Question

(cos75+isin75)÷(cos15-isin15)
simplify

Answer 1

cos75 =sin15 ,sin 75=cos15

Answer 2

simplify plz

Answer 3

(sin 15+i cos 15)/(cos15-i sin15)

Answer 4

Start by multiplying by the conjugate of the denominator,
\[\Large\rm \frac{\cos75+\mathcal i \sin 75}{\cos15-\mathcal i \sin 15}\color{royalblue}{\left(\frac{\cos15+\mathcal i \sin 15}{\cos15+\mathcal i \sin 15}\right)}\]
\[\Large\rm =\frac{(\cos75+\mathcal i \sin 75)(\cos15+\mathcal i \sin15)}{\cos^215+\sin^215}\]Apply our Pythagorean Identity for Sine and Cosine,\[\Large\rm =\frac{(\cos75+\mathcal i \sin 75)(\cos15+\mathcal i \sin15)}{1}\]So it turns into a multiplication problem now.\[\Large\rm =\cos(75+15)+\mathcal i \sin(75+15)\]I forget the name of the property that allows us to do that...
if you convert to exponentials, the addition becomes a lot clearer.

Answer 5

Use Euler's formula:
\[\Large\boxed{e^{i\theta}=\cos\theta+i\sin\theta}\]
And some properties of (even) cosine and (odd) sine:
\[\boxed{\cos\theta =\cos(-\theta)}\]\[\boxed{\sin\theta =-\sin(-\theta)}\]

\[\frac{\cos(75°)+i\sin(75°)}{\cos(15°)-i\sin(15°)}\\
=\frac{\cos(75°)+i\sin(75°)}{\cos(-15°)+i\sin(-15°)}\\
=\frac{e^{75°i}}{e^{-15°i}}\\
=e^{(75°+15°)i}\\
%=e^{90°i}\\
%=\cos(90°)+i\sin(90°)\\
%=i
=\dots \]