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asdfasdfasdfasdfasdf1 Group Title

How do you know that tan(1/2) = pi/4 and tan(1/4) = 1.. etc. etc?

  • 2 years ago
  • 2 years ago

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  1. lgbasallote Group Title
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    hmm tan (1/2) = pi/4?

    • 2 years ago
  2. asdfasdfasdfasdfasdf1 Group Title
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    my bad, tan(1) = pi/4. are there any tricks to remembering them? and I mean arctan!

    • 2 years ago
  3. lgbasallote Group Title
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    ahh are you familiar with the 45-45-90 triangle?

    • 2 years ago
  4. lgbasallote Group Title
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    |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?

    • 2 years ago
  5. asdfasdfasdfasdfasdf1 Group Title
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    yes.. great :D is there a way to remember them for like cos/sin?

    • 2 years ago
  6. lgbasallote Group Title
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    yes. this is how you remember it. using triangles. sin and cosine are also from triangles

    • 2 years ago
  7. lgbasallote Group Title
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    just remember these two triangles |dw:1346929906503:dw| and |dw:1346929946985:dw|

    • 2 years ago
  8. sauravshakya Group Title
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    A 0 pi/6 pi/4 pi/3 pi/2 SinA

    • 2 years ago
  9. sauravshakya Group Title
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    First make table like that

    • 2 years ago
  10. lgbasallote Group Title
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    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...

    • 2 years ago
  11. lgbasallote Group Title
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    i think that's what you meant? \[\tan = \frac{opposite}{hypotenuse} \frac{\sin}{\cos}\] @asdfasdfasdfasdfasdf1 ?

    • 2 years ago
  12. lgbasallote Group Title
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    there should be equal sign there ^

    • 2 years ago
  13. sauravshakya Group Title
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    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

    • 2 years ago
  14. sauravshakya Group Title
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    Now, once we know SinA we will know CosA too

    • 2 years ago
  15. sauravshakya Group Title
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    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

    • 2 years ago
  16. sauravshakya Group Title
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    Now, Find Tan A by using formula Tan A = Sin A/ Cos A

    • 2 years ago
  17. sauravshakya Group Title
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    I hope that helps

    • 2 years ago
  18. asdfasdfasdfasdfasdf1 Group Title
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    Thank you both! the pictures i think is easier to remember though.

    • 2 years ago
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