Question

Sin10+sin20+sin40+sin50-sin70-sin80 =how much

Answer 1

Taking the LHS, we have:
sin10+sin40+sin50+sin20 
Arranging 
=(sin50+sin10)+(sin40+sin20) 
Applying the above formula. 
=2sin(50+10/2)cos(50-10/2)+2sin(40+20/... 
=2sin(30)cos(20)+2sin(30)cos(10) 
=2sin30{cos20+cos10} 
Again using the formula 
cosA+cosB= 2cos(A+B/2)cos(A-B/2) 
=2sin30{2cos(20+10/2).cos(20-10/2)} 
=2sin30{2cos(15).cos(5)} 
=2(1/2){2cos15.cos5} as sin30=1/2 
=2cos15.cos5 





Taking right side of equation 
sin70+ sin80 
Using the formula 
sinA+sinB=2sin(A+B/2)cos(A-B/2) 
=2sin(70+80/2)cos(70-80/2) 
=2sin75cos5 
=2sin(90-15)cos5 
=2cos15.cos5 
Hence proved

Answer 2

sin10+sin20+sin40+sin50=sin120
-sin70-sin80=-sin150
sin120-sin150=-sin30
-sin30=-0.5