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

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

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

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

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
First make table like that
 2 years 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...
 2 years 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 ?
 2 years ago

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

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

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

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