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Question Number 152010 by RB95 last updated on 25/Aug/21

Commented by RB95 last updated on 25/Aug/21

$S l t$ $P o u v i e z v o u s m^{'} a i d e r ?$
	Answered by puissant last updated on 25/Aug/21

	$E x e r c i c e 1 :$ $1)$ ${c o s}^{4} x = {(\frac{e^{i x} + e^{- i x}}{2})}^{4}$ $= \frac{1}{16} (e^{i 4 x} + 4 e^{i 3 x} e^{- i x} + 6 e^{i 2 x} e^{- i 2 x} + 4 e^{i x} e^{- i 3 x} + e^{- i 4 x})$ $= \frac{1}{16} ((e^{i 4 x} + e^{- i 4 x}) + 4 (e^{i 2 x} + e^{- i 2 x}) + 6)$ $\Rightarrow {c o s}^{4} x = \frac{1}{8} c o s 4 x + \frac{1}{2} c o s 2 x + \frac{3}{8}$ $\Rightarrow c o s 4 x = 8 ({c o s}^{4} x - \frac{1}{2} c o s 2 x - \frac{3}{8})$ $\Rightarrow c o s 4 x = 8 {c o s}^{4} x - 4 (2 {c o s}^{2} x - 1) - 3$ $\Rightarrow c o s 4 x = 8 {c o s}^{4} x - 8 {c o s}^{2} x + 1 . .$ $2)$ $l i n e a r i s o n s {s i n}^{5} x$ ${s i n}^{5} x = {(\frac{e^{i x} - e^{- i x}}{2 i})}^{5}$ $= \frac{1}{32 i} (e^{i 5 x} - 5 e^{i 4 x} e^{- i x} + 10 e^{i 3 x} e^{- i 2 x} - 10 e^{i 2 x} e^{- i 3 x} + 5 e^{i x} e^{- i 4 x} - e^{- i 5 x})$ $= \frac{1}{32 i} ((e^{i 5 x} - e^{i 5 x}) - 5 (e^{i 3 x} - e^{- i 3 x}) + 10 (e^{i x} - e^{- i x}))$ $\Rightarrow {s i n}^{5} x = \frac{1}{16} s i n 5 x - \frac{5}{16} s i n 3 x + \frac{5}{8} s i n x$ $3)$ $\to s i n (a + b) = s i n a c o s b + c o s a s i n b$ $\to c o s (a + b) = c o s a c o s b - s i n a s i n b$ $\to t a n (a + b) = \frac{s i n (a + b)}{c o s (a + b)} = \frac{s i n a c o s b + c o s a c o s b}{c o s a c o s b - s i n a s i n b}$ $e n d i v i s a n t p a r c o s a c o s b, o n$ $t a n (a + b) = \frac{t a n a + t a n b}{1 - t a n a t a n b} . .$ $\to s i n (a - b) = s i n a c o s b - c o s a c o s b$ $\to d e f a c o n a n a l o g u e, o n t r o u v e q u e$ $t a n (a - b) = \frac{t a n a - t a n b}{1 + t a n a t a n b} . .$ $E x e r c i c e 2 :$ $s e l o n M O I V R E, o n s a i t q u e {(e^{i θ})}^{n} = e^{i n θ}$ $\Rightarrow {(c o s θ + i s i n θ)}^{n} = (c o s n θ + i s i n n θ)$ $s e r t t o i d e c a p o u r r e p o n d r e a u x q u e s t i o n s d e l^{'} e x e r c i e . .$ $E x e r c i c e 3 :$ $1)$ $j e n e v o i s p a s .$ $m a i s o n s a i t q u e a r g (z^{n}) \equiv n a r g (z) [2 π]$ $2)$ $E n u t i l i s a n t l a f a c t o r i s a t i o n d e s a n g l e s$ $m o i t i e s, o n t r o u v e :$ $z = {[1 + e^{i (\frac{π}{2} + α)}]}^{n} = 2^{n} {c o s}^{n} (\frac{π}{4} + \frac{α}{2}) e^{i n (\frac{π}{4} + \frac{α}{2})}$ $e t d o n c R e (z) = 2^{n} {c o s}^{n} (\frac{π}{4} + \frac{α}{2}) c o s (\frac{n π}{4} + \frac{n α}{2})$ $R e (z) = 0 \Leftrightarrow α = \frac{π}{2} + 2 k π o u α = \frac{(2 k - 1) π}{n} - \frac{π}{2} . .$ $E x e r c i c e 4 :$ $1) z = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}$ $n o m m o n s w e t - w l e s r a c i n e s c a r r \overset{´}{e e s} d e z .$ $w = \sqrt{\frac{\sqrt{2} + 1}{2 \sqrt{2}}} + i \sqrt{\frac{\sqrt{2} - 1}{2 \sqrt{2}}}$ $\Rightarrow w = \frac{1}{2} \sqrt{2 + \sqrt{2}} + i \frac{1}{2} \sqrt{2 - \sqrt{2}}$ $d^{'} a b o r d z = e^{i \frac{π}{4}} o n r e m a r q u e q u e : {(e^{i \frac{π}{8}})}^{2} = (e^{i \frac{2 π}{8}}) = e^{i \frac{π}{4}}$ $d o n c c o s (\frac{π}{8}) = \frac{1}{2} \sqrt{2 + \sqrt{2}} e t s i n (\frac{π}{8}) = \frac{1}{2} \sqrt{2 - \sqrt{2}} (p a r i d e n t i f i c a t i o n)$ $c a r e^{i \frac{π}{8}} = c o s (\frac{π}{8}) + i s i n (\frac{π}{8}) . .$ $2)$ $\to z = {(\frac{1 + i \sqrt{3}}{1 - i})}^{n}$ $o n m o n t r e t r i v i a l e m e n t q u e$ $1 + i \sqrt{3} = 2 e^{i \frac{π}{3}} e t 1 - i = \sqrt{2} e^{- i \frac{π}{4}}$ $\Rightarrow z = {(\frac{2}{\sqrt{2}} e^{i \frac{7 π}{12}})}^{n} = {(\sqrt{2})}^{n} e^{i \frac{7 n π}{12}}$ $l e m o d u l e e s t {(\sqrt{2})}^{n} e t l^{'} a r g u m e n t e s t \frac{7 n π}{12}; θ \in] - π; π [.$ $\to z = {(1 + c o s θ + i s i n θ)}^{n} = {(1 + e^{i θ})}^{n}$ $= {(e^{i \frac{θ}{2}} (e^{i \frac{θ}{2}} + e^{- i \frac{θ}{2}}))}^{n} ≪ d^{'} a p r e s l a f a c t o r i s a t i o n$ $d e s a n g l e s m o i t i \overset{´}{e s} ≫ .$ $\Rightarrow z = {(2 c o s (\frac{θ}{2}) e^{i \frac{θ}{2}})}^{n} = 2^{n} {c o s}^{n} (\frac{θ}{2}) e^{i \frac{n θ}{2}}$ $a l o r s l e m o d u l e d e z e s t 2^{n} {c o s}^{n} (\frac{θ}{2}) e t$ $l^{'} a r g u m e n t e s t a r g (z) = \frac{n θ}{2}, θ \in] - π; π [$
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