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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\$ 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\$ You deserve to use\$\$ an(alpha-eta)=frac analpha- aneta1+ analpha aneta\$\$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\$\$ 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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