not approximately, but precisely, and also what does one need to understand to have the ability to do that?

For instance, I recognize that $ an(45°)$ provides the incline the a heat which encloses $45°$ v the $x$-axis, namely $1$, therefore $arctan(1)=45°$ or $fracpi4$. Is over there a way to use this relation?

Regards




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Well, you can go the other way. That"s easier.

$ an 30^circ = frac1sqrt 3$

Let $t = an 15^circ$.

So $displaystyle frac2t1-t^2 = frac1sqrt 3$

Solving,

$t^2 + 2sqrt 3 t - 1 = 0$

$displaystyle t = -frac-2sqrt 3 + sqrt12+42$ (only admissible root)

$t = 2 -sqrt 3$

(edited together the question seems to have been amended to ask for $ an (-15^circ)$:

$ an (-15^circ) = - an 15^circ = sqrt 3 - 2$


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You deserve to use$$ an(alpha-eta)=frac analpha- aneta1+ analpha aneta$$with $alpha=45^circ$ and $eta=60^circ$, therefore $ an45^circ=1$ and also $ an60^circ=sqrt3$:$$ an(-15^circ)=frac1-sqrt31+sqrt3=frac(1-sqrt3)^21-3=frac1-2sqrt3+3-2=sqrt3-2$$


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Here is yet another way.$$2-sqrt3=frac(sqrt3-1)(sqrt3-1)2=fracsqrt3-11+sqrt3$$

Now we have $$arctan left(fraca-b1+ab ight)=arctan a-arctan b$$

So $$arctan 2-sqrt3=arctan sqrt3-arctan 1=60^circ-45^circ=15^circ.$$


The edge occurs in the dodecagon, which is a ingredient of triangles and also squares. This gives you a name: coordinates system, native which ye find

sin(15) = (sqrt (3)+1)/sqrt(2)cos(15) = (sqrt (3)-1)/sqrt(2)

Whence the proportion is (sqrt(3)-1)̃²/(sqrt(3)+1)(sqrt(3)-1) = 2-sqrt(3).


$$sqrt3-2=cot30^circ-csc30^circ=dfraccos30^circ-1sin30^circ=dfrac-2sin^215^circ2sin15^circcos15^circ=- an15^circ= an(-15^circ)$$

Now usage $-90^circ
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