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asdfasdfasdfasdfasdf1 3 years ago How do you know that tan(1/2) = pi/4 and tan(1/4) = 1.. etc. etc?

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1. lgbasallote

hmm tan (1/2) = pi/4?

2. asdfasdfasdfasdfasdf1

my bad, tan(1) = pi/4. are there any tricks to remembering them? and I mean arctan!

3. lgbasallote

ahh are you familiar with the 45-45-90 triangle?

4. 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?

5. asdfasdfasdfasdfasdf1

yes.. great :D is there a way to remember them for like cos/sin?

6. lgbasallote

yes. this is how you remember it. using triangles. sin and cosine are also from triangles

7. lgbasallote

just remember these two triangles |dw:1346929906503:dw| and |dw:1346929946985:dw|

8. sauravshakya

A 0 pi/6 pi/4 pi/3 pi/2 SinA

9. sauravshakya

First make table like that

10. 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...

11. lgbasallote

i think that's what you meant? $\tan = \frac{opposite}{hypotenuse} \frac{\sin}{\cos}$ @asdfasdfasdfasdfasdf1 ?

12. lgbasallote

there should be equal sign there ^

13. sauravshakya

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

14. sauravshakya

Now, once we know SinA we will know CosA too

15. sauravshakya

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

16. sauravshakya

Now, Find Tan A by using formula Tan A = Sin A/ Cos A

17. sauravshakya

I hope that helps

18. asdfasdfasdfasdfasdf1

Thank you both! the pictures i think is easier to remember though.

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