Question

Rewrite sin 15 degree in terms of a 75 degree angle and in terms of the reciprocal of a trigonometric function.

Answer 1

$\bf csc(\theta)=\cfrac{1}{sin(\theta)} \\\\\\ sin({{ \alpha}} - {{ \beta}})=sin({{ \alpha}})cos({{ \beta}})- cos({{ \alpha}})sin({{ \beta}})\\\\ -------------------------------\\\\ sin(15^o)\implies sin(45^o-30^o) \\\\\\ sin(45^o)cos(30^o)-cos(45^o)sin(30^o)$

$\bf \cfrac{\sqrt{2}}{2}\cdot \cfrac{\sqrt{3}}{2}-\cfrac{\sqrt{2}}{2}\cdot \cfrac{1}{2}\implies \cfrac{\sqrt{6}}{4}-\cfrac{\sqrt{2}}{4}\implies \cfrac{\sqrt{6}-\sqrt{2}}{4}\\\\ -------------------------------\\\\ csc(15^o)=\cfrac{1}{sin(15^o)}\implies csc(15^o)=\cfrac{1}{\frac{\sqrt{6}-\sqrt{2}}{4}} \\\\\\ csc(15^o)=\cfrac{4}{\sqrt{6}-\sqrt{2}}$

and now, we can rationalize the denominator by using its conjugate and difference of squares.

$\bf \cfrac{4}{\sqrt{6}-\sqrt{2}}\cdot \cfrac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}}\implies \cfrac{4(\sqrt{6}+\sqrt{2})}{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})} \\\\\\ \cfrac{4(\sqrt{6}+\sqrt{2})}{(\sqrt{6})^2-(\sqrt{2})^2}\implies \cfrac{4(\sqrt{6}+\sqrt{2})}{6-2}\implies \boxed{\sqrt{6}+\sqrt{2}}$