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

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lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.3hmm tan (1/2) = pi/4?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0my bad, tan(1) = pi/4. are there any tricks to remembering them? and I mean arctan!

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.3ahh are you familiar with the 454590 triangle?

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.3dw: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
 4 years ago
Best ResponseYou've already chosen the best response.0yes.. great :D is there a way to remember them for like cos/sin?

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.3yes. this is how you remember it. using triangles. sin and cosine are also from triangles

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.3just remember these two triangles dw:1346929906503:dw and dw:1346929946985:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0A 0 pi/6 pi/4 pi/3 pi/2 SinA

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0First make table like that

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.3so 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
 4 years ago
Best ResponseYou've already chosen the best response.3i think that's what you meant? \[\tan = \frac{opposite}{hypotenuse} \frac{\sin}{\cos}\] @asdfasdfasdfasdfasdf1 ?

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.3there should be equal sign there ^

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Now, 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
 4 years ago
Best ResponseYou've already chosen the best response.0Now, once we know SinA we will know CosA too

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0A 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
 4 years ago
Best ResponseYou've already chosen the best response.0Now, Find Tan A by using formula Tan A = Sin A/ Cos A

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thank you both! the pictures i think is easier to remember though.
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