Question

Sin10 + sin20 + sin40 + sin50 = sin70 + sin80
Prove it

Answer 1

Here we will use the formula
sinA+sinB=2sin(A+B/2)cos(A-B/2)

LHS
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

RHS

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

LHS = RHS .

this is your answer..

Answer 2

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 [ since, sin75 = cos15 ]



Mrinalini Kalla answered this

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 [ since, sin75 = cos15 ]