Prove-sin-a-sin-a-14pi-3-sin-a-8pi-3-0-

Question Number 1808 by alib last updated on 04/Oct/15

P r o v e \sin (a) + \sin (a + \frac{14 π}{3}) + \sin (a - \frac{8 π}{3}) = 0 /

Answered by 112358 last updated on 05/Oct/15

U s i n g t h e c o m p o u n d a n g l e f o r m u l a

s i n (x \pm y) = s i n x c o s y \pm c o s x s i n y w e g e t

s i n (a + \frac{14 π}{3}) = s i n (a) c o s (\frac{14 π}{3}) + c o s (a) s i n (\frac{14 π}{3})

\frac{14 π}{3} = 4 π + \frac{2 π}{3} s o t h a t

s i n (\frac{14 π}{3}) = s i n 4 π c o s \frac{2 π}{3} + s i n \frac{2 π}{3} c o s 4 π

= 0 + \frac{\sqrt{3}}{2}

s i n (14 π / 3) = \sqrt{3} / 2

U s i n g c o s (x \pm y) = c o s x c o s y \mp s i n x s i n y

w e o b t a i n

c o s \frac{14 π}{3} = c o s 4 π c o s \frac{2 π}{3} - s i n 4 π s i n \frac{2 π}{3}

= - \frac{1}{2} - 0

c o s (14 π / 3) = 1 / 2

∴ s i n (a + \frac{14 π}{3}) = \frac{1}{2} (- s i n a + \sqrt{3} c o s a)

N e x t w e g e t

s i n (a - \frac{8 π}{3}) = s i n (a) c o s \frac{8 π}{3} - s i n \frac{8 π}{3} c o s (a)

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

= - \frac{1}{2} s i n a - \frac{\sqrt{3}}{2} c o s a

= - \frac{1}{2} (s i n a + \sqrt{3} c o s a)

T h e l e f t - h a n d e x p r e s s i o n b e c o m e s

s i n (a) + \frac{1}{2} (- s i n a + \sqrt{3} c o s a) - \frac{1}{2} (s i n a + \sqrt{3} c o s a) = s i n a - s i n a = 0

Commented by alib last updated on 05/Oct/15

s o t h e e q u a t i o n i s n o t e q u a l t o z e r o ?

Commented by 112358 last updated on 05/Oct/15

M i s t a k e c o r r e c t e d .

Facebook Tweet Pin