a) If z=1+i(√3) prove that prove that z^(14) =2^(13) (−1+i(√3) ) b)prove that in triangle ABC a^2 −(b−c)^2 cos^2 (A/2)=(b+c)^2 sin^2 (A/2)


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Question Number 75272 by peter frank last updated on 09/Dec/19

$a) I f z = 1 + i \sqrt{3} p r o v e t h a t$ $p r o v e t h a t$ $z^{14} = 2^{13} (- 1 + i \sqrt{3})$ $b) p r o v e t h a t i n t r i a n g l e A B C$ $a^{2} - {(b - c)}^{2} \cos^{2} \frac{A}{2} = {(b + c)}^{2} \sin^{2} \frac{A}{2}$
	Answered by mind is power last updated on 09/Dec/19

	$z = 1 + i \sqrt{3} = 2 (\frac{1}{2} + \frac{i \sqrt{3}}{2}) \Rightarrow z = 2 (\cos (\frac{π}{3}) + isin (\frac{π}{3})) = {2 e}^{\frac{i π}{3}}$ $z^{14} = 2^{14} (e^{14 \frac{i π}{3}}) = 2^{13} . 2 . e^{i 4 π + \frac{i 2 π}{3}} = 2^{13} . 2 (\cos (\frac{2 π}{3}) + isin (\frac{2 π}{3}))$ $= 2^{13} . 2 (- \frac{1}{2} + \frac{i \sqrt{3}}{2}) = 2^{13} (- 1 + i \sqrt{3})$ $b)$ $a^{2} = b^{2} + c^{2} - 2 cbcos (A) = b^{2} + c^{2} - 2 cb (\cos^{2} (\frac{A}{2}) - \sin^{2} (\frac{A}{2}))$ $= b^{2} (\cos^{2} (\frac{A}{2}) + \sin^{2} (\frac{A}{2})) + c^{2} (\cos^{2} (\frac{A}{2}) + \sin^{2} (\frac{A}{2})) - 2 cb (\cos^{2} (\frac{A}{2}) + \sin^{2} (\frac{A}{2}))$ $= \cos^{2} (\frac{A}{2}) (b^{2} + c^{2} - 2 cb) + \sin^{2} (\frac{A}{2}) (b^{2} + c^{2} + 2 cb)$ $= {(b - c)}^{2} \cos^{2} (\frac{A}{2}) + {(b + c)}^{2} \sin^{2} (\frac{A}{2})$ $\Leftrightarrow a^{2} - {(b - c)}^{2} \cos^{2} (\frac{A}{2}) = {(b + c)}^{2} \sin^{2} (\frac{A}{2})$
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