Question

If $\cos {18^ \circ } - \sin {18^ \circ } = \sqrt n \sin {27^ \circ }$, then $n = $

Answer 1

Hint: First we will convert consent into sine by using formula $\sin A = \cos \left( {\dfrac{\pi }{2} - A} \right)$. After this conversion we will use $\cos A - \cos B = - 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$ and on simplifying this we get the value of n.

 Complete step-by-step answer:
$\cos {18^ \circ } - \sin {18^ \circ }$
We know that $\sin A = \cos \left( {\dfrac{\pi }{2} - A} \right)$, so on converting we get
$ \Rightarrow \cos {18^ \circ } - \cos {\left( {90 - 18} \right)^ \circ }$
$ \Rightarrow \cos {18^ \circ } - \cos {72^ \circ }$
Now using $\cos A - \cos B = - 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$
Therefore, $ \Rightarrow - 2\sin {\left( {\dfrac{{72 + 18}}{2}} \right)^ \circ }\sin {\left( {\dfrac{{18 - 72}}{2}} \right)^ \circ }$
$ \Rightarrow 2\sin {45^ \circ }\sin {27^ \circ }$
Substituting $\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$ we get
$ \Rightarrow \sqrt 2 \sin {27^ \circ }$
Therefore, \[n = 2\]

Note: This could be generalized as $\cos A - \sin A = \sqrt 2 \sin \left( {\dfrac{\pi }{4} - A} \right)$. Formula used to prove this are $\sin A = \cos \left( {\dfrac{\pi }{2} - A} \right)$ and $\cos A - \cos B = - 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$.
Generalized form
$\cos A - \sin A$
\[
   \Rightarrow \cos A - \cos \left( {90 - A} \right) \\
   \Rightarrow - 2\sin {\left( {45} \right)^ \circ }\sin \left( {A - \dfrac{\pi }{4}} \right) \\
   \Rightarrow \sqrt 2 \sin \left( {\dfrac{\pi }{4} - A} \right) \\
 \]