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Question: If \(\cos {18^ \circ } - \sin {18^ \circ } = \sqrt n \sin {27^ \circ }\), then \(n = \)...

If cos⁡18∘−sin⁡18∘=nsin⁡27∘\cos {18^ \circ } - \sin {18^ \circ } = \sqrt n \sin {27^ \circ }, then n=n =

Explanation

Solution

First we will convert consent into sine by using formula sin⁡A=cos⁡(π2−A)\sin A = \cos \left( {\dfrac{\pi }{2} - A} \right). After this conversion we will use cos⁡A−cos⁡B=−2sin⁡(A+B2)sin⁡(A−B2)\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∘−sin⁡18∘\cos {18^ \circ } - \sin {18^ \circ }
We know that sin⁡A=cos⁡(π2−A)\sin A = \cos \left( {\dfrac{\pi }{2} - A} \right), so on converting we get
⇒cos⁡18∘−cos⁡(90−18)∘\Rightarrow \cos {18^ \circ } - \cos {\left( {90 - 18} \right)^ \circ }
⇒cos⁡18∘−cos⁡72∘\Rightarrow \cos {18^ \circ } - \cos {72^ \circ }
Now using cos⁡A−cos⁡B=−2sin⁡(A+B2)sin⁡(A−B2)\cos A - \cos B = - 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)
Therefore, ⇒−2sin⁡(72+182)∘sin⁡(18−722)∘ \Rightarrow - 2\sin {\left( {\dfrac{{72 + 18}}{2}} \right)^ \circ }\sin {\left( {\dfrac{{18 - 72}}{2}} \right)^ \circ }
⇒2sin⁡45∘sin⁡27∘\Rightarrow 2\sin {45^ \circ }\sin {27^ \circ }
Substituting sin⁡45∘=12\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }} we get
⇒2sin⁡27∘\Rightarrow \sqrt 2 \sin {27^ \circ }
Therefore, n=2n = 2

Note: This could be generalized as cos⁡A−sin⁡A=2sin⁡(π4−A)\cos A - \sin A = \sqrt 2 \sin \left( {\dfrac{\pi }{4} - A} \right). Formula used to prove this are sin⁡A=cos⁡(π2−A)\sin A = \cos \left( {\dfrac{\pi }{2} - A} \right) and cos⁡A−cos⁡B=−2sin⁡(A+B2)sin⁡(A−B2)\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\cos A - \sin A

⇒cos⁡A−cos⁡(90−A) ⇒−2sin⁡(45)∘sin⁡(A−π4) ⇒2sin⁡(π4−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) \\\