Is-it-possible-to-find-any-value-for-a-b-c-from-below-system-of-equetions-sina-sinb-sinc-cosa-cosb-cosc-

Question Number 68414 by behi83417@gmail.com last updated on 10/Sep/19

Is it possible to find any value for

a, b, c from below system of equetions ?

{\begin{cases} sina + sinb = sinc \\ cosa + cosb = cosc \end{cases}

Commented by kaivan.ahmadi last updated on 10/Sep/19

{\begin{cases} s i n a c o s a + s i n b c o s a = s i n c c o s a \\ - s i n a c o s a - s i n a c o s b = - s i n a c o s c \end{cases} \Rightarrow

s i n b c o s a - s i n a c o s b = s i n c c o s a - s i n a c o s c \Rightarrow

s i n (b - a) = s i n (c - a) \Rightarrow

{\begin{cases} b - a = 2 k π + (c - a) \\ b - a = (2 k + 1) π + (a - c) \end{cases}

\Rightarrow {\begin{cases} b - c = 2 k π \\ b - 2 a + c = (2 k + 1) π \end{cases}

Answered by Tanmay chaudhury last updated on 10/Sep/19

2 s i n (\frac{a + b}{2}) c o s (\frac{a - b}{2}) = s i n c

2 c o s (\frac{a + b}{2}) c o s (\frac{a - b}{2}) = c o s c

t a n (\frac{a + b}{2}) = t a n c

\frac{a + b}{c} = n π + c

a + b = 2 n π + 2 c \dots . (1)

4 {s i n}^{2} (\frac{a + b}{2}) {c o s}^{2} (\frac{a - b}{2}) = {s i n}^{2} c

4 {c o s}^{2} (\frac{a + b}{2}) {c o s}^{2} (\frac{a - b}{2}) = {c o s}^{2} c

a d d t h e m

4 {c o s}^{2} (\frac{a - b}{2}) = 1

c o s (\frac{a - b}{2}) = \pm \frac{1}{2}

c o s i d e r i n g + s i g n

c o s (\frac{a - b}{2}) = c o s (\frac{π}{3})

\frac{a - b}{2} = 2 n π + \frac{π}{3}

c o s (\frac{a - b}{2}) = - \frac{1}{2} = c o s (π + \frac{π}{3})

(\frac{a - b}{2}) = 2 n π + π + \frac{π}{3}

o r c o s (\frac{a - b}{2}) = - \frac{1}{2} = c o s (π - \frac{π}{3})

\frac{a - b}{2} = 2 n π + \frac{2 π}{3}

t h u s w e c a n f i n d v a l u e o f a a n d b i n t e r m s

o f c

Answered by mr W last updated on 10/Sep/19

\sin^{2} a + \sin^{2} b + 2 \sin a \sin b = \sin^{2} c

\cos^{2} a + \cos^{2} b + 2 \cos a \cos b = \cos^{2} c

2 + 2 (\sin a \sin b + \cos a \cos b) = 1

\sin a \sin b + \cos a \cos b = - \frac{1}{2}

\cos (a - b) = - \frac{1}{2}

\Rightarrow a - b = (2 n + 1) π \pm \frac{π}{3}

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