Prove That :
Sinx.Sin(60-x).Sin(60+x) = 1/4Sin3x
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Answered by
27
Heya!!! ✌️✌️
Here is your answer dear..... sinA . sin(60 – A) . sin(60 + A) = sinA . (sin60 cos A – cos60 sin A) . (sin60 cos A + cos60 sin A)
= sinA . (sin260 cos2A – cos260 sin2A)
= sinA . {(3/4) cos2A – (1/4) sin2A)
= sinA . {(3/4) – (3/4) sin2A – (1/4) sin2A)
= sinA .{(3/4) – sin2A
= (3/4) sinA – sin3A
= 1/4 (3 sinA – 4sin3A)
=1/4 sin3A
I HOPE THIS WILL HELP YOU OUT.....
HAVE A GREAT DAY DEAR.....
#Bhavana ☺️
Here is your answer dear..... sinA . sin(60 – A) . sin(60 + A) = sinA . (sin60 cos A – cos60 sin A) . (sin60 cos A + cos60 sin A)
= sinA . (sin260 cos2A – cos260 sin2A)
= sinA . {(3/4) cos2A – (1/4) sin2A)
= sinA . {(3/4) – (3/4) sin2A – (1/4) sin2A)
= sinA .{(3/4) – sin2A
= (3/4) sinA – sin3A
= 1/4 (3 sinA – 4sin3A)
=1/4 sin3A
I HOPE THIS WILL HELP YOU OUT.....
HAVE A GREAT DAY DEAR.....
#Bhavana ☺️
vatsalsinghkv:
Aapne bahut lengthy kiya h.... kyo didi, sin(60-x).sin(60+x) ko humlog 2sinC.sinD waali identity se nhi kr sakte ? Aur aapke ans me jab 1/4[2Sin2x.Cosx - Sinx] ho gya tha go aap 2Sin2x.Cosx ko directly Sin3x + Sinx likh sakte the to seedhe hi ans aa jaaya
Yes....
Is it not helpful to you???
TBH it is helpful but I'm trying to solve this like this sinx/2[2sin(60-x).sin(60+x)] » sinx/2 [ cos(2x) - cos(120) ] Can we solve this like this ? also?
Sry I can't yr
and i cant understand your ans, after this step : 1/4[2Sin2x.Cosx - Sinx] this can be solve by this formula 2Sin2x.Cosx = Sin(x+x) + Sin(x-x)… i can't understand why u made it so lengthy and complicated
See I have edited it
didi, can u plz explain again how sinA.(sin260.cos2A - cos260.sin2a). ! how
Didi aap rehne do
actually this one is totally wrong, its sin2A and u r using it as a Sin²A
Answered by
1
Answer:
sin x.sin(60-x)=1/4sin 3x
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