Question

Find the value of $(cos\frac{\pi }{8}+isin\frac{\pi }{8})\times (cos\frac{\pi }{12}+isin\frac{\pi }{12})\times (cos\frac{\pi }{24}+isin\frac{\pi }{24})\times (cos\frac{\pi }{4}+isin\frac{\pi }{4})$
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Answer 1

$(cos\frac{\pi }{8}+isin\frac{\pi }{8})\times (cos\frac{\pi }{12}+isin\frac{\pi }{12})\times (cos\frac{\pi }{24}+isin\frac{\pi }{24})\times (cos\frac{\pi }{4}+isin\frac{\pi }{4})\dots \dots$ (1)

We know, $e^{\frac{i\pi }{8}}$ = $cos\frac{i\pi }{8}+isin\frac{i\pi }{8}$

Similarly, $e^{\frac{i\pi }{12}}=cos\frac{\pi }{12}+isin\frac{\pi }{12},e^{\frac{i\pi }{24}}=cos\frac{\pi }{24}+isin\frac{\pi }{24},e^{\frac{i\pi }{4}}=cos\frac{\pi }{4}+isin\frac{\pi }{4}$

From equation (1)

=$e^{\frac{i\pi }{8}}$ $\times$ $e^{\frac{i\pi }{12}}$ $\times$$e^{\frac{i\pi }{24}}$ $\times$ $e^{\frac{i\pi }{4}}$

=$e^{(\frac{i\pi }{8}+\frac{i\pi }{12}+\frac{i\pi }{24}+\frac{i\pi }{4})}$

=$e^{i(\frac{\pi }{8}+\frac{\pi }{12}+\frac{\pi }{24}+\frac{\pi }{4})}$

Convert this Euler's form to polar form

$=cos(\frac{\pi }{8}+\frac{\pi }{12}+\frac{\pi }{24}+\frac{\pi }{4})+isin(\frac{\pi }{8}+\frac{\pi }{12}+\frac{\pi }{24}+\frac{\pi }{4})$

=$cos\frac{\pi }{2}+isin\frac{\pi }{2}$

= 0 + i = i