A) \[\cos n\varphi -i\sin n\varphi \]
B) \[\cos n\varphi +i\sin n\varphi \]
C) \[\sin n\varphi +i\cos n\varphi \]
D) \[\sin n\varphi -i\cos n\varphi \]
Correct Answer: B
Solution :
L.H.S. \[={{\left[ \frac{2{{\cos }^{2}}(\varphi /2)+2i\sin (\varphi /2)\cos (\varphi /2)}{2{{\cos }^{2}}\,(\varphi /2)-2i\sin (\varphi /2)\cos (\varphi /2)} \right]}^{n}}\] \[={{\left[ \frac{\cos \,(\varphi /2)+i\sin (\varphi /2)}{\cos (\varphi /2)-i\sin (\varphi /2)} \right]}^{n}}\]\[={{\left[ \frac{{{e}^{i(\varphi /2)}}}{{{e}^{-i(\varphi /2)}}} \right]}^{n}}={{({{e}^{i\varphi }})}^{n}}\] \[=\cos n\varphi +i\sin n\varphi \].
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