Question

show that sin 10+ sin 20+ sin 30 + sin 40 = sin 70 + sin 80​

Answer 1

Answer:

L.H.S

= 2sin15cos5+2sin45cos5 [ using sin C+sin D= 2sin C+D/2 cos C-D/2 for sin10+sin20 & sin40+sin50]

= 2cos5 (sin15+sin45)

= 2cos5 (2sin30cos15) [ using sin C+sin D= 2sin C+D/2 cos C-D/2 ]

= 2cos5 (2 x 1/2 x cos15)

= 2cos5 cos15

R.H.S.

= sin70+sin80

= 2sin75cos5 [ using sin C+sin D= 2sin C+D/2 cos C-D/2 ]

sin75 = sin(90-15) = cos 15

L.H.S = 2cos5 cos15

R.H.S. = 2cos15 cos5

Answer 2

Step-by-step explanation:

$\huge\underline\blue{Answer}$

sin70+sin80

Formula used:

1. sinC+sinD

$= 2 sin ((C+D)/2) cos((C-D)/2)</p><p></p><p></p><p>2.cosA = sin(90°-A)</p><p></p><p></p><p>sin10+sin20+sin40+sin50sin10+sin20+sin40+sin50</p><p></p><p>=(sin50+sin10)+(sin40+sin20)</p><p></p><p>[tex]=[2\:sin(\frac{(50+10)}{2})\:cos(\frac{(50-10)}{2})]+[2\:sin(\frac{(40+20)}{2})\:cos(\frac{(40-20)}{2})]=[2sin( </p><p>2</p><p>(50+10)</p><p> </p><p> )cos( </p><p>2</p><p>(50−10)</p><p> </p><p> )]+[2sin( </p><p>2</p><p>(40+20)</p><p> </p><p> )cos( </p><p>2</p><p>(40−20)</p><p> </p><p> )]</p><p></p><p>=[2\:sin60\:cos20]+[2\:sin60\:cos10]=[2sin60cos20]+[2sin60cos10]</p><p></p><p>=[2\:(\frac{1}{2})\:cos20]+[2\:(\frac{1}{2})\:cos10]$=[2(2 1)cos20]+[2( 21 )cos10]

=cos20+cos10=cos20+cos10

=sin70+sin80=sin70+sin80

Hence proved

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