If a=cos α +isin α , b=cos β +isin β , c=cos γ +isin γ and b c + c a + a b =1,Then cos( β - γ )+cos( γ - α )+cos( α - β ) is equal to Options: A . &nbsp 3 2 B . &nbsp - 3 2 C . &nbsp 0 D . &nbsp 1

Answer: Option D : D b c = c o s β + i s i n β c o s γ + i s i n γ × c o s γ − i s i n γ c o s γ − i s i n γ ⇒ b c = c o s ( β − γ ) + i s i n ( β − γ ) . . . . . ( i ) Similarly, c a = c o s ( γ − α ) + i s i n ( γ − α ) . . . . . . . ( i i ) And a b = c o s ( α − β ) + i s i n ( α − β ) . . . . . . . ( i i ) Equating real and imaginary parts, c o s ( β − γ ) + c o s ( γ − α ) + c o s ( α − β ) = 1

If a=cosα+isinα, b=cosβ+isinβ, c=cosγ+isin;

Discussion Forum : Complex Numbers

Question -

If a=cos $α$ +isin $α$ , b=cos $β$ +isin $β$ ,

c=cos $γ$ +isin $γ$ and $\frac{b}{c}$ + $\frac{c}{a}$ + $\frac{a}{b}$ =1,Then

cos( $β$ - $γ$ )+cos( $γ$ - $α$ )+cos( $α$ - $β$ ) is equal to

Options:

A .

\frac{3}{2}

B . -

\frac{3}{2}

C . 0

D . 1

Answer: Option D
:
D

$\frac{b}{c}$ = $\frac{c o s β + i s i n β}{c o s γ + i s i n γ} \times \frac{c o s γ - i s i n γ}{c o s γ - i s i n γ}$

$\Rightarrow \frac{b}{c} = c o s (β - γ) + i s i n (β - γ) . . . . . (i)$

Similarly, $\frac{c}{a} = c o s (γ - α) + i s i n (γ - α) . . . . . . . (i i)$

And $\frac{a}{b} = c o s (α - β) + i s i n (α - β) . . . . . . . (i i)$

Equating real and imaginary parts,

$c o s (β - γ) + c o s (γ - α) + c o s (α - β) = 1$

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Discussion Forum : Complex Numbers

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