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asdfasdfasdfasdfasdf1
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How do you know that tan(1/2) = pi/4 and tan(1/4) = 1.. etc. etc?
 one year ago
 one year ago
asdfasdfasdfasdfasdf1 Group Title
How do you know that tan(1/2) = pi/4 and tan(1/4) = 1.. etc. etc?
 one year ago
 one year ago

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lgbasallote Group TitleBest ResponseYou've already chosen the best response.3
hmm tan (1/2) = pi/4?
 one year ago

asdfasdfasdfasdfasdf1 Group TitleBest ResponseYou've already chosen the best response.0
my bad, tan(1) = pi/4. are there any tricks to remembering them? and I mean arctan!
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.3
ahh are you familiar with the 454590 triangle?
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.3
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?
 one year ago

asdfasdfasdfasdfasdf1 Group TitleBest ResponseYou've already chosen the best response.0
yes.. great :D is there a way to remember them for like cos/sin?
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.3
yes. this is how you remember it. using triangles. sin and cosine are also from triangles
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.3
just remember these two triangles dw:1346929906503:dw and dw:1346929946985:dw
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
A 0 pi/6 pi/4 pi/3 pi/2 SinA
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
First make table like that
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.3
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...
 one year ago

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

lgbasallote Group TitleBest ResponseYou've already chosen the best response.3
there should be equal sign there ^
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
Now, once we know SinA we will know CosA too
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
Now, Find Tan A by using formula Tan A = Sin A/ Cos A
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
I hope that helps
 one year ago

asdfasdfasdfasdfasdf1 Group TitleBest ResponseYou've already chosen the best response.0
Thank you both! the pictures i think is easier to remember though.
 one year ago
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