Question

sin10-sin110+sin130=

Answer 1

Answer:

Sin10-sin110+sin130 = 0

Step-by-step explanation:

Formula used:

$sinC+sinD=2\:sin(\frac{C+D}{2})\:cos(\frac{C-D}{2})$

$sinC-sinD=2\:cos(\frac{C+D}{2})\:sin(\frac{C-D}{2})$

Now

Sin10-sin110+sin130

=sin10+sin130-sin110

$=2\:sin(\frac{10+130}{2})cos(\frac{10-130}{2})-sin110$

$=2\:sin70\:cos(-60)-sin110$

$=2\:sin70\:cos60-sin110$

$=2\:sin70(\frac{1}{2})-sin110$

$=sin70-sin110$

$=2\:cos(\frac{70+110}{2})\:sin(\frac{70-110}{2})$

$=2\:cos90\:sin(-20)$

$=2\:(0)\:sin(-20)$

$=0$

Answer 2

Answer:

Step-by-step explanation:

acording to formulae

sina+sin(120+a)-sin(120-a)=0

a=10

sin10+sin(120+10)-sin(120-10)=0

ANS=0