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{√(2)}{2}\cdot \cfrac{√(3)}{2}-\cfrac{√(2)}{2}\cdot \cfrac{1}{2}\implies \cfrac{√(6)}{4}-\cfrac{√(2)}{4}\implies \cfrac{√(6)-√(2)}{4}\\\\ -------------------------------\\\\ csc(15^o)=\cfrac{1}{sin(15^o)}\implies csc(15^o)=\cfrac{1}{(√(6)-√(2))/(4)} \\\\\\ csc(15^o)=\cfrac{4}{√(6)-√(2)}`

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


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