If-tan-i-cos-isin-prove-that-n-2-4-and-1-2-log-tan-4-2-

Question Number 168118 by Mastermind last updated on 03/Apr/22

I f t a n (θ + i ϕ) = c o s α + i s i n α,

p r o v e t h a t : θ = \frac{n Π}{2} + \frac{Π}{4} a n d ϕ = \frac{1}{2} l o g t a n (\frac{Π}{4} + \frac{α}{2})

Answered by mathsmine last updated on 05/Apr/22

t a n (θ + i \emptyset) = \frac{t a n (θ) + t a n (i \emptyset)}{1 - t a n (θ) t a n (i \emptyset)}

t a n (i x) = i t h (x)

\Leftrightarrow \frac{t a n (θ) + i t h (\emptyset)}{1 - i t a n (θ) t h (\emptyset)} = c o s (a) + i s i n (a)

\Leftrightarrow c o s (a) + s i n (a) t a n (θ) t h (\emptyset) + i (s i n (a) - t a n (θ) t h (\emptyset) c o s (a))

= t a n (θ) + i t h (\emptyset)

t a n (θ) = \frac{c o s (a)}{1 - s i n (a) t h (\emptyset)}

t h (\emptyset) = \frac{s i n (a)}{1 + t g (θ) c o s (a)}

\Rightarrow t g (θ) = c o s (a) (1 + t g (θ) c o s (a)) . \frac{1}{(1 + t g (θ) c o s (a) - {s i n}^{2} (a)}

= \frac{1 + t g (θ) c o s (a)}{t g (θ) + c o s (a)} \Leftrightarrow

{t g}^{2} (θ) = 1 \Rightarrow t g (θ) = \underline{+} 1 \Rightarrow

θ = \frac{π}{4} + n \frac{π}{2}

t h (\emptyset) = \frac{s i n (a)}{1 \underline{+} c o s (a)},

\frac{s i n (a)}{1 + c o s (a)} = t g (\frac{a}{2}) \dots . . E

\frac{s i n (a)}{1 - c o s (a)} = c o t (\frac{a}{2})

t h (x) = a \Rightarrow x = a r g t h (a) = \frac{1}{2} l n (\frac{a + 1}{1 - a})

E \Rightarrow \emptyset = \frac{1}{2} l n (\frac{1 + t g (\frac{a}{2})}{1 - t g (\frac{a}{2}))}) = \frac{1}{2} l n (t g (\frac{π}{4} + \frac{a}{2}))

⩽ \frac{1 + t g (x)}{1 - t g (x)} = t g (x + \frac{π}{4}) ⩾

S o θ = \frac{n π}{2} + \frac{π}{4}

\emptyset = \frac{1}{2} l n (t g (\frac{a}{2} + \frac{π}{4}))

Commented by Mastermind last updated on 05/Apr/22

T h a n k s m a n, y o u d i d a g r e a t J o b!

C o u l d y o u p l e a s e d r o p y o u r W h a t s a p p n u m b e r ?