Question

Prove that sin 10° + sin 20° + sin 40° + sin 50° = sin 70° + sin 80°.

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)

$sin10+sin20+sin40+sin50$

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

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

$=[2\:sin60\:cos20]+[2\:sin60\:cos10]$

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

$=cos20+cos10$

$=sin70+sin80$

Answer 2

Answer:

refer to the attachment for the solution.

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