Could somebody explain to me why the following statement is true, " arctan(1) = pi/4"
Do you know what arctan is?
What is pi/4? It is 90 degrees. Now check if \(\tan 90^{\circ} = 1\)
Yes, I understand that it is the inverse of the tangent function.
Is \(\tan90^{\circ} = 1\)?
ParthKohli can put it much more simply than I can T.T
haha
Isn't tan at 90 undefined?
In other words, \(\tan = \large {\sin \over \cos}\)
Yes it is undefined then ^_^
pi/4 is 45 degrees o.o
** tan 45, he means.
OMG SORRY
\( \color{Black}{\Rightarrow \tan45^{\circ} = 1}\)
This is awkward...
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I guess you already know the important ratios, do you?
Don't you guys think we're complicating things a bit?
Haha no. \(\sin 90 = 1\) \( \color{Black}{\Rightarrow\cos 0 = 1 }\) \( \color{Black}{\Rightarrow \tan 45 = 1 }\)
\[\arctan\left(\frac{\text {opposite}}{\text{adjacent}}\right) = \theta\] \[\arctan\left(\frac 11\right) =45°= \pi/4\]
The biggest issue is that I am completely knew to trig and like when I learned algebra, I needed steps to find my solutions. Could somebody give me some step-by-step instructions to solving the problem?
\( \color{Black}{\Rightarrow \tan 45 = \large {\sin 45 \over \cos 45 } = {\Large {1 \over \sqrt2} \over {1 \over \sqrt2}} }\)
Well, you know the relationship between functions and their inverses?
That leads us to the statement tan 45 = 1
Yes, I understand that (for lack of a better explanation) the domain and range is flipped.
Well you can refer to @UnkleRhaukus 's diagram
(90,1) becomes (1,90) where 90 = x and f(x) is sin.
Similarly, (45,1) becomes (1,45) that's it ^_^
other than that, for instance, Let f be a bijective* function, such that g = f^-1 then if a is in the range of f, then what's f(g(a)) ?
if your arc tan is one it means the opposite and adjacent sides are equal. the sum of the angles in a triangle is 180° the triangle has a right angle , because we are using trig ratios , and the angles are equal because the opp and adjacent are equal 180°-90°=2x45°
\[\tan45 = 1\] Or,\[\tan(\pi /4) = 1\] Because, \[45 = \pi /4\] So, \[\tan^{-1} (1) = \tan^{-1} (\tan \pi /4) = \pi /4\] is the answer...
Long story, short, if you compose a function with its inverse, they'd cancel each other out :) so since tan (pi/4) = 1 then arctan[tan (pi/4)] = arctan 1 since arctan is the inverse of tan, they'd cancel out so... pi/4 = arctan 1
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