Sequence And Series Question 525
Question: If $ i=\sqrt{-1} $ , then $ \frac{e^{xi}+{e^{-xi}}}{2}= $
Options:
A) $ 1+\frac{x^{2}}{2,!}+\frac{x^{4}}{4,!}+…..\infty $
B) $ 1-\frac{x^{2}}{2,!}+\frac{x^{4}}{4,!}-…..\infty $
C) $ x+\frac{x^{3}}{3,!}+\frac{x^{5}}{5,!}+….\infty $
D) $ i,[ x-\frac{x^{3}}{3,!}+\frac{x^{5}}{5,!}-…..\infty ] $
Show Answer
Answer:
Correct Answer: B
Solution:
We know that $ e^{x}=1+\frac{x}{1\ !}+\frac{x^{2}}{2\ !}+\frac{x^{3}}{3\ !}+\frac{x^{4}}{4\ !}+…..\infty $ $ e^{ix}=1+\frac{(ix)}{1\ !}+\frac{{{(xi)}^{2}}}{2\ !}+\frac{x^{3}i^{3}}{3\ !}+……. $ $ =1+\frac{ix}{1\ !}-\frac{x^{2}}{2\ !}-\frac{ix^{3}}{3\ !}+\frac{x^{4}}{4\ !}+\frac{ix^{5}}{5\ !}-\frac{x^{6}}{6\ !}+……\infty $ ……(i) and $ ={\log_{e}}( \frac{1+\frac{1}{x}}{1-\frac{1}{x}} )=2{ \frac{1}{x}+\frac{1}{3x^{3}}+\frac{1}{5x^{5}}+……. } $ $ =1-\frac{xi}{1\ !}-\frac{x^{2}}{2\ !}+\frac{ix^{3}}{3\ !}+\frac{x^{4}}{4\ !}-\frac{ix^{5}}{5\ !}-\frac{x^{6}}{6\ !}+……\infty $ ……(ii) Now from (i) and (ii), we have $ \frac{e^{ix}+{e^{-ix}}}{2}=( 1-\frac{x^{2}}{2\ !}+\frac{x^{4}}{4\ !}-\frac{x^{6}}{6\ !}+…. ) $ .
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