Question

show that sin 10°+sin 40° sin 20°+ sin 50° = sin 70°+sin 80°

plz solve it ​

Answer 1

Answer:

sin70+sin80

Step-by-step explanation:

Formula used:

1. sinC+sinD

= 2 sin ((C+D)/2) cos((C-D)/2)

2.cosA = sin(90°-A)

sin10+sin20+sin40+sin50sin10+sin20+sin40+sin50

=(sin50+sin10)+(sin40+sin20)

=[2\:sin(\frac{(50+10)}{2})\:cos(\frac{(50-10)}{2})]+[2\:sin(\frac{(40+20)}{2})\:cos(\frac{(40-20)}{2})]=[2sin(2(50+10))cos(2(50−10))]+[2sin(2(40+20))cos(2(40−20))]

=[2\:sin60\:cos20]+[2\:sin60\:cos10]=[2sin60cos20]+[2sin60cos10]

=[2\:(\frac{1}{2})\:cos20]+[2\:(\frac{1}{2})\:cos10]=[2(21)cos20]+[2(21)cos10]

=cos20+cos10=cos20+cos10

=sin70+sin80=sin70+sin80