymous
First make table like that
lgbasallote
  • lgbasallote
so you can find out that \[\sin (\frac \pi 3) = \frac{opposite}{hypotenuse} = \frac {\sqrt 3} 2\] or\[\tan (\frac \pi 6) = \frac{opposite}{adjacent} = \frac{1}{\sqrt 3}\] and so on...
lgbasallote
  • lgbasallote
i think that's what you meant? \[\tan = \frac{opposite}{hypotenuse} \frac{\sin}{\cos}\] @asdfasdfasdfasdfasdf1 ?
lgbasallote
  • lgbasallote
there should be equal sign there ^
anonymous
  • anonymous
Now, A 0 pi/6 pi/4 pi/3 pi/2 SinA ( 0/4)^1/2 (1/4)^1/2 (2/4)^1/2 (3/4)^1/2 (4/4)^1/2
anonymous
  • anonymous
Now, once we know SinA we will know CosA too
anonymous
  • anonymous
A 0 pi/6 pi/4 pi/3 pi/2 SinA ( 0/4)^1/2 (1/4)^1/2 (2/4)^1/2 (3/4)^1/2 (4/4)^1/2 CosA (4/4)^1/2 (3/4)^1/2 (2/4)^1/2 (1/4)^1/2 (0/4)^1/2
anonymous
  • anonymous
Now, Find Tan A by using formula Tan A = Sin A/ Cos A
anonymous
  • anonymous
I hope that helps
anonymous
  • anonymous
Thank you both! the pictures i think is easier to remember though.

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